OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors Exercise 22 (B)

S Chand Class 12 ICSE Maths Solutions Chapter 22 Vectors (Continued) Ex 22(b)

Question 1. Find the required vector products and verify perpendicularity as stated:
(i) Find \( \vec{a} \times \vec{b} \) when \( \vec{a} = 3 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 2 \hat{i} + \hat{j} – \hat{k} \).
(ii) Find \( \vec{a} \times \vec{b} \) when \( \vec{a} = 2 \hat{i} – 3 \hat{j} – \hat{k} \) and \( \vec{b} = \hat{i} + 4 \hat{j} – 2 \hat{k} \).
(iii) Show that \( \vec{a} \times \vec{b} \neq \vec{b} \times \vec{a} \) for the vectors given in part (ii).
(iv) Given \( \vec{a} = 2 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 3 \hat{i} + 4 \hat{j} – \hat{k} \), verify that \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).
(v) Compute \( \vec{a} \times (\vec{b} \times \vec{c}) \), if \( \vec{a} = 7 \hat{i} – 2 \hat{j} + 3 \hat{k} \), \( \vec{b} = 2 \hat{i} + 8 \hat{k} \), \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \).
Answer:
(i) Given vectors are \( \vec{a} = 3 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 2 \hat{i} + \hat{j} – \hat{k} \).
We calculate the cross product as a determinant:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \)
\( = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((3)(-1) - (1)(2)) + \hat{k}((3)(1) - (-1)(2)) \)
\( = \hat{i}(1 - 1) - \hat{j}(-3 - 2) + \hat{k}(3 + 2) \)
\( = 0 \hat{i} - \hat{j}(-5) + \hat{k}(5) \)
\( = 0 \hat{i} + 5 \hat{j} + 5 \hat{k} \).

(ii) Given vectors are \( \vec{a} = 2 \hat{i} – 3 \hat{j} – \hat{k} \) and \( \vec{b} = \hat{i} + 4 \hat{j} – 2 \hat{k} \).
We calculate the cross product as a determinant:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix} \)
\( = \hat{i}((-3)(-2) - (-1)(4)) - \hat{j}((2)(-2) - (-1)(1)) + \hat{k}((2)(4) - (-3)(1)) \)
\( = \hat{i}(6 + 4) - \hat{j}(-4 + 1) + \hat{k}(8 + 3) \)
\( = \hat{i}(10) - \hat{j}(-3) + \hat{k}(11) \)
\( = 10 \hat{i} + 3 \hat{j} + 11 \hat{k} \).

(iii) For the vectors from part (ii), \( \vec{a} = 2 \hat{i} – 3 \hat{j} – \hat{k} \) and \( \vec{b} = \hat{i} + 4 \hat{j} – 2 \hat{k} \).
Now we calculate \( \vec{b} \times \vec{a} \):
\( \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -2 \\ 2 & -3 & -1 \end{vmatrix} \)
\( = \hat{i}((4)(-1) - (-2)(-3)) - \hat{j}((1)(-1) - (-2)(2)) + \hat{k}((1)(-3) - (4)(2)) \)
\( = \hat{i}(-4 - 6) - \hat{j}(-1 + 4) + \hat{k}(-3 - 8) \)
\( = \hat{i}(-10) - \hat{j}(3) + \hat{k}(-11) \)
\( = -10 \hat{i} - 3 \hat{j} - 11 \hat{k} \).
Since \( 10 \hat{i} + 3 \hat{j} + 11 \hat{k} \neq -10 \hat{i} - 3 \hat{j} - 11 \hat{k} \), we can see that \( \vec{a} \times \vec{b} \neq \vec{b} \times \vec{a} \). This shows that the order matters for vector cross products.

(iv) Given \( \vec{a} = 2 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 3 \hat{i} + 4 \hat{j} – \hat{k} \).
First, calculate the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1 \end{vmatrix} \)
\( = \hat{i}((-1)(-1) - (1)(4)) - \hat{j}((2)(-1) - (1)(3)) + \hat{k}((2)(4) - (-1)(3)) \)
\( = \hat{i}(1 - 4) - \hat{j}(-2 - 3) + \hat{k}(8 + 3) \)
\( = -3 \hat{i} + 5 \hat{j} + 11 \hat{k} \).
To verify perpendicularity, the dot product with both original vectors must be zero.
\( (\vec{a} \times \vec{b}) \cdot \vec{a} = (-3 \hat{i} + 5 \hat{j} + 11 \hat{k}) \cdot (2 \hat{i} – \hat{j} + \hat{k}) \)
\( = (-3)(2) + (5)(-1) + (11)(1) = -6 - 5 + 11 = 0 \).
\( (\vec{a} \times \vec{b}) \cdot \vec{b} = (-3 \hat{i} + 5 \hat{j} + 11 \hat{k}) \cdot (3 \hat{i} + 4 \hat{j} – \hat{k}) \)
\( = (-3)(3) + (5)(4) + (11)(-1) = -9 + 20 - 11 = 0 \).
Since both dot products are zero, \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \). This is a fundamental property of the cross product.

(v) Given \( \vec{a} = 7 \hat{i} – 2 \hat{j} + 3 \hat{k} \), \( \vec{b} = 2 \hat{i} + 8 \hat{k} \), and \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \).
First, we find the sum of vectors \( \vec{b} \) and \( \vec{c} \):
\( \vec{b} + \vec{c} = (2 \hat{i} + 0 \hat{j} + 8 \hat{k}) + (\hat{i} + \hat{j} + \hat{k}) = 3 \hat{i} + \hat{j} + 9 \hat{k} \).
Now, we compute the cross product of \( \vec{a} \) with \( (\vec{b} + \vec{c}) \):
\( \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -2 & 3 \\ 3 & 1 & 9 \end{vmatrix} \)
\( = \hat{i}((-2)(9) - (3)(1)) - \hat{j}((7)(9) - (3)(3)) + \hat{k}((7)(1) - (-2)(3)) \)
\( = \hat{i}(-18 - 3) - \hat{j}(63 - 9) + \hat{k}(7 + 6) \)
\( = -21 \hat{i} - 54 \hat{j} + 13 \hat{k} \).
In simple words: To find the cross product of two vectors, we set up a special grid (a determinant) with the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the numbers from the vectors. Then we do a specific calculation to get a new vector. When we want to check if the new vector is "perpendicular" to the original ones, we multiply them using the dot product, and if the answer is zero, they are perpendicular. This is like finding a line that stands straight up from a flat surface.

🎯 Exam Tip: Remember that \( \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) \) and the cross product \( \vec{a} \times \vec{b} \) is always perpendicular to both \( \vec{a} \) and \( \vec{b} \). This property is key for solving verification problems.

Question 2. Find \( |\vec{a} \times \vec{b}| \) when
(i) \( \vec{a} = \hat{i} + 3 \hat{j} – 2 \hat{k} \), \( \vec{b} = -\hat{i} + 3 \hat{k} \)
(ii) \( \vec{a} = 2 \hat{i} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \).
Answer:
(i) Given vectors are \( \vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k} \) and \( \vec{b} = -\hat{i} + 0 \hat{j} + 3 \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix} \)
\( = \hat{i}((3)(3) - (-2)(0)) - \hat{j}((1)(3) - (-2)(-1)) + \hat{k}((1)(0) - (3)(-1)) \)
\( = \hat{i}(9 - 0) - \hat{j}(3 - 2) + \hat{k}(0 + 3) \)
\( = 9 \hat{i} - \hat{j} + 3 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{9^2 + (-1)^2 + 3^2} \)
\( = \sqrt{81 + 1 + 9} \)
\( = \sqrt{91} \).

(ii) Given vectors are \( \vec{a} = 2 \hat{i} + 0 \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( = \hat{i}((0)(1) - (1)(1)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(1) - (0)(1)) \)
\( = \hat{i}(0 - 1) - \hat{j}(2 - 1) + \hat{k}(2 - 0) \)
\( = -\hat{i} - \hat{j} + 2 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-1)^2 + 2^2} \)
\( = \sqrt{1 + 1 + 4} \)
\( = \sqrt{6} \).
In simple words: To find the length (magnitude) of the cross product of two vectors, first calculate the cross product itself using the determinant method. Once you have the resulting vector, use the distance formula (square root of the sum of the squares of its components) to find its length. This length tells us the area of the parallelogram formed by the two original vectors.

🎯 Exam Tip: Always be careful with the signs when calculating determinants. A common mistake is to forget the minus sign for the \( \hat{j} \) component in the expansion.

Question 3. If \( \vec{a} = 2 \hat{i} + 3 \hat{j} \), \( \vec{b} = -\hat{i} + 3 \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} + 2 \hat{j} + 5 \hat{k} \) be three vectors, find
(i) \( \vec{a} \times \vec{b} \)
(ii) \( \vec{b} \times \vec{c} \)
(iii) \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) \).
Answer:
Given vectors are \( \vec{a} = 2 \hat{i} + 3 \hat{j} + 0 \hat{k} \), \( \vec{b} = -\hat{i} + 3 \hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} + 2 \hat{j} + 5 \hat{k} \).

(i) Find \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ -1 & 3 & 1 \end{vmatrix} \)
\( = \hat{i}((3)(1) - (0)(3)) - \hat{j}((2)(1) - (0)(-1)) + \hat{k}((2)(3) - (3)(-1)) \)
\( = \hat{i}(3 - 0) - \hat{j}(2 - 0) + \hat{k}(6 + 3) \)
\( = 3 \hat{i} - 2 \hat{j} + 9 \hat{k} \).

(ii) Find \( \vec{b} \times \vec{c} \):
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 1 \\ 1 & 2 & 5 \end{vmatrix} \)
\( = \hat{i}((3)(5) - (1)(2)) - \hat{j}((-1)(5) - (1)(1)) + \hat{k}((-1)(2) - (3)(1)) \)
\( = \hat{i}(15 - 2) - \hat{j}(-5 - 1) + \hat{k}(-2 - 3) \)
\( = 13 \hat{i} + 6 \hat{j} - 5 \hat{k} \).

(iii) The question asks for \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) \). However, following the provided solution steps, we will calculate \( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) \).
First, calculate \( \vec{a} - \vec{b} \):
\( \vec{a} - \vec{b} = (2 \hat{i} + 3 \hat{j}) - (-\hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (2 - (-1))\hat{i} + (3 - 3)\hat{j} + (0 - 1)\hat{k} \)
\( = 3 \hat{i} + 0 \hat{j} - \hat{k} \).
Next, calculate \( \vec{c} - \vec{b} \):
\( \vec{c} - \vec{b} = (\hat{i} + 2 \hat{j} + 5 \hat{k}) - (-\hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (1 - (-1))\hat{i} + (2 - 3)\hat{j} + (5 - 1)\hat{k} \)
\( = 2 \hat{i} - \hat{j} + 4 \hat{k} \).
Now, find the cross product of these two new vectors:
\( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 2 & -1 & 4 \end{vmatrix} \)
\( = \hat{i}((0)(4) - (-1)(-1)) - \hat{j}((3)(4) - (-1)(2)) + \hat{k}((3)(-1) - (0)(2)) \)
\( = \hat{i}(0 - 1) - \hat{j}(12 + 2) + \hat{k}(-3 - 0) \)
\( = -\hat{i} - 14 \hat{j} - 3 \hat{k} \).
In simple words: This problem involves finding cross products of different combinations of given vectors. The cross product of two vectors results in a new vector that is perpendicular to both original vectors. It's important to keep track of the signs and the order of operations when calculating the determinant for each cross product.

🎯 Exam Tip: When working with multiple vector operations, calculate each intermediate vector or cross product carefully before combining them. Double-check your determinant expansions for sign errors.

Question 4. \( |\vec{a}| = \sqrt{26} \), \( |\vec{b}| = 7 \) and \( |\vec{a} \times \vec{b}| = 35 \), find \( \vec{a} \cdot \vec{b} \).
Answer:
Given are the magnitudes \( |\vec{a}| = \sqrt{26} \), \( |\vec{b}| = 7 \), and the magnitude of their cross product \( |\vec{a} \times \vec{b}| = 35 \).
We know the formula for the magnitude of the cross product: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).
Substituting the given values:
\( 35 = (\sqrt{26})(7) \sin \theta \)
\( \sin \theta = \frac{35}{7\sqrt{26}} \)
\( \sin \theta = \frac{5}{\sqrt{26}} \).
Now we need to find \( \cos \theta \) to calculate the dot product. We use the identity \( \cos \theta = \sqrt{1 - \sin^2 \theta} \).
\( \cos \theta = \sqrt{1 - \left(\frac{5}{\sqrt{26}}\right)^2} \)
\( = \sqrt{1 - \frac{25}{26}} \)
\( = \sqrt{\frac{26 - 25}{26}} \)
\( = \sqrt{\frac{1}{26}} \)
\( = \frac{1}{\sqrt{26}} \).
Finally, we find the dot product \( \vec{a} \cdot \vec{b} \) using the formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
\( \vec{a} \cdot \vec{b} = (\sqrt{26})(7)\left(\frac{1}{\sqrt{26}}\right) \)
\( = 7 \).
In simple words: We used two main vector formulas here. The first formula connects the magnitude of the cross product with the sine of the angle between vectors. The second formula connects the dot product with the cosine of the angle. By finding sine first, we could then find cosine and calculate the dot product. This shows how cross and dot products are related through the angle between the vectors.

🎯 Exam Tip: Remember the fundamental relationships: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \) and \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \). These are essential for solving problems involving angles and magnitudes of vectors.

Question 5. If \( \vec{a} = 3 \hat{i} – \hat{j} – 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k} \) find \( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) \).
Answer:
Given vectors are \( \vec{a} = 3 \hat{i} – \hat{j} – 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k} \).
First, calculate the vector \( \vec{a} + 2 \vec{b} \):
\( \vec{a} + 2 \vec{b} = (3 \hat{i} – \hat{j} – 2 \hat{k}) + 2(2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (3 \hat{i} – \hat{j} – 2 \hat{k}) + (4 \hat{i} + 6 \hat{j} + 2 \hat{k}) \)
\( = (3+4)\hat{i} + (-1+6)\hat{j} + (-2+2)\hat{k} \)
\( = 7 \hat{i} + 5 \hat{j} + 0 \hat{k} \).
Next, calculate the vector \( 2 \vec{a} – \vec{b} \):
\( 2 \vec{a} – \vec{b} = 2(3 \hat{i} – \hat{j} – 2 \hat{k}) – (2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (6 \hat{i} – 2 \hat{j} – 4 \hat{k}) – (2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (6-2)\hat{i} + (-2-3)\hat{j} + (-4-1)\hat{k} \)
\( = 4 \hat{i} – 5 \hat{j} – 5 \hat{k} \).
Now, find the cross product of these two resulting vectors, \( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) \):
\( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{vmatrix} \)
\( = \hat{i}((5)(-5) - (0)(-5)) - \hat{j}((7)(-5) - (0)(4)) + \hat{k}((7)(-5) - (5)(4)) \)
\( = \hat{i}(-25 - 0) - \hat{j}(-35 - 0) + \hat{k}(-35 - 20) \)
\( = -25 \hat{i} + 35 \hat{j} - 55 \hat{k} \).
In simple words: This problem asks us to first combine the given vectors \( \vec{a} \) and \( \vec{b} \) in two different ways using scalar multiplication and addition/subtraction. After finding these two new vectors, we then calculate their cross product using the determinant method. It's like building new vectors from existing ones and then finding a vector perpendicular to both of them.

🎯 Exam Tip: Break down complex vector problems into smaller, manageable steps. First, perform scalar multiplication and vector addition/subtraction to get the component vectors, then apply the cross product formula accurately.

Question 6. Find a unit vector perpendicular to the plane of the vectors.
(i) \( -3 \hat{i} + 4 \hat{k} \) and \( 4 \hat{i} + 3 \hat{j} \)
(ii) \( \vec{a} = 2 \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \).
Answer:
(i) Let the given vectors be \( \vec{a} = -3 \hat{i} + 0 \hat{j} + 4 \hat{k} \) and \( \vec{b} = 4 \hat{i} + 3 \hat{j} + 0 \hat{k} \).
A vector perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is given by their cross product \( \vec{a} \times \vec{b} \).
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 4 \\ 4 & 3 & 0 \end{vmatrix} \)
\( = \hat{i}((0)(0) - (4)(3)) - \hat{j}((-3)(0) - (4)(4)) + \hat{k}((-3)(3) - (0)(4)) \)
\( = \hat{i}(0 - 12) - \hat{j}(0 - 16) + \hat{k}(-9 - 0) \)
\( = -12 \hat{i} + 16 \hat{j} - 9 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(-12)^2 + (16)^2 + (-9)^2} \)
\( = \sqrt{144 + 256 + 81} \)
\( = \sqrt{481} \).
The unit vector perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is \( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \).
Thus, the required unit vector is \( \frac{-12 \hat{i} + 16 \hat{j} - 9 \hat{k}}{\sqrt{481}} \).
We can write this as: \( -\frac{12}{\sqrt{481}} \hat{i} + \frac{16}{\sqrt{481}} \hat{j} - \frac{9}{\sqrt{481}} \hat{k} \).

(ii) Given vectors are \( \vec{a} = 2 \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} \)
\( = \hat{i}((1)(1) - (1)(2)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(2) - (1)(1)) \)
\( = \hat{i}(1 - 2) - \hat{j}(2 - 1) + \hat{k}(4 - 1) \)
\( = -\hat{i} - \hat{j} + 3 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-1)^2 + 3^2} \)
\( = \sqrt{1 + 1 + 9} \)
\( = \sqrt{11} \).
The unit vector perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \) is \( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \).
Thus, the required unit vector is \( \frac{-\hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{11}} \).
We can write this as: \( \frac{1}{\sqrt{11}}(-\hat{i} - \hat{j} + 3 \hat{k}) \). This vector represents the direction that is orthogonal to both given vectors.
In simple words: To find a unit vector that points straight out from the flat surface created by two other vectors, we first calculate their cross product. This cross product gives us a vector perpendicular to both. Then, to make it a 'unit' vector (meaning its length is 1), we divide this vector by its own length (magnitude).

🎯 Exam Tip: A unit vector always has a magnitude of 1. Remember to divide the cross product by its magnitude to normalize it into a unit vector. Also, the cross product can point in two opposite directions, so the unit vector can be \( \pm \) this result.

Question 7.
(i) Find the vector of magnitude 9, which is perpendicular to both the vectors \( 4 \hat{i} - \hat{j} + 3 \hat{k} \) and \( 2 \hat{i} + \hat{j} – 2 \hat{k} \)
(ii) Find a vector whose length is 3 and which is perpendicular to both the vectors \( \vec{a} = 3 \hat{i} + \hat{j} + 3 \hat{k} \) and \( -2 \hat{i} + \hat{j} – 2 \hat{k} \).
Answer:
(i) Let the given vectors be \( \vec{a} = 4 \hat{i} - \hat{j} + 3 \hat{k} \) and \( \vec{b} = 2 \hat{i} + \hat{j} – 2 \hat{k} \).
First, find a vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) by calculating their cross product:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ 2 & 1 & -2 \end{vmatrix} \)
\( = \hat{i}((-1)(-2) - (3)(1)) - \hat{j}((4)(-2) - (3)(2)) + \hat{k}((4)(1) - (-1)(2)) \)
\( = \hat{i}(2 - 3) - \hat{j}(-8 - 6) + \hat{k}(4 + 2) \)
\( = -\hat{i} + 14 \hat{j} + 6 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (14)^2 + (6)^2} \)
\( = \sqrt{1 + 196 + 36} \)
\( = \sqrt{233} \).
The unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is \( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-\hat{i} + 14 \hat{j} + 6 \hat{k}}{\sqrt{233}} \).
We need a vector of magnitude 9, so multiply the unit vector by 9:
Required vector \( = 9 \hat{n} = 9 \left( \frac{-\hat{i} + 14 \hat{j} + 6 \hat{k}}{\sqrt{233}} \right) = \frac{-9 \hat{i} + 126 \hat{j} + 54 \hat{k}}{\sqrt{233}} \).

(ii) Let the given vectors be \( \vec{A} = 3 \hat{i} + \hat{j} + 3 \hat{k} \) and \( \vec{B} = -2 \hat{i} + \hat{j} – 2 \hat{k} \).
First, find a vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) by calculating their cross product:
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \)
\( = \hat{i}((1)(-2) - (3)(1)) - \hat{j}((3)(-2) - (3)(-2)) + \hat{k}((3)(1) - (1)(-2)) \)
\( = \hat{i}(-2 - 3) - \hat{j}(-6 + 6) + \hat{k}(3 + 2) \)
\( = -5 \hat{i} + 0 \hat{j} + 5 \hat{k} = -5 \hat{i} + 5 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{(-5)^2 + (0)^2 + (5)^2} \)
\( = \sqrt{25 + 0 + 25} \)
\( = \sqrt{50} = 5\sqrt{2} \).
The unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is \( \hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \frac{-5 \hat{i} + 5 \hat{k}}{5\sqrt{2}} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}} \).
We need a vector of magnitude 3, so multiply the unit vector by 3:
Required vector \( = 3 \hat{n} = 3 \left( \frac{-\hat{i} + \hat{k}}{\sqrt{2}} \right) = \frac{-3 \hat{i} + 3 \hat{k}}{\sqrt{2}} \).
In simple words: To find a vector with a specific length that is also perpendicular to two given vectors, we first calculate the cross product of the two vectors. This gives us the direction of the perpendicular vector. Then, we find the unit vector in that direction by dividing by its magnitude. Finally, we multiply this unit vector by the desired length to get the final answer. This helps us scale the vector to any required size while keeping its orientation.

🎯 Exam Tip: A vector perpendicular to two vectors \( \vec{a} \) and \( \vec{b} \) is \( \vec{a} \times \vec{b} \) (or \( \vec{b} \times \vec{a} \)). To get a specific magnitude, calculate the unit vector in that direction first, then scale it by the desired magnitude.

Question 8.
(i) If the position vectors of the three points A, B, C are respectively \( \hat{i} + \hat{j} + \hat{k} \), \( \hat{i} + 3 \hat{j} – 4 \hat{k} \) and \( 7 \hat{i} + 4 \hat{j} + 9 \hat{k} \), find the unit vector perpendicular to the plane of the triangle ABC.
(ii) Find the unit vectors perpendicular to the plane A B C, when the position vectors of A, B and C are \( 2 \hat{i} – \hat{j} + \hat{k} \), \( \hat{i} + \hat{j} + 2 \hat{k} \) and \( \hat{i} + 3 \hat{k} \) respectively.
(iii) Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Answer:
(i) Given position vectors (P.V.) are:
\( \text{P.V. of A } (\vec{a}) = \hat{i} + \hat{j} + \hat{k} \)
\( \text{P.V. of B } (\vec{b}) = 2 \hat{i} + 3 \hat{j} – 4 \hat{k} \)
\( \text{P.V. of C } (\vec{c}) = 7 \hat{i} + 4 \hat{j} + 9 \hat{k} \).
To find a vector perpendicular to the plane of triangle ABC, we can find the cross product of two vectors formed by its sides, e.g., \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \).
Calculate \( \overrightarrow{BA} = \text{P.V. of A} - \text{P.V. of B} \):
\( \overrightarrow{BA} = (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) \)
\( = (1-2)\hat{i} + (1-3)\hat{j} + (1-(-4))\hat{k} \)
\( = -\hat{i} – 2 \hat{j} + 5 \hat{k} \).
Calculate \( \overrightarrow{BC} = \text{P.V. of C} - \text{P.V. of B} \):
\( \overrightarrow{BC} = (7 \hat{i} + 4 \hat{j} + 9 \hat{k}) - (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) \)
\( = (7-2)\hat{i} + (4-3)\hat{j} + (9-(-4))\hat{k} \)
\( = 5 \hat{i} – \hat{j} + 13 \hat{k} \).
Now, find the cross product \( \overrightarrow{BA} \times \overrightarrow{BC} \), which is a vector perpendicular to the plane of triangle ABC:
\( \overrightarrow{BA} \times \overrightarrow{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & 5 \\ 5 & -1 & 13 \end{vmatrix} \)
\( = \hat{i}((-2)(13) - (5)(-1)) - \hat{j}((-1)(13) - (5)(5)) + \hat{k}((-1)(-1) - (-2)(5)) \)
\( = \hat{i}(-26 + 5) - \hat{j}(-13 - 25) + \hat{k}(1 + 10) \)
\( = -21 \hat{i} + 38 \hat{j} + 11 \hat{k} \).
Find the magnitude of this vector:
\( |\overrightarrow{BA} \times \overrightarrow{BC}| = \sqrt{(-21)^2 + (38)^2 + (11)^2} \)
\( = \sqrt{441 + 1444 + 121} \)
\( = \sqrt{2006} \).
The required unit vector is \( \frac{\overrightarrow{BA} \times \overrightarrow{BC}}{|\overrightarrow{BA} \times \overrightarrow{BC}|} = \frac{-21 \hat{i} + 38 \hat{j} + 11 \hat{k}}{\sqrt{2006}} \).

(ii) Given position vectors (P.V.) are:
\( \text{P.V. of A } (\vec{a}) = 2 \hat{i} – \hat{j} + \hat{k} \)
\( \text{P.V. of B } (\vec{b}) = \hat{i} + \hat{j} + 2 \hat{k} \)
\( \text{P.V. of C } (\vec{c}) = \hat{i} + 3 \hat{k} \).
Calculate \( \overrightarrow{BA} = \text{P.V. of A} - \text{P.V. of B} \):
\( \overrightarrow{BA} = (2 \hat{i} – \hat{j} + \hat{k}) - (\hat{i} + \hat{j} + 2 \hat{k}) \)
\( = (2-1)\hat{i} + (-1-1)\hat{j} + (1-2)\hat{k} \)
\( = \hat{i} – 2 \hat{j} – \hat{k} \).
Calculate \( \overrightarrow{BC} = \text{P.V. of C} - \text{P.V. of B} \):
\( \overrightarrow{BC} = (\hat{i} + 0 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + 2 \hat{k}) \)
\( = (1-1)\hat{i} + (0-1)\hat{j} + (3-2)\hat{k} \)
\( = 0 \hat{i} – \hat{j} + \hat{k} \).
Now, find the cross product \( \overrightarrow{BA} \times \overrightarrow{BC} \):
\( \overrightarrow{BA} \times \overrightarrow{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 0 & -1 & 1 \end{vmatrix} \)
\( = \hat{i}((-2)(1) - (-1)(-1)) - \hat{j}((1)(1) - (-1)(0)) + \hat{k}((1)(-1) - (-2)(0)) \)
\( = \hat{i}(-2 - 1) - \hat{j}(1 - 0) + \hat{k}(-1 - 0) \)
\( = -3 \hat{i} - \hat{j} - \hat{k} \).
Find the magnitude of this vector:
\( |\overrightarrow{BA} \times \overrightarrow{BC}| = \sqrt{(-3)^2 + (-1)^2 + (-1)^2} \)
\( = \sqrt{9 + 1 + 1} \)
\( = \sqrt{11} \).
The required unit vector is \( \frac{\overrightarrow{BA} \times \overrightarrow{BC}}{|\overrightarrow{BA} \times \overrightarrow{BC}|} = \frac{-3 \hat{i} - \hat{j} - \hat{k}}{\sqrt{11}} = \frac{1}{\sqrt{11}}(-3 \hat{i} - \hat{j} - \hat{k}) \).

(iii) Given points are P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Their position vectors are:
\( \vec{p} = \hat{i} - \hat{j} + 2 \hat{k} \)
\( \vec{q} = 2 \hat{i} + 0 \hat{j} - \hat{k} \)
\( \vec{r} = 0 \hat{i} + 2 \hat{j} + \hat{k} \).
We form two vectors in the plane, for example, \( \overrightarrow{QP} \) and \( \overrightarrow{QR} \).
Calculate \( \overrightarrow{QR} = \vec{r} - \vec{q} \):
\( \overrightarrow{QR} = (0 \hat{i} + 2 \hat{j} + \hat{k}) - (2 \hat{i} + 0 \hat{j} - \hat{k}) \)
\( = (0-2)\hat{i} + (2-0)\hat{j} + (1-(-1))\hat{k} \)
\( = -2 \hat{i} + 2 \hat{j} + 2 \hat{k} \).
Calculate \( \overrightarrow{QP} = \vec{p} - \vec{q} \):
\( \overrightarrow{QP} = (\hat{i} - \hat{j} + 2 \hat{k}) - (2 \hat{i} + 0 \hat{j} - \hat{k}) \)
\( = (1-2)\hat{i} + (-1-0)\hat{j} + (2-(-1))\hat{k} \)
\( = -\hat{i} - \hat{j} + 3 \hat{k} \).
Now, find the cross product \( \overrightarrow{QR} \times \overrightarrow{QP} \):
\( \overrightarrow{QR} \times \overrightarrow{QP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 2 \\ -1 & -1 & 3 \end{vmatrix} \)
\( = \hat{i}((2)(3) - (2)(-1)) - \hat{j}((-2)(3) - (2)(-1)) + \hat{k}((-2)(-1) - (2)(-1)) \)
\( = \hat{i}(6 + 2) - \hat{j}(-6 + 2) + \hat{k}(2 + 2) \)
\( = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} \).
Find the magnitude of this vector:
\( |\overrightarrow{QR} \times \overrightarrow{QP}| = \sqrt{8^2 + 4^2 + 4^2} \)
\( = \sqrt{64 + 16 + 16} \)
\( = \sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6} \).
The required unit vector is \( \frac{\overrightarrow{QR} \times \overrightarrow{QP}}{|\overrightarrow{QR} \times \overrightarrow{QP}|} = \frac{8 \hat{i} + 4 \hat{j} + 4 \hat{k}}{4\sqrt{6}} \).
\( = \frac{4(2 \hat{i} + \hat{j} + \hat{k})}{4\sqrt{6}} = \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \).
In simple words: To find a unit vector that is perpendicular to a triangle's flat surface, we first define two vectors that form two sides of the triangle. Then, we find the cross product of these two side vectors; this new vector will be perpendicular to the triangle's plane. Finally, we make it a unit vector by dividing it by its own length. The resulting vector will always have a length of one.

🎯 Exam Tip: When finding a vector perpendicular to a plane defined by points, first convert the points to position vectors. Then, create two vectors from these points (e.g., \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)) and compute their cross product. Finally, normalize it to a unit vector.

Question 9. Find the sine of the angle between the vectors.
(i) \( \vec{a} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \), \( \vec{b} = \hat{i} – \hat{j} + \hat{k} \)
(ii) \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \).
Answer:
(i) Given vectors are \( \vec{a} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} + \hat{k} \).
To find \( \sin \theta \), we use the formula \( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ 1 & -1 & 1 \end{vmatrix} \)
\( = \hat{i}((-4)(1) - (5)(-1)) - \hat{j}((3)(1) - (5)(1)) + \hat{k}((3)(-1) - (-4)(1)) \)
\( = \hat{i}(-4 + 5) - \hat{j}(3 - 5) + \hat{k}(-3 + 4) \)
\( = \hat{i} + 2 \hat{j} + \hat{k} \).
Now, find the magnitude of \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \).
Next, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \).
\( |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \).
Finally, calculate \( \sin \theta \):
\( \sin \theta = \frac{\sqrt{6}}{5\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{6}}{5\sqrt{6}} = \frac{1}{5} \).

(ii) Let the given vectors be \( \vec{a} = \hat{i} + \hat{j} + 0 \hat{k} \) and \( \vec{b} = 0 \hat{i} + \hat{j} + \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \)
\( = \hat{i}((1)(1) - (0)(1)) - \hat{j}((1)(1) - (0)(0)) + \hat{k}((1)(1) - (1)(0)) \)
\( = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) \)
\( = \hat{i} - \hat{j} + \hat{k} \).
Now, find the magnitude of \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \).
Next, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \).
\( |\vec{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \).
Finally, calculate \( \sin \theta \):
\( \sin \theta = \frac{\sqrt{3}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{4}} \).
In simple words: To find the sine of the angle between two vectors, we use a special formula that involves the magnitude of their cross product and the magnitudes of the individual vectors. First, calculate the cross product, then find the length of all vectors, and finally, plug these lengths into the formula to get the sine value. This value helps us understand how "spread out" the vectors are from each other.

🎯 Exam Tip: Always remember that \( \sin \theta \) gives you the angle in the range \( [0, \pi] \). Ensure all magnitudes are correctly calculated before substituting them into the formula for \( \sin \theta \).

Question 10. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \), if \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \).
Answer:
Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
We are given that \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \).
We know the formulas for the magnitude of the cross product and the dot product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
Substitute these into the given condition:
\( |\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| \cos \theta \).
Since \( \vec{a} \) and \( \vec{b} \) are non-zero vectors, \( |\vec{a}| \neq 0 \) and \( |\vec{b}| \neq 0 \).
We can divide both sides by \( |\vec{a}| |\vec{b}| \):
\( \sin \theta = \cos \theta \).
If \( \cos \theta \neq 0 \), we can divide both sides by \( \cos \theta \):
\( \frac{\sin \theta}{\cos \theta} = 1 \)
\( \tan \theta = 1 \).
The angle \( \theta \) for which \( \tan \theta = 1 \) is \( \frac{\pi}{4} \) (or 45 degrees) in the range \( [0, \pi] \).
Therefore, \( \theta = \frac{\pi}{4} \). This specific relationship between the cross product magnitude and the dot product value means the vectors are at a 45-degree angle.
In simple words: When the 'strength' of how two vectors are perpendicular (cross product magnitude) is equal to their 'strength' of being in the same direction (dot product), it means they are at a 45-degree angle to each other. We use the formulas for cross and dot products to figure this out.

🎯 Exam Tip: This problem is a direct application of the definitions of the cross product and dot product magnitudes. Always recall these definitions and how they relate to the angle \( \theta \) between the vectors.

Question 11. Given that \( |\vec{a}| = 2 \), \( |\vec{b}| = 7 \), and \( \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \), find the angle between \( \vec{a} \) and \( \vec{b} \).
Answer:
Given \( |\vec{a}| = 2 \), \( |\vec{b}| = 7 \), and \( \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \).
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \).
We use the formula \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).
First, find the magnitude of the cross product \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{3^2 + 2^2 + 6^2} \)
\( = \sqrt{9 + 4 + 36} \)
\( = \sqrt{49} \)
\( = 7 \).
Now substitute the magnitudes into the formula:
\( 7 = (2)(7) \sin \theta \)
\( 7 = 14 \sin \theta \)
\( \sin \theta = \frac{7}{14} \)
\( \sin \theta = \frac{1}{2} \).
For \( \sin \theta = \frac{1}{2} \), the angle \( \theta \) in the range \( [0, \pi] \) is \( \frac{\pi}{6} \) (or 30 degrees).
Therefore, \( \theta = \frac{\pi}{6} \). This means the vectors are at a 30-degree angle to each other.
In simple words: We are given the lengths of two vectors and the vector that results from their cross product. To find the angle between the original vectors, we first calculate the length of the cross product vector. Then, we use a formula that connects these lengths with the sine of the angle, allowing us to find the angle itself.

🎯 Exam Tip: Always remember that the magnitude of the cross product \( |\vec{a} \times \vec{b}| \) relates to the sine of the angle between vectors, while the dot product \( \vec{a} \cdot \vec{b} \) relates to the cosine. Choose the correct formula based on the given information.

Question 12. If \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), then prove that \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \).
Answer:
Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
We know the definition of the dot product:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) --- (1)
From this, we can write \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \).
We also know the definition of the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \) --- (2)
From this, we can write \( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} \).
To prove \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \), we know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Substitute the expressions for \( \sin \theta \) and \( \cos \theta \):
\( \tan \theta = \frac{\frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}}{\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}} \).
We can cancel out \( |\vec{a}| |\vec{b}| \) from the numerator and the denominator (assuming \( |\vec{a}| \neq 0 \) and \( |\vec{b}| \neq 0 \)):
\( \tan \theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \).
This proves the identity. This formula is very useful for finding the angle when both the dot product and cross product magnitude are known.
In simple words: We want to show that the tangent of the angle between two vectors is equal to the length of their cross product divided by their dot product. We do this by writing down the known formulas for sine and cosine of the angle using dot and cross products, then dividing sine by cosine. This makes the lengths of the vectors cancel out, leaving us with the desired relationship.

🎯 Exam Tip: This proof relies on the definitions of dot and cross products. Start by writing these definitions and then use the trigonometric identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to derive the result. Make sure to state assumptions like non-zero vectors.

Question 13. If \( \vec{a} = 4 \hat{i} – \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} – \hat{j} + \hat{k} \), verify that, \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) \).
Answer:
Given vectors are \( \vec{a} = 4 \hat{i} – \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \), and \( \vec{c} = \hat{i} – \hat{j} + \hat{k} \).
The question asks to verify \( \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) \).
However, the calculation provided in the source verifies the distributive property: \( \vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \). We will proceed with this verification following the solution steps.

**Calculate the Left-Hand Side (L.H.S.): \( \vec{a} \times (\vec{b} + \vec{c}) \)**
First, find \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} – \hat{j} + \hat{k}) \)
\( = (1+1)\hat{i} + (1-1)\hat{j} + (1+1)\hat{k} \)
\( = 2 \hat{i} + 0 \hat{j} + 2 \hat{k} \).
Now, calculate \( \vec{a} \times (\vec{b} + \vec{c}) \):
\( \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 2 & 0 & 2 \end{vmatrix} \)
\( = \hat{i}((-1)(2) - (1)(0)) - \hat{j}((4)(2) - (1)(2)) + \hat{k}((4)(0) - (-1)(2)) \)
\( = \hat{i}(-2 - 0) - \hat{j}(8 - 2) + \hat{k}(0 + 2) \)
\( = -2 \hat{i} - 6 \hat{j} + 2 \hat{k} \) --- (1)

**Calculate the Right-Hand Side (R.H.S.): \( (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \)**
First, find \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( = \hat{i}((-1)(1) - (1)(1)) - \hat{j}((4)(1) - (1)(1)) + \hat{k}((4)(1) - (-1)(1)) \)
\( = \hat{i}(-1 - 1) - \hat{j}(4 - 1) + \hat{k}(4 + 1) \)
\( = -2 \hat{i} - 3 \hat{j} + 5 \hat{k} \).
Next, find \( \vec{a} \times \vec{c} \):
\( \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \)
\( = \hat{i}((-1)(1) - (1)(-1)) - \hat{j}((4)(1) - (1)(1)) + \hat{k}((4)(-1) - (-1)(1)) \)
\( = \hat{i}(-1 + 1) - \hat{j}(4 - 1) + \hat{k}(-4 + 1) \)
\( = 0 \hat{i} - 3 \hat{j} - 3 \hat{k} \).
Now, add these two cross products:
\( (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = (-2 \hat{i} - 3 \hat{j} + 5 \hat{k}) + (0 \hat{i} - 3 \hat{j} - 3 \hat{k}) \)
\( = (-2+0)\hat{i} + (-3-3)\hat{j} + (5-3)\hat{k} \)
\( = -2 \hat{i} - 6 \hat{j} + 2 \hat{k} \) --- (2)
Comparing (1) and (2), we see that L.H.S. = R.H.S.
Thus, the distributive property of vector cross product is verified.
In simple words: This problem asks us to prove a rule called the distributive property for vector cross products. We do this by calculating both sides of the equation separately. On one side, we add two vectors first, then take their cross product with a third vector. On the other side, we take the cross product of the third vector with each of the first two separately, and then add those results. If both final answers are the same, the rule is proven.

🎯 Exam Tip: The distributive property \( \vec{A} \times (\vec{B} + \vec{C}) = (\vec{A} \times \vec{B}) + (\vec{A} \times \vec{C}) \) is a valid vector identity. When verifying, calculate each side independently and show they are equal. Be careful not to confuse it with other triple product formulas.

Question 14. Find the area of the parallelogram whose adjacent sides are
(i) \( \hat{i} - 3 \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} + \hat{k} \)
(ii) \( \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( 3 \hat{i} – 2 \hat{j} + \hat{k} \)
(iii) \( 2 \hat{i} + \hat{j} + 3 \hat{k} \) and \( \hat{i} – \hat{j} \)
(iv) \( 2 \hat{i} \) and \( 3 \hat{j} \).
Answer:
The area of a parallelogram with adjacent sides \( \vec{a} \) and \( \vec{b} \) is given by \( |\vec{a} \times \vec{b}| \).

(i) Let \( \vec{a} = \hat{i} - 3 \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( = \hat{i}((-3)(1) - (1)(1)) - \hat{j}((1)(1) - (1)(1)) + \hat{k}((1)(1) - (-3)(1)) \)
\( = \hat{i}(-3 - 1) - \hat{j}(1 - 1) + \hat{k}(1 + 3) \)
\( = -4 \hat{i} + 0 \hat{j} + 4 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(-4)^2 + 0^2 + 4^2} \)
\( = \sqrt{16 + 0 + 16} \)
\( = \sqrt{32} = 4\sqrt{2} \).
Thus, the required area of the parallelogram is \( 4\sqrt{2} \) sq. units.

(ii) Let \( \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} – 2 \hat{j} + \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix} \)
\( = \hat{i}((2)(1) - (3)(-2)) - \hat{j}((1)(1) - (3)(3)) + \hat{k}((1)(-2) - (2)(3)) \)
\( = \hat{i}(2 + 6) - \hat{j}(1 - 9) + \hat{k}(-2 - 6) \)
\( = 8 \hat{i} + 8 \hat{j} - 8 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{8^2 + 8^2 + (-8)^2} \)
\( = \sqrt{64 + 64 + 64} \)
\( = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \).
Thus, the required area of the parallelogram is \( 8\sqrt{3} \) sq. units.

(iii) Let \( \vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} + 0 \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{vmatrix} \)
\( = \hat{i}((1)(0) - (3)(-1)) - \hat{j}((2)(0) - (3)(1)) + \hat{k}((2)(-1) - (1)(1)) \)
\( = \hat{i}(0 + 3) - \hat{j}(0 - 3) + \hat{k}(-2 - 1) \)
\( = 3 \hat{i} + 3 \hat{j} - 3 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{3^2 + 3^2 + (-3)^2} \)
\( = \sqrt{9 + 9 + 9} \)
\( = \sqrt{27} = 3\sqrt{3} \).
Thus, the required area of the parallelogram is \( 3\sqrt{3} \) sq. units.

(iv) Let \( \vec{a} = 2 \hat{i} + 0 \hat{j} + 0 \hat{k} \) and \( \vec{b} = 0 \hat{i} + 3 \hat{j} + 0 \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{vmatrix} \)
\( = \hat{i}((0)(0) - (0)(3)) - \hat{j}((2)(0) - (0)(0)) + \hat{k}((2)(3) - (0)(0)) \)
\( = \hat{i}(0) - \hat{j}(0) + \hat{k}(6) \)
\( = 6 \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{0^2 + 0^2 + 6^2} \)
\( = \sqrt{36} = 6 \).
Thus, the required area of the parallelogram is \( 6 \) sq. units.
In simple words: The area of a parallelogram can be found by taking the length (magnitude) of the cross product of its two adjacent side vectors. First, form the cross product using a determinant, then calculate the square root of the sum of the squares of its components. This single number represents the area.

🎯 Exam Tip: The area of a parallelogram is simply the magnitude of the cross product of its adjacent sides. Ensure you correctly identify the adjacent side vectors and perform the cross product calculation accurately.

Question 15. Find the area of a parallelogram whose diagonals are \( 3 \hat{i} + 4 \hat{j} \) and \( \hat{i} + \hat{j} + \hat{k} \).
Answer:
Let the diagonals of the parallelogram be \( \vec{d_1} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k} \) and \( \vec{d_2} = \hat{i} + \hat{j} + \hat{k} \).
The area of a parallelogram with diagonals \( \vec{d_1} \) and \( \vec{d_2} \) is given by the formula \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \).
First, find the cross product of the diagonals \( \vec{d_1} \times \vec{d_2} \):
\( \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{vmatrix} \)
\( = \hat{i}((4)(1) - (0)(1)) - \hat{j}((3)(1) - (0)(1)) + \hat{k}((3)(1) - (4)(1)) \)
\( = \hat{i}(4 - 0) - \hat{j}(3 - 0) + \hat{k}(3 - 4) \)
\( = 4 \hat{i} - 3 \hat{j} - \hat{k} \).
Now, find the magnitude of this cross product:
\( |\vec{d_1} \times \vec{d_2}| = \sqrt{4^2 + (-3)^2 + (-1)^2} \)
\( = \sqrt{16 + 9 + 1} \)
\( = \sqrt{26} \).
Finally, calculate the area of the parallelogram:
Area \( = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| = \frac{1}{2} \sqrt{26} \) sq. units.
In simple words: When you know the diagonal vectors of a parallelogram, you can find its area by first calculating the cross product of these diagonal vectors. Then, find the length of this cross product vector, and finally, divide that length by two. This is a special formula for finding the area when only the diagonals are known.

🎯 Exam Tip: Remember the special formula for the area of a parallelogram when given its diagonals: \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \). Do not confuse this with the formula for adjacent sides which is simply \( |\vec{a} \times \vec{b}| \).

Question 16. If \( \vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \), \( \vec{b} = -\hat{i} + \hat{k} \), \( \vec{c} = 2 \hat{j} - \hat{k} \), find the area of the parallelogram having diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \).
Answer:
Given the vectors:
\( \vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \)
\( \vec{b} = -\hat{i} + \hat{k} \)
\( \vec{c} = 2 \hat{j} - \hat{k} \)

First, find the diagonal \( \vec{d_1} = \vec{a} + \vec{b} \):
\( \vec{d_1} = (2 \hat{i} - 3 \hat{j} + \hat{k}) + (-\hat{i} + \hat{k}) \)
\( \vec{d_1} = (2-1)\hat{i} + (-3)\hat{j} + (1+1)\hat{k} \)
\( \vec{d_1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \)

Next, find the diagonal \( \vec{d_2} = \vec{b} + \vec{c} \):
\( \vec{d_2} = (-\hat{i} + \hat{k}) + (2 \hat{j} - \hat{k}) \)
\( \vec{d_2} = -\hat{i} + 2 \hat{j} + (1-1)\hat{k} \)
\( \vec{d_2} = -\hat{i} + 2 \hat{j} + 0 \hat{k} \)

Now, calculate the cross product of the diagonals \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} \] \( \implies \vec{d_1} \times \vec{d_2} = \hat{i}((-3)(0) - (2)(2)) - \hat{j}((1)(0) - (2)(-1)) + \hat{k}((1)(2) - (-3)(-1)) \)
\( \implies \vec{d_1} \times \vec{d_2} = \hat{i}(0 - 4) - \hat{j}(0 + 2) + \hat{k}(2 - 3) \)
\( \implies \vec{d_1} \times \vec{d_2} = -4 \hat{i} - 2 \hat{j} - \hat{k} \)

Find the magnitude of the cross product \( |\vec{d_1} \times \vec{d_2}| \):
\( |\vec{d_1} \times \vec{d_2}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} \)
\( = \sqrt{16 + 4 + 1} \)
\( = \sqrt{21} \)

The area of a parallelogram with diagonals \( \vec{d_1} \) and \( \vec{d_2} \) is given by \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \).
Area \( = \frac{1}{2} \sqrt{21} \) sq. units.
In simple words: First, we add the given vectors to find the two diagonal vectors of the parallelogram. Then, we find the cross product of these two diagonal vectors. The length of this cross product, divided by two, gives us the area of the parallelogram. This is a special formula for finding the area when you know the diagonals.

🎯 Exam Tip: Remember that the area of a parallelogram when its diagonals \( \vec{d_1} \) and \( \vec{d_2} \) are given is \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \). Do not forget to divide by 2.

Question 17. If \( \vec{p} \) and \( \vec{q} \) are unit vectors forming an angle of 30°; find the area of the parallelogram having \( \vec{a} = \vec{p} + 2 \vec{q} \) and \( \vec{b} = 2 \vec{p} + \vec{q} \) as its diagonals.
Answer:
Given that \( \vec{p} \) and \( \vec{q} \) are unit vectors, so \( |\vec{p}| = 1 \) and \( |\vec{q}| = 1 \). The angle between them is 30°, so \( \theta = 30^\circ \).
We are given the diagonals of the parallelogram:
\( \vec{a} = \vec{p} + 2 \vec{q} \)
\( \vec{b} = 2 \vec{p} + \vec{q} \)

First, find the cross product of the diagonals, \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = (\vec{p} + 2 \vec{q}) \times (2\vec{p} + \vec{q}) \)
\( = (\vec{p} \times 2\vec{p}) + (\vec{p} \times \vec{q}) + (2\vec{q} \times 2\vec{p}) + (2\vec{q} \times \vec{q}) \)
We know that \( \vec{x} \times \vec{x} = \overrightarrow{0} \) for any vector \( \vec{x} \), and \( \vec{y} \times \vec{x} = -(\vec{x} \times \vec{y}) \).
\( = 2(\vec{p} \times \vec{p}) + (\vec{p} \times \vec{q}) + 4(\vec{q} \times \vec{p}) + 2(\vec{q} \times \vec{q}) \)
\( = 2(\overrightarrow{0}) + (\vec{p} \times \vec{q}) - 4(\vec{p} \times \vec{q}) + 2(\overrightarrow{0}) \)
\( = -3(\vec{p} \times \vec{q}) \)

Now, find the magnitude of \( |\vec{p} \times \vec{q}| \):
\( |\vec{p} \times \vec{q}| = |\vec{p}| |\vec{q}| \sin \theta \)
\( = (1)(1) \sin 30^\circ \)
\( = \sin 30^\circ \)
\( = \frac{1}{2} \)

Substitute this back into the expression for \( |\vec{a} \times \vec{b}| \):
\( |\vec{a} \times \vec{b}| = |-3(\vec{p} \times \vec{q})| \)
\( = |-3| |\vec{p} \times \vec{q}| \)
\( = 3 \times \frac{1}{2} \)
\( = \frac{3}{2} \)

The area of a parallelogram with diagonals \( \vec{a} \) and \( \vec{b} \) is \( \frac{1}{2} |\vec{a} \times \vec{b}| \).
Area \( = \frac{1}{2} \times \frac{3}{2} \)
Area \( = \frac{3}{4} \) sq. units.
In simple words: First, we find the cross product of the given diagonal vectors. This helps us simplify the expression using the properties of vector cross products. Then, we calculate the magnitude of the cross product of the unit vectors, using the sine of the angle between them. Finally, we use a formula that says the area of a parallelogram is half the magnitude of the cross product of its diagonals.

🎯 Exam Tip: Remember the property \( \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) \) and \( \vec{u} \times \vec{u} = \overrightarrow{0} \). Also, the formula for the area of a parallelogram given diagonals \( \vec{d_1} \) and \( \vec{d_2} \) is \( \frac{1}{2}|\vec{d_1} \times \vec{d_2}| \).

Question 18. Find the area of the triangle whose adjacent sides are determined by the vectors.
(i) \( \overrightarrow{\mathrm{OA}} = 3 \hat{i} + 2 \hat{j} - \hat{k} \), \( \overrightarrow{\mathrm{OB}} = \hat{i} + 3 \hat{j} + \hat{k} \)
(ii) \( 3 \hat{i} + 4 \hat{j} \) and \( -5 \hat{i} + 7 \hat{j} \)
Answer:
(i) Given adjacent sides of the triangle as position vectors from the origin O:
\( \vec{a} = \overrightarrow{\mathrm{OA}} = 3 \hat{i} + 2 \hat{j} - \hat{k} \)
\( \vec{b} = \overrightarrow{\mathrm{OB}} = \hat{i} + 3 \hat{j} + \hat{k} \)

To find the area of the triangle, we first calculate the cross product of the two adjacent side vectors:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & 3 & 1 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i}((2)(1) - (-1)(3)) - \hat{j}((3)(1) - (-1)(1)) + \hat{k}((3)(3) - (2)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2 + 3) - \hat{j}(3 + 1) + \hat{k}(9 - 2) \)
\( \implies \vec{a} \times \vec{b} = 5 \hat{i} - 4 \hat{j} + 7 \hat{k} \)

Next, find the magnitude of the cross product \( |\vec{a} \times \vec{b}| \):
\( |\vec{a} \times \vec{b}| = \sqrt{5^2 + (-4)^2 + 7^2} \)
\( = \sqrt{25 + 16 + 49} \)
\( = \sqrt{90} \)

The area of the triangle formed by two adjacent vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{1}{2} |\vec{a} \times \vec{b}| \).
Area \( = \frac{1}{2} \sqrt{90} \)
\( = \frac{1}{2} \sqrt{9 \times 10} \)
\( = \frac{1}{2} \times 3 \sqrt{10} \)
\( = \frac{3}{2} \sqrt{10} \) sq. units.

(ii) Given adjacent sides of the triangle as vectors:
\( \vec{u} = 3 \hat{i} + 4 \hat{j} \)
\( \vec{v} = -5 \hat{i} + 7 \hat{j} \)

To find the area of the triangle, we calculate the cross product of these vectors:
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ -5 & 7 & 0 \end{vmatrix} \] \( \implies \vec{u} \times \vec{v} = \hat{i}((4)(0) - (0)(7)) - \hat{j}((3)(0) - (0)(-5)) + \hat{k}((3)(7) - (4)(-5)) \)
\( \implies \vec{u} \times \vec{v} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(21 + 20) \)
\( \implies \vec{u} \times \vec{v} = 0 \hat{i} - 0 \hat{j} + 41 \hat{k} \)

Next, find the magnitude of the cross product \( |\vec{u} \times \vec{v}| \):
\( |\vec{u} \times \vec{v}| = \sqrt{0^2 + 0^2 + 41^2} \)
\( = \sqrt{41^2} \)
\( = 41 \)

The area of the triangle formed by two adjacent vectors \( \vec{u} \) and \( \vec{v} \) is \( \frac{1}{2} |\vec{u} \times \vec{v}| \).
Area \( = \frac{1}{2} \times 41 \)
Area \( = \frac{41}{2} \) sq. units.
In simple words: For a triangle with two sides given as vectors, we first find the cross product of these vectors. The length of this cross product, cut in half, gives us the area of the triangle. This works because the magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by them, and a triangle is half a parallelogram.

🎯 Exam Tip: Remember the formula for the area of a triangle given two adjacent sides \( \vec{u} \) and \( \vec{v} \) is \( \frac{1}{2} |\vec{u} \times \vec{v}| \). If position vectors are given, convert them to actual side vectors by subtraction first.

Question 19. Find by vector method, the area of triangle A B C whose vertices are
(i) A(1, 3, 2), B(2, -1, 1) and C(-1, 2, 3)
(ii) A(1, 2, 3), B(2, 5, -1) and C(-1, 1, 2).
Answer:
(i) Given the vertices of triangle ABC:
A(1, 3, 2), B(2, -1, 1), C(-1, 2, 3)

First, find the position vectors of the vertices relative to the origin O:
\( \overrightarrow{\mathrm{OA}} = \hat{i} + 3 \hat{j} + 2 \hat{k} \)
\( \overrightarrow{\mathrm{OB}} = 2 \hat{i} - \hat{j} + \hat{k} \)
\( \overrightarrow{\mathrm{OC}} = -\hat{i} + 2 \hat{j} + 3 \hat{k} \)

Next, find two adjacent side vectors of the triangle, for example, \( \overrightarrow{\mathrm{AB}} \) and \( \overrightarrow{\mathrm{AC}} \).
\( \overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{OB}} - \overrightarrow{\mathrm{OA}} \)
\( = (2 \hat{i} - \hat{j} + \hat{k}) - (\hat{i} + 3 \hat{j} + 2 \hat{k}) \)
\( = (2-1)\hat{i} + (-1-3)\hat{j} + (1-2)\hat{k} \)
\( = \hat{i} - 4 \hat{j} - \hat{k} \)

\( \overrightarrow{\mathrm{AC}} = \overrightarrow{\mathrm{OC}} - \overrightarrow{\mathrm{OA}} \)
\( = (-\hat{i} + 2 \hat{j} + 3 \hat{k}) - (\hat{i} + 3 \hat{j} + 2 \hat{k}) \)
\( = (-1-1)\hat{i} + (2-3)\hat{j} + (3-2)\hat{k} \)
\( = -2 \hat{i} - \hat{j} + \hat{k} \)

Now, calculate the cross product \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} \):
\[ \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & -1 \\ -2 & -1 & 1 \end{vmatrix} \] \( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}((-4)(1) - (-1)(-1)) - \hat{j}((1)(1) - (-1)(-2)) + \hat{k}((1)(-1) - (-4)(-2)) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}(-4 - 1) - \hat{j}(1 - 2) + \hat{k}(-1 - 8) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = -5 \hat{i} + \hat{j} - 9 \hat{k} \)

Find the magnitude of the cross product \( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \):
\( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{(-5)^2 + 1^2 + (-9)^2} \)
\( = \sqrt{25 + 1 + 81} \)
\( = \sqrt{107} \)

The area of triangle ABC is \( \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \).
Area \( = \frac{1}{2} \sqrt{107} \) sq. units.

(ii) Given the vertices of triangle ABC:
A(1, 2, 3), B(2, 5, -1), C(-1, 1, 2)

Position vectors of the vertices:
\( \overrightarrow{\mathrm{OA}} = \hat{i} + 2 \hat{j} + 3 \hat{k} \)
\( \overrightarrow{\mathrm{OB}} = 2 \hat{i} + 5 \hat{j} - \hat{k} \)
\( \overrightarrow{\mathrm{OC}} = -\hat{i} + \hat{j} + 2 \hat{k} \)

Side vectors \( \overrightarrow{\mathrm{AB}} \) and \( \overrightarrow{\mathrm{AC}} \):
\( \overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{OB}} - \overrightarrow{\mathrm{OA}} \)
\( = (2 \hat{i} + 5 \hat{j} - \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
\( = (2-1)\hat{i} + (5-2)\hat{j} + (-1-3)\hat{k} \)
\( = \hat{i} + 3 \hat{j} - 4 \hat{k} \)

\( \overrightarrow{\mathrm{AC}} = \overrightarrow{\mathrm{OC}} - \overrightarrow{\mathrm{OA}} \)
\( = (-\hat{i} + \hat{j} + 2 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
\( = (-1-1)\hat{i} + (1-2)\hat{j} + (2-3)\hat{k} \)
\( = -2 \hat{i} - \hat{j} - \hat{k} \)

Now, calculate the cross product \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} \):
\[ \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -4 \\ -2 & -1 & -1 \end{vmatrix} \] \( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}((3)(-1) - (-4)(-1)) - \hat{j}((1)(-1) - (-4)(-2)) + \hat{k}((1)(-1) - (3)(-2)) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}(-3 - 4) - \hat{j}(-1 - 8) + \hat{k}(-1 + 6) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = -7 \hat{i} + 9 \hat{j} + 5 \hat{k} \)

Find the magnitude of the cross product \( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \):
\( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{(-7)^2 + 9^2 + 5^2} \)
\( = \sqrt{49 + 81 + 25} \)
\( = \sqrt{155} \)

The area of triangle ABC is \( \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \).
Area \( = \frac{1}{2} \sqrt{155} \) sq. units.
In simple words: To find the area of a triangle when you know its corner points (vertices), first find the vectors for two sides that meet at one corner. You do this by subtracting the position vector of the starting point from the position vector of the ending point. Then, find the cross product of these two side vectors. The length of this cross product, divided by two, gives you the area of the triangle.

🎯 Exam Tip: When finding side vectors from vertices, make sure to subtract the position vector of the initial point from the position vector of the terminal point (e.g., \( \overrightarrow{\mathrm{AB}} = \vec{B} - \vec{A} \)). Consistency in the starting point for both side vectors is crucial for the cross product to correctly represent the triangle's area.

Question 20. Show that the points whose position vectors are given below, are collinear.
(i) \( \vec{\alpha} = \vec{a} - 2 \vec{b} + 3 \vec{c} \), \( \vec{\beta} = 2 \vec{a} + 3 \vec{b} - 4 \vec{c} \) and \( \vec{\gamma} = -7 \vec{b} + 10 \vec{c} \)
(ii) \( 5 \hat{i} + 6 \hat{j} + 7 \hat{k} \), \( 7 \hat{i} - 8 \hat{j} + 9 \hat{k} \) and \( 3 \hat{i} + 20 \hat{j} + 5 \hat{k} \)
Answer:
(i) Let the given position vectors be \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \):
\( \vec{\alpha} = \vec{a} - 2 \vec{b} + 3 \vec{c} \)
\( \vec{\beta} = 2 \vec{a} + 3 \vec{b} - 4 \vec{c} \)
\( \vec{\gamma} = -7 \vec{b} + 10 \vec{c} \)

For three points to be collinear, the vector cross product of any two vectors formed by these points must be a zero vector (i.e., \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \overrightarrow{0} \)). Another condition is that \( \vec{\alpha} \times \vec{\beta} + \vec{\beta} \times \vec{\gamma} + \vec{\gamma} \times \vec{\alpha} = \overrightarrow{0} \). Let's use the latter for this general case.

Calculate \( \vec{\alpha} \times \vec{\beta} \):
\( \vec{\alpha} \times \vec{\beta} = (\vec{a} - 2 \vec{b} + 3 \vec{c}) \times (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) \)
\( = (\vec{a} \times 2\vec{a}) + (\vec{a} \times 3\vec{b}) + (\vec{a} \times -4\vec{c}) + (-2\vec{b} \times 2\vec{a}) + (-2\vec{b} \times 3\vec{b}) + (-2\vec{b} \times -4\vec{c}) + (3\vec{c} \times 2\vec{a}) + (3\vec{c} \times 3\vec{b}) + (3\vec{c} \times -4\vec{c}) \)
Using \( \vec{x} \times \vec{x} = \overrightarrow{0} \) and \( \vec{y} \times \vec{x} = -(\vec{x} \times \vec{y}) \):
\( = \overrightarrow{0} + 3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{c}) + 4(\vec{a} \times \vec{b}) + \overrightarrow{0} - 8(\vec{b} \times \vec{c}) + 6(\vec{c} \times \vec{a}) + 9(\vec{c} \times \vec{b}) + \overrightarrow{0} \)
\( = 7(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{c}) - 8(\vec{b} \times \vec{c}) + 6(\vec{c} \times \vec{a}) - 9(\vec{b} \times \vec{c}) \)
\( = 7(\vec{a} \times \vec{b}) + 4(\vec{c} \times \vec{a}) - 8(\vec{b} \times \vec{c}) + 6(\vec{c} \times \vec{a}) - 9(\vec{b} \times \vec{c}) \)
\( = 7(\vec{a} \times \vec{b}) + 10(\vec{c} \times \vec{a}) - 17(\vec{b} \times \vec{c}) \) (Equation 1)

Calculate \( \vec{\beta} \times \vec{\gamma} \):
\( \vec{\beta} \times \vec{\gamma} = (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) \times (-7 \vec{b} + 10 \vec{c}) \)
\( = (2\vec{a} \times -7\vec{b}) + (2\vec{a} \times 10\vec{c}) + (3\vec{b} \times -7\vec{b}) + (3\vec{b} \times 10\vec{c}) + (-4\vec{c} \times -7\vec{b}) + (-4\vec{c} \times 10\vec{c}) \)
\( = -14(\vec{a} \times \vec{b}) + 20(\vec{a} \times \vec{c}) + \overrightarrow{0} + 30(\vec{b} \times \vec{c}) - 28(\vec{c} \times \vec{b}) + \overrightarrow{0} \)
\( = -14(\vec{a} \times \vec{b}) - 20(\vec{c} \times \vec{a}) + 30(\vec{b} \times \vec{c}) + 28(\vec{b} \times \vec{c}) \)
\( = -14(\vec{a} \times \vec{b}) - 20(\vec{c} \times \vec{a}) + 58(\vec{b} \times \vec{c}) \) (Equation 2)

Calculate \( \vec{\gamma} \times \vec{\alpha} \):
\( \vec{\gamma} \times \vec{\alpha} = (-7 \vec{b} + 10 \vec{c}) \times (\vec{a} - 2 \vec{b} + 3 \vec{c}) \)
\( = (-7\vec{b} \times \vec{a}) + (-7\vec{b} \times -2\vec{b}) + (-7\vec{b} \times 3\vec{c}) + (10\vec{c} \times \vec{a}) + (10\vec{c} \times -2\vec{b}) + (10\vec{c} \times 3\vec{c}) \)
\( = 7(\vec{a} \times \vec{b}) + \overrightarrow{0} - 21(\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a}) + 20(\vec{b} \times \vec{c}) + \overrightarrow{0} \)
\( = 7(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a}) \) (Equation 3)

Now, add Equations 1, 2, and 3:
\( (\vec{\alpha} \times \vec{\beta}) + (\vec{\beta} \times \vec{\gamma}) + (\vec{\gamma} \times \vec{\alpha}) \)
\( = (7(\vec{a} \times \vec{b}) + 10(\vec{c} \times \vec{a}) - 17(\vec{b} \times \vec{c})) + (-14(\vec{a} \times \vec{b}) - 20(\vec{c} \times \vec{a}) + 58(\vec{b} \times \vec{c})) + (7(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a})) \)
Combine terms for \( (\vec{a} \times \vec{b}) \): \( 7 - 14 + 7 = 0 \)
Combine terms for \( (\vec{c} \times \vec{a}) \): \( 10 - 20 + 10 = 0 \)
Combine terms for \( (\vec{b} \times \vec{c}) \): \( -17 + 58 - 1 = 40 \)
This result is \( 40(\vec{b} \times \vec{c}) \), which is not \( \overrightarrow{0} \).

Let's re-evaluate for collinearity using the condition that \( \overrightarrow{\mathrm{AB}} = k \overrightarrow{\mathrm{BC}} \) for some scalar \( k \).
Let P, Q, R be the points corresponding to \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \).
\( \overrightarrow{\mathrm{PQ}} = \vec{\beta} - \vec{\alpha} = (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) - (\vec{a} - 2 \vec{b} + 3 \vec{c}) \)
\( = (2-1)\vec{a} + (3-(-2))\vec{b} + (-4-3)\vec{c} \)
\( = \vec{a} + 5 \vec{b} - 7 \vec{c} \)

\( \overrightarrow{\mathrm{QR}} = \vec{\gamma} - \vec{\beta} = (-7 \vec{b} + 10 \vec{c}) - (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) \)
\( = -2 \vec{a} + (-7-3)\vec{b} + (10-(-4))\vec{c} \)
\( = -2 \vec{a} - 10 \vec{b} + 14 \vec{c} \)

Notice that \( \overrightarrow{\mathrm{QR}} = -2(\vec{a} + 5 \vec{b} - 7 \vec{c}) \).
So, \( \overrightarrow{\mathrm{QR}} = -2 \overrightarrow{\mathrm{PQ}} \).
Since \( \overrightarrow{\mathrm{QR}} \) is a scalar multiple of \( \overrightarrow{\mathrm{PQ}} \), the vectors are parallel. Also, since they share a common point Q, the points P, Q, R are collinear.

(ii) Let the position vectors of the three points be A, B, C:
\( \vec{A} = 5 \hat{i} + 6 \hat{j} + 7 \hat{k} \)
\( \vec{B} = 7 \hat{i} - 8 \hat{j} + 9 \hat{k} \)
\( \vec{C} = 3 \hat{i} + 20 \hat{j} + 5 \hat{k} \)

We need to show that \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \overrightarrow{0} \) or that one vector formed by the points is a scalar multiple of another.

Calculate \( \overrightarrow{\mathrm{AB}} = \vec{B} - \vec{A} \):
\( \overrightarrow{\mathrm{AB}} = (7 \hat{i} - 8 \hat{j} + 9 \hat{k}) - (5 \hat{i} + 6 \hat{j} + 7 \hat{k}) \)
\( = (7-5)\hat{i} + (-8-6)\hat{j} + (9-7)\hat{k} \)
\( = 2 \hat{i} - 14 \hat{j} + 2 \hat{k} \)

Calculate \( \overrightarrow{\mathrm{AC}} = \vec{C} - \vec{A} \):
\( \overrightarrow{\mathrm{AC}} = (3 \hat{i} + 20 \hat{j} + 5 \hat{k}) - (5 \hat{i} + 6 \hat{j} + 7 \hat{k}) \)
\( = (3-5)\hat{i} + (20-6)\hat{j} + (5-7)\hat{k} \)
\( = -2 \hat{i} + 14 \hat{j} - 2 \hat{k} \)

Notice that \( \overrightarrow{\mathrm{AC}} = -1 \times (2 \hat{i} - 14 \hat{j} + 2 \hat{k}) \).
So, \( \overrightarrow{\mathrm{AC}} = - \overrightarrow{\mathrm{AB}} \).
Since \( \overrightarrow{\mathrm{AC}} \) is a scalar multiple of \( \overrightarrow{\mathrm{AB}} \), the vectors are parallel. Because they share a common point A, the points A, B, C are collinear.
In simple words: To show that three points are in a straight line (collinear), we can make vectors from these points. If one vector is just a scaled version of another vector (like being double or negative), and they share a common point, then all three points must lie on the same line. We check if \( \overrightarrow{\mathrm{AB}} = k \overrightarrow{\mathrm{BC}} \) or \( \overrightarrow{\mathrm{AB}} = k \overrightarrow{\mathrm{AC}} \).

🎯 Exam Tip: To prove collinearity of three points A, B, C using vectors, show that \( \overrightarrow{\mathrm{AB}} \) is a scalar multiple of \( \overrightarrow{\mathrm{BC}} \) (i.e., \( \overrightarrow{\mathrm{AB}} = k \overrightarrow{\mathrm{BC}} \)) or \( \overrightarrow{\mathrm{AB}} \) is a scalar multiple of \( \overrightarrow{\mathrm{AC}} \). This means the vectors are parallel and share a common point, thus they are collinear.

Question 21. Find \( \lambda \) such that \( \vec{a} = \hat{i} + \lambda \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) are parallel.
Answer:
Given the vectors:
\( \vec{a} = \hat{i} + \lambda \hat{j} + 3 \hat{k} \)
\( \vec{b} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \)

If two vectors are parallel, their cross product is the zero vector ( \( \vec{a} \times \vec{b} = \overrightarrow{0} \) ).

Calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \lambda & 3 \\ 3 & 2 & 9 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i}((\lambda)(9) - (3)(2)) - \hat{j}((1)(9) - (3)(3)) + \hat{k}((1)(2) - (\lambda)(3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(9\lambda - 6) - \hat{j}(9 - 9) + \hat{k}(2 - 3\lambda) \)
\( \implies \vec{a} \times \vec{b} = (9\lambda - 6) \hat{i} + 0 \hat{j} + (2 - 3\lambda) \hat{k} \)

Since \( \vec{a} \) and \( \vec{b} \) are parallel, \( \vec{a} \times \vec{b} = \overrightarrow{0} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \).
This means the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) must all be zero.
For \( \hat{i} \): \( 9\lambda - 6 = 0 \)
\( \implies 9\lambda = 6 \)
\( \implies \lambda = \frac{6}{9} \)
\( \implies \lambda = \frac{2}{3} \)

For \( \hat{j} \): \( 0 = 0 \) (this is already true)

For \( \hat{k} \): \( 2 - 3\lambda = 0 \)
\( \implies 2 = 3\lambda \)
\( \implies \lambda = \frac{2}{3} \)

Both conditions give \( \lambda = \frac{2}{3} \). Therefore, the value of \( \lambda \) for which the vectors are parallel is \( \frac{2}{3} \).
In simple words: When two vectors are parallel, their cross product is always zero. So, we calculate the cross product of the given vectors and set each component (i, j, k) to zero. Solving these equations for lambda will give us the value that makes the vectors parallel.

🎯 Exam Tip: For vectors to be parallel, either their cross product is the zero vector, or their components are proportional (e.g., if \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) are parallel, then \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \)). Both methods will yield the same result.

Question 22. Find the values of a for which the vectors \( 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) and \( \hat{i} + a \hat{j} + 3 \hat{k} \) are
(i) perpendicular
(ii) parallel.
Answer:
Let the given vectors be \( \vec{u} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) and \( \vec{v} = \hat{i} + a \hat{j} + 3 \hat{k} \).

(i) For vectors to be perpendicular, their dot product must be zero ( \( \vec{u} \cdot \vec{v} = 0 \) ).
\( (3 \hat{i} + 2 \hat{j} + 9 \hat{k}) \cdot (\hat{i} + a \hat{j} + 3 \hat{k}) = 0 \)
\( (3)(1) + (2)(a) + (9)(3) = 0 \)
\( 3 + 2a + 27 = 0 \)
\( 2a + 30 = 0 \)<
\( 2a = -30 \)
\( a = -15 \)

(ii) For vectors to be parallel, their cross product must be the zero vector ( \( \vec{u} \times \vec{v} = \overrightarrow{0} \) ).
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 9 \\ 1 & a & 3 \end{vmatrix} \] \( \implies \vec{u} \times \vec{v} = \hat{i}((2)(3) - (9)(a)) - \hat{j}((3)(3) - (9)(1)) + \hat{k}((3)(a) - (2)(1)) \)
\( \implies \vec{u} \times \vec{v} = \hat{i}(6 - 9a) - \hat{j}(9 - 9) + \hat{k}(3a - 2) \)
\( \implies \vec{u} \times \vec{v} = (6 - 9a) \hat{i} + 0 \hat{j} + (3a - 2) \hat{k} \)

Set each component to zero:
For \( \hat{i} \): \( 6 - 9a = 0 \)
\( \implies 6 = 9a \)
\( \implies a = \frac{6}{9} = \frac{2}{3} \)

For \( \hat{j} \): \( 0 = 0 \) (already true)

For \( \hat{k} \): \( 3a - 2 = 0 \)
\( \implies 3a = 2 \)
\( \implies a = \frac{2}{3} \)

Both conditions give \( a = \frac{2}{3} \).
In simple words: If vectors are perpendicular, their dot product is zero, which helps us find the value of 'a'. If they are parallel, their cross product is the zero vector, and setting each component of the cross product to zero will also help us find 'a'. These are fundamental rules for how vectors interact in space.

🎯 Exam Tip: Distinguish clearly between the conditions for perpendicular vectors (dot product is zero) and parallel vectors (cross product is the zero vector, or components are proportional). Applying the correct condition is key to solving these types of problems.

Question 23. Find a unit vector parallel to the x y plane and perpendicular to the vector \( 4 \hat{i} - 3 \hat{j} + \hat{k} \).
Answer:
A vector parallel to the x y plane has its z-component equal to 0. So, we can represent such a vector as \( \vec{c} = \alpha \hat{i} + \beta \hat{j} \).
We are given that this vector \( \vec{c} \) is perpendicular to \( \vec{u} = 4 \hat{i} - 3 \hat{j} + \hat{k} \).

If two vectors are perpendicular, their dot product is zero:
\( \vec{c} \cdot \vec{u} = 0 \)
\( (\alpha \hat{i} + \beta \hat{j}) \cdot (4 \hat{i} - 3 \hat{j} + \hat{k}) = 0 \)
\( (\alpha)(4) + (\beta)(-3) + (0)(1) = 0 \)
\( 4\alpha - 3\beta = 0 \)
\( 4\alpha = 3\beta \)
\( \beta = \frac{4\alpha}{3} \)

So, the vector \( \vec{c} \) can be written as:
\( \vec{c} = \alpha \hat{i} + \frac{4\alpha}{3} \hat{j} \)
Take \( \alpha \) common:
\( \vec{c} = \alpha (\hat{i} + \frac{4}{3} \hat{j}) \)
This can also be written as \( \vec{c} = \frac{\alpha}{3} (3 \hat{i} + 4 \hat{j}) \).

We need to find a unit vector. Let \( K = \frac{\alpha}{3} \). Then \( \vec{c} = K (3 \hat{i} + 4 \hat{j}) \).
The unit vector in the direction of \( \vec{c} \) is given by \( \hat{c} = \frac{\vec{c}}{|\vec{c}|} \).
Let's find the magnitude of \( (3 \hat{i} + 4 \hat{j}) \):
\( |3 \hat{i} + 4 \hat{j}| = \sqrt{3^2 + 4^2} \)
\( = \sqrt{9 + 16} \)
\( = \sqrt{25} \)
\( = 5 \)

So, the unit vector is:
\( \hat{c} = \pm \frac{1}{5} (3 \hat{i} + 4 \hat{j}) \)
The \( \pm \) sign means there are two such unit vectors, pointing in opposite directions.
In simple words: First, we define a vector that lies flat on the x y plane, meaning it has no up or down (z-component is zero). Then, we use the rule that if two vectors are at a right angle, their dot product is zero, to find a relationship between the x and y parts of our unknown vector. After finding this relationship, we simplify the vector and then divide it by its length to make it a unit vector (a vector with length 1).

🎯 Exam Tip: A vector parallel to the x y plane will always have a zero z-component. Remember to include the \( \pm \) sign for unit vectors, as there are always two opposite directions that satisfy the conditions.

Question 24. By vector method obtain the perpendicular distance of the point (6,-4,4) from the line passing through the points (2,1,2) and (3,-1,4).
Answer:
Let P be the given point (6,-4,4). Let A be (2,1,2) and B be (3,-1,4) which are points on the line.

Position vectors:
\( \overrightarrow{\mathrm{P}} = 6 \hat{i} - 4 \hat{j} + 4 \hat{k} \)
\( \overrightarrow{\mathrm{A}} = 2 \hat{i} + \hat{j} + 2 \hat{k} \)
\( \overrightarrow{\mathrm{B}} = 3 \hat{i} - \hat{j} + 4 \hat{k} \)

First, find the vector \( \overrightarrow{\mathrm{AP}} \):
\( \overrightarrow{\mathrm{AP}} = \overrightarrow{\mathrm{P}} - \overrightarrow{\mathrm{A}} \)
\( = (6 \hat{i} - 4 \hat{j} + 4 \hat{k}) - (2 \hat{i} + \hat{j} + 2 \hat{k}) \)
\( = (6-2)\hat{i} + (-4-1)\hat{j} + (4-2)\hat{k} \)
\( = 4 \hat{i} - 5 \hat{j} + 2 \hat{k} \)

Next, find the vector \( \overrightarrow{\mathrm{AB}} \) (the direction vector of the line):
\( \overrightarrow{\mathrm{AB}} = \overrightarrow{\mathrm{B}} - \overrightarrow{\mathrm{A}} \)
\( = (3 \hat{i} - \hat{j} + 4 \hat{k}) - (2 \hat{i} + \hat{j} + 2 \hat{k}) \)
\( = (3-2)\hat{i} + (-1-1)\hat{j} + (4-2)\hat{k} \)
\( = \hat{i} - 2 \hat{j} + 2 \hat{k} \)

Now, calculate the cross product \( \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} \):
\[ \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & 2 \\ 1 & -2 & 2 \end{vmatrix} \] \( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \hat{i}((-5)(2) - (2)(-2)) - \hat{j}((4)(2) - (2)(1)) + \hat{k}((4)(-2) - (-5)(1)) \)
\( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \hat{i}(-10 + 4) - \hat{j}(8 - 2) + \hat{k}(-8 + 5) \)
\( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = -6 \hat{i} - 6 \hat{j} - 3 \hat{k} \)

Find the magnitude of the cross product \( |\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}| \):
\( |\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}| = \sqrt{(-6)^2 + (-6)^2 + (-3)^2} \)
\( = \sqrt{36 + 36 + 9} \)
\( = \sqrt{81} \)
\( = 9 \)

Find the magnitude of the direction vector \( |\overrightarrow{\mathrm{AB}}| \):
\( |\overrightarrow{\mathrm{AB}}| = \sqrt{1^2 + (-2)^2 + 2^2} \)
\( = \sqrt{1 + 4 + 4} \)
\( = \sqrt{9} \)
\( = 3 \)

The perpendicular distance (d) from point P to the line passing through A and B is given by:
\( d = \frac{|\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{AB}}|} \)
\( d = \frac{9}{3} \)
\( d = 3 \) units.
In simple words: To find the shortest distance from a point to a line, we first create a vector from a point on the line to the given external point. Then, we find the direction vector of the line itself. We take the cross product of these two vectors and find its length. Finally, we divide this length by the length of the line's direction vector. This formula comes from the geometric meaning of the cross product as the area of a parallelogram.

🎯 Exam Tip: The perpendicular distance from a point P to a line passing through A and B is given by \( \frac{|\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{AB}}|} \). Remember to correctly calculate the vectors \( \overrightarrow{\mathrm{AP}} \) and \( \overrightarrow{\mathrm{AB}} \) first.

Question 25. Given \( \vec{a} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \), \( \vec{b} = \frac{1}{7}(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \), \( \vec{c} = \frac{1}{7}(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \), \( \hat{i}, \hat{j}, \hat{k} \) being a right handed orthogonal system of unit vectors in space, show that \( \vec{a}, \vec{b}, \vec{c} \) is also another system of unit vectors mutually perpendicular.
Answer:
Given the vectors:
\( \vec{a} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \)
\( \vec{b} = \frac{1}{7}(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \)
\( \vec{c} = \frac{1}{7}(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \)

To show they form a system of unit vectors mutually perpendicular, we need to prove two things:
1. Each vector is a unit vector (its magnitude is 1).
2. Each pair of vectors is perpendicular (their dot product is 0).

**Step 1: Check if they are unit vectors.**
Calculate \( |\vec{a}| \):
\( |\vec{a}| = \left| \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \right| = \frac{1}{7} \sqrt{2^2 + 3^2 + 6^2} \)
\( = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1 \)

Calculate \( |\vec{b}| \):
\( |\vec{b}| = \left| \frac{1}{7}(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \right| = \frac{1}{7} \sqrt{3^2 + (-6)^2 + 2^2} \)
\( = \frac{1}{7} \sqrt{9 + 36 + 4} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1 \)

Calculate \( |\vec{c}| \):
\( |\vec{c}| = \left| \frac{1}{7}(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \right| = \frac{1}{7} \sqrt{6^2 + 2^2 + (-3)^2} \)
\( = \frac{1}{7} \sqrt{36 + 4 + 9} = \frac{1}{7} \sqrt{49} = \frac{1}{7} \times 7 = 1 \)
All three vectors are unit vectors.

**Step 2: Check if they are mutually perpendicular.**
Calculate \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \cdot \frac{1}{7}(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \)
\( = \frac{1}{49} ((2)(3) + (3)(-6) + (6)(2)) \)
\( = \frac{1}{49} (6 - 18 + 12) = \frac{1}{49} (0) = 0 \)

Calculate \( \vec{b} \cdot \vec{c} \):
\( \vec{b} \cdot \vec{c} = \frac{1}{7}(3 \hat{i} - 6 \hat{j} + 2 \hat{k}) \cdot \frac{1}{7}(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \)
\( = \frac{1}{49} ((3)(6) + (-6)(2) + (2)(-3)) \)
\( = \frac{1}{49} (18 - 12 - 6) = \frac{1}{49} (0) = 0 \)

Calculate \( \vec{c} \cdot \vec{a} \):
\( \vec{c} \cdot \vec{a} = \frac{1}{7}(6 \hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \)
\( = \frac{1}{49} ((6)(2) + (2)(3) + (-3)(6)) \)
\( = \frac{1}{49} (12 + 6 - 18) = \frac{1}{49} (0) = 0 \)
All pairs of vectors are perpendicular.

**Step 3: Check for right-handed system (optional but good practice).**
A right-handed system means \( \vec{a} \times \vec{b} = \vec{c} \), \( \vec{b} \times \vec{c} = \vec{a} \), and \( \vec{c} \times \vec{a} = \vec{b} \).
Let's check \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \frac{1}{49} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{vmatrix} \] \( \implies = \frac{1}{49} [\hat{i}((3)(2) - (6)(-6)) - \hat{j}((2)(2) - (6)(3)) + \hat{k}((2)(-6) - (3)(3))] \)
\( \implies = \frac{1}{49} [\hat{i}(6 + 36) - \hat{j}(4 - 18) + \hat{k}(-12 - 9)] \)
\( \implies = \frac{1}{49} [42 \hat{i} + 14 \hat{j} - 21 \hat{k}] \)
\( \implies = \frac{7}{49} [6 \hat{i} + 2 \hat{j} - 3 \hat{k}] \)
\( \implies = \frac{1}{7} [6 \hat{i} + 2 \hat{j} - 3 \hat{k}] = \vec{c} \)
This confirms \( \vec{a} \times \vec{b} = \vec{c} \). The other cross products will follow similarly.

Thus, \( \vec{a}, \vec{b}, \vec{c} \) form a system of unit vectors that are mutually perpendicular and constitute a right-handed system.
In simple words: To show that these three given vectors form a special set of 'mutually perpendicular unit vectors', we need to check two main things. First, we confirm that each vector has a length of exactly one (a unit vector). Second, we check that each pair of vectors is at a right angle to each other by making sure their dot product is zero. If both conditions are met, they form such a system.

🎯 Exam Tip: For a set of vectors to be an orthonormal basis (mutually perpendicular unit vectors), each vector must have a magnitude of 1, and the dot product of any two distinct vectors must be 0. A right-handed system additionally requires the cross product to follow a specific cyclic pattern.

Question 26. If \( \vec{a}, \vec{b}, \vec{c} \) represent the lengths of the sides opposite to vertices A, B, C respectively, of a triangle ABC, and \( |\vec{a}| = 13, |\vec{b}|= 14, |\vec{c}|= 15 \), find angle C.
Answer:
Given the magnitudes of the side vectors of a triangle:
\( |\vec{a}| = 13 \implies a = 13 \)
\( |\vec{b}| = 14 \implies b = 14 \)
\( |\vec{c}| = 15 \implies c = 15 \)

We need to find angle C. We can use the cosine rule for triangles, which relates the sides and angles:
\( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)

First, calculate the squares of the side lengths:
\( a^2 = 13^2 = 169 \)
\( b^2 = 14^2 = 196 \)
\( c^2 = 15^2 = 225 \)

Substitute these values into the cosine rule formula:
\( \cos C = \frac{169 + 196 - 225}{2 \times 13 \times 14} \)
\( = \frac{365 - 225}{364} \)
\( = \frac{140}{364} \)

Simplify the fraction:
Divide by 2: \( \frac{70}{182} \)
Divide by 2 again: \( \frac{35}{91} \)
Divide by 7: \( \frac{5}{13} \)
So, \( \cos C = \frac{5}{13} \)

To find angle C, take the inverse cosine (arc cos):
\( C = \cos^{-1}\left(\frac{5}{13}\right) \)
In simple words: We are given the lengths of all three sides of a triangle. To find one of the angles, like angle C, we use a special formula called the cosine rule. This formula helps us connect the lengths of the sides to the cosine of the angle. After putting in the side lengths and doing the math, we can find the angle using a calculator.

🎯 Exam Tip: The cosine rule \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \) is essential for finding angles when all three side lengths of a triangle are known. Ensure accurate calculation of squares and products to avoid errors.

Question 27. If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \) and \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \), where \( \vec{a} \neq \overrightarrow{0} \), prove that \( \vec{b} = \vec{c} \).
Answer:
Given the conditions:
1. \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \)
2. \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \)
3. \( \vec{a} \neq \overrightarrow{0} \)

**From condition 1:**
\( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \)
\( \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0 \)
\( \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \) (Equation A)
This implies that either \( \vec{a} = \overrightarrow{0} \) (which is ruled out by condition 3), or \( \vec{b} - \vec{c} = \overrightarrow{0} \), or \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).
So, \( \vec{b} = \vec{c} \) or \( \vec{a} \perp (\vec{b} - \vec{c}) \).

**From condition 2:**
\( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \)
\( \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = \overrightarrow{0} \)
\( \vec{a} \times (\vec{b} - \vec{c}) = \overrightarrow{0} \) (Equation B)
This implies that either \( \vec{a} = \overrightarrow{0} \) (ruled out by condition 3), or \( \vec{b} - \vec{c} = \overrightarrow{0} \), or \( \vec{a} \) is parallel to \( (\vec{b} - \vec{c}) \).
So, \( \vec{b} = \vec{c} \) or \( \vec{a} \parallel (\vec{b} - \vec{c}) \).

Now, consider the possibilities from Equations A and B, given \( \vec{a} \neq \overrightarrow{0} \):
* If \( \vec{b} - \vec{c} = \overrightarrow{0} \), then \( \vec{b} = \vec{c} \). This satisfies both equations.
* If \( \vec{b} - \vec{c} \neq \overrightarrow{0} \), then from Equation A, \( \vec{a} \) must be perpendicular to \( (\vec{b} - \vec{c}) \).
And from Equation B, \( \vec{a} \) must be parallel to \( (\vec{b} - \vec{c}) \).
A non-zero vector \( \vec{a} \) cannot be both perpendicular and parallel to another non-zero vector \( (\vec{b} - \vec{c}) \) simultaneously.
The only way a vector can be both parallel and perpendicular to another vector is if one of them is the zero vector. Since \( \vec{a} \neq \overrightarrow{0} \), this means \( \vec{b} - \vec{c} \) must be the zero vector.

Therefore, the only consistent conclusion is that \( \vec{b} - \vec{c} = \overrightarrow{0} \), which means \( \vec{b} = \vec{c} \).
In simple words: We are given two conditions about vectors: one with dot products and one with cross products. Both conditions, when rearranged, tell us something about the vector \( (\vec{b} - \vec{c}) \). The first condition says \( \vec{a} \) is either zero or perpendicular to \( (\vec{b} - \vec{c}) \). The second says \( \vec{a} \) is either zero or parallel to \( (\vec{b} - \vec{c}) \). Since \( \vec{a} \) is not zero, the only way for \( \vec{a} \) to be both perpendicular and parallel to \( (\vec{b} - \vec{c}) \) at the same time is if \( (\vec{b} - \vec{c}) \) itself is the zero vector. This means \( \vec{b} \) and \( \vec{c} \) must be the same vector.

🎯 Exam Tip: This proof relies on the key properties of dot and cross products: if \( \vec{u} \cdot \vec{v} = 0 \), then \( \vec{u} \perp \vec{v} \) (or one is zero); if \( \vec{u} \times \vec{v} = \overrightarrow{0} \), then \( \vec{u} \parallel \vec{v} \) (or one is zero). A non-zero vector cannot be simultaneously parallel and perpendicular to another non-zero vector.

Question 28. If \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) then show that
(i) \( (\vec{r} \times \hat{i})^2 = y^2 + z^2 \)
(ii) \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + x y = 0 \)
Answer:
Given \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).

(i) Show that \( (\vec{r} \times \hat{i})^2 = y^2 + z^2 \).
First, calculate the cross product \( \vec{r} \times \hat{i} \):
\( \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} \)
\( = x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i}) \)
We know \( \hat{i} \times \hat{i} = \overrightarrow{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), and \( \hat{k} \times \hat{i} = \hat{j} \).
\( = x(\overrightarrow{0}) + y(-\hat{k}) + z(\hat{j}) \)
\( = z \hat{j} - y \hat{k} \)

Now, find the square of its magnitude, \( (\vec{r} \times \hat{i})^2 = | \vec{r} \times \hat{i} |^2 \):
\( | z \hat{j} - y \hat{k} |^2 = (\sqrt{z^2 + (-y)^2})^2 \)
\( = z^2 + y^2 \)
Thus, \( (\vec{r} \times \hat{i})^2 = y^2 + z^2 \). (Proved)

(ii) Show that \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + x y = 0 \).
From part (i), we have \( \vec{r} \times \hat{i} = z \hat{j} - y \hat{k} \).

Next, calculate the cross product \( \vec{r} \times \hat{j} \):
\( \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{j} \)
\( = x(\hat{i} \times \hat{j}) + y(\hat{j} \times \hat{j}) + z(\hat{k} \times \hat{j}) \)
We know \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \overrightarrow{0} \), and \( \hat{k} \times \hat{j} = -\hat{i} \).
\( = x(\hat{k}) + y(\overrightarrow{0}) + z(-\hat{i}) \)
\( = x \hat{k} - z \hat{i} \)

Now, calculate the dot product \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) \):
\( (z \hat{j} - y \hat{k}) \cdot (x \hat{k} - z \hat{i}) \)
\( = (z \hat{j} - y \hat{k}) \cdot (-z \hat{i} + 0 \hat{j} + x \hat{k}) \)
\( = (z)(0) + (-y)(x) + (0)(-z) \) (This step seems incorrect from source. Let's re-calculate: \( (z \hat{j} - y \hat{k}) \cdot (-z \hat{i} + x \hat{k}) \) )
The dot product is the sum of the products of corresponding components:
\( = (0)(-z) + (z)(0) + (-y)(x) \)
\( = 0 + 0 - yx \)
\( = -xy \)

So, \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = -xy \).
Add \( xy \) to this result:
\( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = -xy + xy = 0 \)
Thus, \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + x y = 0 \). (Proved)
In simple words: For the first part, we take the cross product of the vector \( \vec{r} \) with the \( \hat{i} \) unit vector. Then, we find the square of the length of this new vector, which should match the sum of the squares of the y and z components of \( \vec{r} \). For the second part, we calculate two cross products: \( \vec{r} \) with \( \hat{i} \), and \( \vec{r} \) with \( \hat{j} \). Then, we find the dot product of these two results. When we add \( xy \) to this dot product, the answer should be zero, showing the relationship between the components.

🎯 Exam Tip: Remember the cyclic properties of cross products for unit vectors: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \). Also, \( \hat{i} \times \hat{i} = \overrightarrow{0} \), and if the order is reversed, the sign changes (e.g., \( \hat{j} \times \hat{i} = -\hat{k} \)). The square of a vector's cross product is equal to the square of its magnitude.

Question 29. For any vector \( \vec{a} \), prove that \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \).
Answer:
Let \( \vec{a} \) be an arbitrary vector, which can be written in component form as:
\( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \)

The magnitude of \( \vec{a} \) is \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).
So, \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \).

First, calculate \( \vec{a} \times \hat{i} \):
\( \vec{a} \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} \)
\( = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) \)
Using \( \hat{i} \times \hat{i} = \overrightarrow{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), and \( \hat{k} \times \hat{i} = \hat{j} \):
\( = a_1(\overrightarrow{0}) + a_2(-\hat{k}) + a_3(\hat{j}) \)
\( = a_3 \hat{j} - a_2 \hat{k} \)
Now find its magnitude squared: \( |\vec{a} \times \hat{i}|^2 = |a_3 \hat{j} - a_2 \hat{k}|^2 = a_3^2 + (-a_2)^2 = a_2^2 + a_3^2 \) (Equation 1)

Next, calculate \( \vec{a} \times \hat{j} \):
\( \vec{a} \times \hat{j} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{j} \)
\( = a_1(\hat{i} \times \hat{j}) + a_2(\hat{j} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) \)
Using \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \overrightarrow{0} \), and \( \hat{k} \times \hat{j} = -\hat{i} \):
\( = a_1(\hat{k}) + a_2(\overrightarrow{0}) + a_3(-\hat{i}) \)
\( = -a_3 \hat{i} + a_1 \hat{k} \)
Now find its magnitude squared: \( |\vec{a} \times \hat{j}|^2 = |-a_3 \hat{i} + a_1 \hat{k}|^2 = (-a_3)^2 + a_1^2 = a_1^2 + a_3^2 \) (Equation 2)

Finally, calculate \( \vec{a} \times \hat{k} \):
\( \vec{a} \times \hat{k} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{k} \)
\( = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) + a_3(\hat{k} \times \hat{k}) \)
Using \( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{j} \times \hat{k} = \hat{i} \), and \( \hat{k} \times \hat{k} = \overrightarrow{0} \):
\( = a_1(-\hat{j}) + a_2(\hat{i}) + a_3(\overrightarrow{0}) \)
\( = a_2 \hat{i} - a_1 \hat{j} \)
Now find its magnitude squared: \( |\vec{a} \times \hat{k}|^2 = |a_2 \hat{i} - a_1 \hat{j}|^2 = a_2^2 + (-a_1)^2 = a_1^2 + a_2^2 \) (Equation 3)

Add the results from Equations 1, 2, and 3:
\( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \)
\( = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \)
\( = a_1^2 + a_1^2 + a_2^2 + a_2^2 + a_3^2 + a_3^2 \)
\( = 2a_1^2 + 2a_2^2 + 2a_3^2 \)
\( = 2(a_1^2 + a_2^2 + a_3^2) \)
Since \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \), we can substitute this in:
\( = 2|\vec{a}|^2 \)
Thus, \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \). (Proved)
In simple words: This problem asks us to prove a vector identity. We start by writing a general vector \( \vec{a} \) with its \( i, j, k \) components. Then, we calculate the cross product of \( \vec{a} \) with each of the standard unit vectors (\( \hat{i}, \hat{j}, \hat{k} \)) and find the square of the length (magnitude) of each result. When we add these three squared magnitudes together, we find that the sum is exactly twice the square of the length of the original vector \( \vec{a} \). This shows how the components relate to the vector's overall size in a specific way.

🎯 Exam Tip: This identity demonstrates a useful relationship between a vector and its cross products with the standard basis vectors. Knowing the cyclic properties of unit vector cross products (e.g., \( \hat{i} \times \hat{j} = \hat{k} \)) and that \( |\vec{v}|^2 = v_x^2 + v_y^2 + v_z^2 \) are essential for this proof.

Question 1. Find \( \vec{a} \times \vec{b} \) when
(i) \( \vec{a} = 3 \hat{i} – \hat{j} + \hat{k} \), \( \vec{b} = 2 \hat{i} + \hat{j} – \hat{k} \)
(ii) \( \vec{a} = 2 \hat{i} – 3 \hat{j} – \hat{k} \), \( \vec{b} = \hat{i} + 4 \hat{j} – 2 \hat{k} \)
(iii) Now, \( \vec{b} \times \vec{a} \)
(iv) \( \vec{a} = 2 \hat{i} – \hat{j} + \hat{k} \), \( \vec{b} = 3 \hat{i} + 4 \hat{j} – \hat{k} \), verify that \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \).
(v) Compute \( \vec{a} \times (\vec{b} \times \vec{c}) \), if \( \vec{a} = 7 \hat{i} – 2 \hat{j} + 3 \hat{k} \), \( \vec{b} = 2 \hat{i} + 8 \hat{k} \), \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \).

Answer:
(i) Given \( \vec{a} = 3 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 2 \hat{i} + \hat{j} – \hat{k} \)
We find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1)(-1) - (1)(1)) - \hat{j}((3)(-1) - (1)(2)) + \hat{k}((3)(1) - (-1)(2)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 - 1) - \hat{j}(-3 - 2) + \hat{k}(3 + 2) \)
\( \implies \vec{a} \times \vec{b} = 0 \hat{i} + 5 \hat{j} + 5 \hat{k} \)
So, \( \vec{a} \times \vec{b} = 5 \hat{j} + 5 \hat{k} \).
(ii) Given \( \vec{a} = 2 \hat{i} – 3 \hat{j} – \hat{k} \) and \( \vec{b} = \hat{i} + 4 \hat{j} – 2 \hat{k} \)
We find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-3)(-2) - (-1)(4)) - \hat{j}((2)(-2) - (-1)(1)) + \hat{k}((2)(4) - (-3)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(6 + 4) - \hat{j}(-4 + 1) + \hat{k}(8 + 3) \)
\( \implies \vec{a} \times \vec{b} = 10 \hat{i} + 3 \hat{j} + 11 \hat{k} \)
(iii) Now, to find \( \vec{b} \times \vec{a} \):
\( \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -2 \\ 2 & -3 & -1 \end{vmatrix} \)
\( \implies \vec{b} \times \vec{a} = \hat{i}((4)(-1) - (-2)(-3)) - \hat{j}((1)(-1) - (-2)(2)) + \hat{k}((1)(-3) - (4)(2)) \)
\( \implies \vec{b} \times \vec{a} = \hat{i}(-4 - 6) - \hat{j}(-1 + 4) + \hat{k}(-3 - 8) \)
\( \implies \vec{b} \times \vec{a} = -10 \hat{i} - 3 \hat{j} - 11 \hat{k} \)
Thus, from (ii), \( \vec{a} \times \vec{b} = 10 \hat{i} + 3 \hat{j} + 11 \hat{k} \) and \( \vec{b} \times \vec{a} = -10 \hat{i} - 3 \hat{j} - 11 \hat{k} \). This shows that \( \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) \), so \( \vec{a} \times \vec{b} \neq \vec{b} \times \vec{a} \). The cross product is anti-commutative.
(iv) Given \( \vec{a} = 2 \hat{i} – \hat{j} + \hat{k} \) and \( \vec{b} = 3 \hat{i} + 4 \hat{j} – \hat{k} \)
First, find \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1)(-1) - (1)(4)) - \hat{j}((2)(-1) - (1)(3)) + \hat{k}((2)(4) - (-1)(3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 - 4) - \hat{j}(-2 - 3) + \hat{k}(8 + 3) \)
\( \implies \vec{a} \times \vec{b} = -3 \hat{i} + 5 \hat{j} + 11 \hat{k} \)
Now, to verify if \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} \), their dot product should be zero:
\( (\vec{a} \times \vec{b}) \cdot \vec{a} = (-3 \hat{i} + 5 \hat{j} + 11 \hat{k}) \cdot (2 \hat{i} – \hat{j} + \hat{k}) \)
\( \implies (\vec{a} \times \vec{b}) \cdot \vec{a} = (-3)(2) + (5)(-1) + (11)(1) \)
\( \implies (\vec{a} \times \vec{b}) \cdot \vec{a} = -6 - 5 + 11 = 0 \)
Since the dot product is 0, \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} \).
Next, verify if \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{b} \):
\( (\vec{a} \times \vec{b}) \cdot \vec{b} = (-3 \hat{i} + 5 \hat{j} + 11 \hat{k}) \cdot (3 \hat{i} + 4 \hat{j} – \hat{k}) \)
\( \implies (\vec{a} \times \vec{b}) \cdot \vec{b} = (-3)(3) + (5)(4) + (11)(-1) \)
\( \implies (\vec{a} \times \vec{b}) \cdot \vec{b} = -9 + 20 - 11 = 0 \)
Since the dot product is 0, \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{b} \). This confirms that the cross product vector is always perpendicular to the plane containing the original two vectors.
(v) Given \( \vec{a} = 7 \hat{i} – 2 \hat{j} + 3 \hat{k} \); \( \vec{b} = 2 \hat{i} + 8 \hat{k} \) and \( \vec{c} = \hat{i} + \hat{j} + \hat{k} \)
First, calculate \( \vec{b} + \vec{c} \) (as the solution proceeds with addition, not cross product for \( \vec{b} \times \vec{c} \) inside the bracket):
\( \vec{b} + \vec{c} = (2 \hat{i} + 0 \hat{j} + 8 \hat{k}) + (\hat{i} + \hat{j} + \hat{k}) \)
\( \implies \vec{b} + \vec{c} = 3 \hat{i} + \hat{j} + 9 \hat{k} \)
Now, compute \( \vec{a} \times (\vec{b} + \vec{c}) \):
\( \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -2 & 3 \\ 3 & 1 & 9 \end{vmatrix} \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = \hat{i}((-2)(9) - (3)(1)) - \hat{j}((7)(9) - (3)(3)) + \hat{k}((7)(1) - (-2)(3)) \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = \hat{i}(-18 - 3) - \hat{j}(63 - 9) + \hat{k}(7 + 6) \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = -21 \hat{i} - 54 \hat{j} + 13 \hat{k} \)
In simple words: To find the cross product of two vectors, we arrange their components like a special 3x3 grid and then calculate the determinant. This gives us a new vector that is at right angles to both original vectors. If we need to find if a cross product is perpendicular to the original vectors, we check if their dot product is zero. For part (v), we first added the vectors inside the bracket before doing the final cross product.

🎯 Exam Tip: Remember that the vector cross product \( \vec{a} \times \vec{b} \) produces a vector perpendicular to both \( \vec{a} \) and \( \vec{b} \). Its direction follows the right-hand rule, and its magnitude is \( |\vec{a}||\vec{b}|\sin\theta \). Always ensure your determinant calculations are precise for each component.

Question 2. Find \( |\vec{a} \times \vec{b}| \) when
(i) \( \vec{a} = \hat{i} + 3 \hat{j} – 2 \hat{k} \), \( \vec{b} = -\hat{i} + 3 \hat{k} \)
(ii) \( \vec{a} = 2 \hat{i} + \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \)

Answer:
(i) Given \( \vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k} \) and \( \vec{b} = -\hat{i} + 0 \hat{j} + 3 \hat{k} \)
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((3)(3) - (-2)(0)) - \hat{j}((1)(3) - (-2)(-1)) + \hat{k}((1)(0) - (3)(-1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(9 - 0) - \hat{j}(3 - 2) + \hat{k}(0 + 3) \)
\( \implies \vec{a} \times \vec{b} = 9 \hat{i} - \hat{j} + 3 \hat{k} \)
Now, find the magnitude \( |\vec{a} \times \vec{b}| \):
\( |\vec{a} \times \vec{b}| = \sqrt{9^2 + (-1)^2 + 3^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{81 + 1 + 9} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{91} \)
(ii) Given \( \vec{a} = 2 \hat{i} + 0 \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \)
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((0)(1) - (1)(1)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(1) - (0)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(0 - 1) - \hat{j}(2 - 1) + \hat{k}(2 - 0) \)
\( \implies \vec{a} \times \vec{b} = -\hat{i} - \hat{j} + 2 \hat{k} \)
Now, find the magnitude \( |\vec{a} \times \vec{b}| \):
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-1)^2 + 2^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 4} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{6} \)
In simple words: To find the length (magnitude) of the cross product of two vectors, we first calculate the cross product itself. This gives us a new vector. Then, we find the length of this new vector using the formula \( \sqrt{x^2 + y^2 + z^2} \), where x, y, and z are its components.

🎯 Exam Tip: Always remember that \( |\vec{a} \times \vec{b}| \) is a scalar value (a number representing length), while \( \vec{a} \times \vec{b} \) is a vector. Pay close attention to signs when calculating the determinant for the cross product.

Question 3. If \( \vec{a} = 2 \hat{i} + 3 \hat{j} \), \( \vec{b} = -\hat{i} + 3 \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} + 2 \hat{j} + 5 \hat{k} \) be three vectors, find
(i) \( \vec{a} \times \vec{b} \)
(ii) \( \vec{b} \times \vec{c} \)
(iii) \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) \)

Answer:
Given \( \vec{a} = 2 \hat{i} + 3 \hat{j} + 0 \hat{k} \); \( \vec{b} = -\hat{i} + 3 \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} + 2 \hat{j} + 5 \hat{k} \)
(i) To find \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ -1 & 3 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((3)(1) - (0)(3)) - \hat{j}((2)(1) - (0)(-1)) + \hat{k}((2)(3) - (3)(-1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(3 - 0) - \hat{j}(2 - 0) + \hat{k}(6 + 3) \)
\( \implies \vec{a} \times \vec{b} = 3 \hat{i} - 2 \hat{j} + 9 \hat{k} \)
(ii) To find \( \vec{b} \times \vec{c} \):
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 1 \\ 1 & 2 & 5 \end{vmatrix} \)
\( \implies \vec{b} \times \vec{c} = \hat{i}((3)(5) - (1)(2)) - \hat{j}((-1)(5) - (1)(1)) + \hat{k}((-1)(2) - (3)(1)) \)
\( \implies \vec{b} \times \vec{c} = \hat{i}(15 - 2) - \hat{j}(-5 - 1) + \hat{k}(-2 - 3) \)
\( \implies \vec{b} \times \vec{c} = 13 \hat{i} + 6 \hat{j} - 5 \hat{k} \)
(iii) The question asks for \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{b}) \). However, the solution calculates \( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) \). We will follow the steps shown in the provided solution.
First, calculate \( \vec{a} - \vec{b} \):
\( \vec{a} - \vec{b} = (2 \hat{i} + 3 \hat{j} + 0 \hat{k}) - (-\hat{i} + 3 \hat{j} + \hat{k}) \)
\( \implies \vec{a} - \vec{b} = (2 - (-1)) \hat{i} + (3 - 3) \hat{j} + (0 - 1) \hat{k} \)
\( \implies \vec{a} - \vec{b} = 3 \hat{i} + 0 \hat{j} - \hat{k} \)
Next, calculate \( \vec{c} - \vec{b} \):
\( \vec{c} - \vec{b} = (\hat{i} + 2 \hat{j} + 5 \hat{k}) - (-\hat{i} + 3 \hat{j} + \hat{k}) \)
\( \implies \vec{c} - \vec{b} = (1 - (-1)) \hat{i} + (2 - 3) \hat{j} + (5 - 1) \hat{k} \)
\( \implies \vec{c} - \vec{b} = 2 \hat{i} - \hat{j} + 4 \hat{k} \)
Now, find the cross product \( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) \):
\( (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 2 & -1 & 4 \end{vmatrix} \)
\( \implies (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \hat{i}((0)(4) - (-1)(-1)) - \hat{j}((3)(4) - (-1)(2)) + \hat{k}((3)(-1) - (0)(2)) \)
\( \implies (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = \hat{i}(0 - 1) - \hat{j}(12 + 2) + \hat{k}(-3 - 0) \)
\( \implies (\vec{a} - \vec{b}) \times (\vec{c} - \vec{b}) = -\hat{i} - 14 \hat{j} - 3 \hat{k} \)
In simple words: This problem asks us to perform cross product calculations with given vectors. For each part, we set up a determinant using the components of the vectors. Calculating this determinant gives us the resulting vector. In part (iii), we needed to calculate new vectors by subtracting the given vectors before performing the final cross product.

🎯 Exam Tip: Be careful with the order of vectors in cross products, as \( \vec{u} \times \vec{v} \neq \vec{v} \times \vec{u} \). Also, pay attention to vector addition or subtraction before performing a cross product, ensuring you combine components correctly.

Question 4. \( |\vec{a}| = \sqrt{26} \), \( |\vec{b}| = 7 \) and \( |\vec{a} \times \vec{b}| = 35 \), find \( \vec{a} \cdot \vec{b} \).

Answer:
Given \( |\vec{a}| = \sqrt{26} \), \( |\vec{b}| = 7 \) and \( |\vec{a} \times \vec{b}| = 35 \)
We know the formula for the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \)
Substitute the given values:
\( 35 = (\sqrt{26})(7) \sin \theta \)
\( \implies \sin \theta = \frac{35}{7 \sqrt{26}} \)
\( \implies \sin \theta = \frac{5}{\sqrt{26}} \)
Now, we need to find \( \cos \theta \) to calculate the dot product. We use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\( \cos \theta = \sqrt{1 - \sin^2 \theta} \)
\( \implies \cos \theta = \sqrt{1 - \left(\frac{5}{\sqrt{26}}\right)^2} \)
\( \implies \cos \theta = \sqrt{1 - \frac{25}{26}} \)
\( \implies \cos \theta = \sqrt{\frac{26 - 25}{26}} \)
\( \implies \cos \theta = \sqrt{\frac{1}{26}} \)
\( \implies \cos \theta = \frac{1}{\sqrt{26}} \)
Finally, we can find the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)
\( \implies \vec{a} \cdot \vec{b} = (\sqrt{26})(7)\left(\frac{1}{\sqrt{26}}\right) \)
\( \implies \vec{a} \cdot \vec{b} = 7 \)
In simple words: We are given the lengths of two vectors and the length of their cross product. We use the cross product formula to first find the sine of the angle between them. Then, we use a basic trigonometric identity to find the cosine of that angle. Finally, we use the dot product formula, along with the lengths and the cosine, to calculate the dot product. This problem combines different vector operations.

🎯 Exam Tip: When solving problems involving both dot and cross products, remember their definitions in terms of magnitude and angle: \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \) and \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \). You can use the identity \( \sin^2\theta + \cos^2\theta = 1 \) to switch between sine and cosine if one is known.

Question 5. If \( \vec{a} = 3 \hat{i} – \hat{j} – 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k} \) find \( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) \).

Answer:
Given \( \vec{a} = 3 \hat{i} – \hat{j} – 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k} \)
First, calculate the vector \( \vec{a} + 2 \vec{b} \):
\( \vec{a} + 2 \vec{b} = (3 \hat{i} – \hat{j} – 2 \hat{k}) + 2(2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( \implies \vec{a} + 2 \vec{b} = (3 \hat{i} – \hat{j} – 2 \hat{k}) + (4 \hat{i} + 6 \hat{j} + 2 \hat{k}) \)
\( \implies \vec{a} + 2 \vec{b} = (3+4) \hat{i} + (-1+6) \hat{j} + (-2+2) \hat{k} \)
\( \implies \vec{a} + 2 \vec{b} = 7 \hat{i} + 5 \hat{j} + 0 \hat{k} \)
Next, calculate the vector \( 2 \vec{a} – \vec{b} \):
\( 2 \vec{a} – \vec{b} = 2(3 \hat{i} – \hat{j} – 2 \hat{k}) – (2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( \implies 2 \vec{a} – \vec{b} = (6 \hat{i} – 2 \hat{j} – 4 \hat{k}) – (2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( \implies 2 \vec{a} – \vec{b} = (6-2) \hat{i} + (-2-3) \hat{j} + (-4-1) \hat{k} \)
\( \implies 2 \vec{a} – \vec{b} = 4 \hat{i} – 5 \hat{j} – 5 \hat{k} \)
Now, find the cross product of these two new vectors \( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) \):
\( (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{vmatrix} \)
\( \implies (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) = \hat{i}((5)(-5) - (0)(-5)) - \hat{j}((7)(-5) - (0)(4)) + \hat{k}((7)(-5) - (5)(4)) \)
\( \implies (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) = \hat{i}(-25 - 0) - \hat{j}(-35 - 0) + \hat{k}(-35 - 20) \)
\( \implies (\vec{a} + 2 \vec{b}) \times (2 \vec{a} – \vec{b}) = -25 \hat{i} + 35 \hat{j} – 55 \hat{k} \)
In simple words: This problem involves a few steps. First, we calculate two new vectors by scaling and adding/subtracting the original vectors \( \vec{a} \) and \( \vec{b} \). Once we have these two new vectors, we perform a cross product on them, using the determinant method. The final result is a single vector.

🎯 Exam Tip: Break down complex vector operations into smaller, manageable steps. First, calculate any vector sums or differences, then proceed with the cross product. Double-check your arithmetic, especially with negative signs, as a small error can propagate through the entire calculation.

Question 6. Find a unit vector perpendicular to the plane of the vectors.
(i) \( -3 \hat{i} + 4 \hat{k} \) and \( 4 \hat{i} + 3 \hat{j} \)
(ii) \( \vec{a} = 2 \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \)

Answer:
(i) Let the given vectors be \( \vec{A} = -3 \hat{i} + 0 \hat{j} + 4 \hat{k} \) and \( \vec{B} = 4 \hat{i} + 3 \hat{j} + 0 \hat{k} \)
A vector perpendicular to the plane of \( \vec{A} \) and \( \vec{B} \) is given by their cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 4 \\ 4 & 3 & 0 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((0)(0) - (4)(3)) - \hat{j}((-3)(0) - (4)(4)) + \hat{k}((-3)(3) - (0)(4)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(0 - 12) - \hat{j}(0 - 16) + \hat{k}(-9 - 0) \)
\( \implies \vec{A} \times \vec{B} = -12 \hat{i} + 16 \hat{j} - 9 \hat{k} \)
Now, find the magnitude of \( \vec{A} \times \vec{B} \):
\( |\vec{A} \times \vec{B}| = \sqrt{(-12)^2 + (16)^2 + (-9)^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{144 + 256 + 81} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{481} \)
The required unit vector is \( \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \):
Unit vector \( = \frac{-12 \hat{i} + 16 \hat{j} - 9 \hat{k}}{\sqrt{481}} \)
\( \implies \) Unit vector \( = \frac{-12}{\sqrt{481}} \hat{i} + \frac{16}{\sqrt{481}} \hat{j} - \frac{9}{\sqrt{481}} \hat{k} \). This vector represents the direction perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \).
(ii) Given \( \vec{a} = 2 \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \)
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((1)(1) - (1)(2)) - \hat{j}((2)(1) - (1)(1)) + \hat{k}((2)(2) - (1)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 - 2) - \hat{j}(2 - 1) + \hat{k}(4 - 1) \)
\( \implies \vec{a} \times \vec{b} = -\hat{i} - \hat{j} + 3 \hat{k} \)
Now, find the magnitude of \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-1)^2 + 3^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 9} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{11} \)
The required unit vector is \( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \):
Unit vector \( = \frac{-\hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{11}} \)
\( \implies \) Unit vector \( = \frac{-1}{\sqrt{11}} (\hat{i} + \hat{j} - 3 \hat{k}) \)
In simple words: To find a unit vector that is perpendicular to the plane containing two given vectors, we first calculate the cross product of these two vectors. This cross product gives us a vector that is already perpendicular to the plane. Then, to make it a "unit" vector (meaning its length is 1), we divide this cross product vector by its own magnitude. This process normalizes the vector.

🎯 Exam Tip: A unit vector perpendicular to a plane defined by vectors \( \vec{A} \) and \( \vec{B} \) can be \( \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \) or \( -\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \). Both directions are valid, so the solution might involve a \( \pm \) sign. Also, simplify the magnitude (e.g., \( \sqrt{544} = 4\sqrt{34} \)) whenever possible.

Question 7.
(i) Find the vector of magnitude 9, which is perpendicular to both the vectors \( 4 \hat{i} - \hat{j} + 3 \hat{k} \) and \( 2 \hat{i} + \hat{j} – 2 \hat{k} \)
(ii) Find a vector whose length is 3 and which is perpendicular to both the vectors \( \vec{a} = 3 \hat{i} + \hat{j} + 3 \hat{k} \) and \( -2 \hat{i} + \hat{j} – 2 \hat{k} \)

Answer:
(i) Let the given vectors be \( \vec{A} = 4 \hat{i} - \hat{j} + 3 \hat{k} \) and \( \vec{B} = -2 \hat{i} + \hat{j} – 2 \hat{k} \).
First, find the cross product \( \vec{A} \times \vec{B} \), which gives a vector perpendicular to both:
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((-1)(-2) - (3)(1)) - \hat{j}((4)(-2) - (3)(-2)) + \hat{k}((4)(1) - (-1)(-2)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(2 - 3) - \hat{j}(-8 + 6) + \hat{k}(4 - 2) \)
\( \implies \vec{A} \times \vec{B} = -\hat{i} + 2 \hat{j} + 2 \hat{k} \)
Next, find the magnitude of this cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{(-1)^2 + 2^2 + 2^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{1 + 4 + 4} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{9} = 3 \)
The unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is \( \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \):
Unit vector \( = \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3} \)
Since we need a vector of magnitude 9, we multiply this unit vector by 9:
Required vector \( = 9 \left( \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3} \right) \)
\( \implies \) Required vector \( = 3(-\hat{i} + 2 \hat{j} + 2 \hat{k}) \)
\( \implies \) Required vector \( = -3 \hat{i} + 6 \hat{j} + 6 \hat{k} \)
(ii) Let the given vectors be \( \vec{a} = 3 \hat{i} + \hat{j} – 4 \hat{k} \) and \( \vec{b} = 6 \hat{i} + 5 \hat{j} – 2 \hat{k} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((1)(-2) - (-4)(5)) - \hat{j}((3)(-2) - (-4)(6)) + \hat{k}((3)(5) - (1)(6)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-2 + 20) - \hat{j}(-6 + 24) + \hat{k}(15 - 6) \)
\( \implies \vec{a} \times \vec{b} = 18 \hat{i} - 18 \hat{j} + 9 \hat{k} \)
Next, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{18^2 + (-18)^2 + 9^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{324 + 324 + 81} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{729} = 27 \)
The required vector has magnitude 3 and is perpendicular to both \( \vec{a} \) and \( \vec{b} \). So, we take the unit vector in the direction of \( \vec{a} \times \vec{b} \) and multiply it by 3:
Required vector \( = 3 \left( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \right) \)
\( \implies \) Required vector \( = 3 \left( \frac{18 \hat{i} - 18 \hat{j} + 9 \hat{k}}{27} \right) \)
\( \implies \) Required vector \( = \frac{1}{9}(18 \hat{i} - 18 \hat{j} + 9 \hat{k}) \)
\( \implies \) Required vector \( = 2 \hat{i} - 2 \hat{j} + \hat{k} \)
In simple words: To find a vector with a specific length that is perpendicular to two other vectors, we first calculate the cross product of those two vectors. This gives us the correct direction. Then, we find the length of this cross product vector. Finally, we make it a unit vector (length 1) by dividing it by its length, and then multiply this unit vector by the desired magnitude (e.g., 9 or 3) to get the final answer.

🎯 Exam Tip: A vector \( \vec{V} \) of magnitude \( k \) perpendicular to two vectors \( \vec{A} \) and \( \vec{B} \) is given by \( \pm k \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} \). Always calculate the cross product and its magnitude carefully before scaling.

Question 8.
(i) If the position vectors of the three points A, B, C are respectively \( \hat{i} + \hat{j} + \hat{k} \), \( \hat{i} + 3 \hat{j} – 4 \hat{k} \) and \( 7 \hat{i} + 4 \hat{j} + 9 \hat{k} \), find the unit vector perpendicular to the plane of the triangle ABC.
(ii) Find the unit vectors perpendicular to the plane A B C, when the position vectors of A, B and C are \( 2 \hat{i} – \hat{j} + \hat{k} \), \( \hat{i} + \hat{j} + 2 \hat{k} \) and \( \hat{i} + 3 \hat{k} \) respectively.
(iii) Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).

Answer:
(i) Given position vectors of A, B, C:
\( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
\( \vec{B} = 2 \hat{i} + 3 \hat{j} – 4 \hat{k} \)
\( \vec{C} = 7 \hat{i} + 4 \hat{j} + 9 \hat{k} \)
To find a vector perpendicular to the plane of \( \triangle ABC \), we can find the cross product of two side vectors, such as \( \overrightarrow{\mathrm{BA}} \) and \( \overrightarrow{\mathrm{BC}} \).
First, calculate \( \overrightarrow{\mathrm{BA}} \):
\( \overrightarrow{\mathrm{BA}} = \vec{A} - \vec{B} = (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{BA}} = (1 - 2) \hat{i} + (1 - 3) \hat{j} + (1 - (-4)) \hat{k} \)
\( \implies \overrightarrow{\mathrm{BA}} = -\hat{i} - 2 \hat{j} + 5 \hat{k} \)
Next, calculate \( \overrightarrow{\mathrm{BC}} \):
\( \overrightarrow{\mathrm{BC}} = \vec{C} - \vec{B} = (7 \hat{i} + 4 \hat{j} + 9 \hat{k}) - (2 \hat{i} + 3 \hat{j} – 4 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{BC}} = (7 - 2) \hat{i} + (4 - 3) \hat{j} + (9 - (-4)) \hat{k} \)
\( \implies \overrightarrow{\mathrm{BC}} = 5 \hat{i} + \hat{j} + 13 \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} \):
\( \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & 5 \\ 5 & 1 & 13 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \hat{i}((-2)(13) - (5)(1)) - \hat{j}((-1)(13) - (5)(5)) + \hat{k}((-1)(1) - (-2)(5)) \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \hat{i}(-26 - 5) - \hat{j}(-13 - 25) + \hat{k}(-1 + 10) \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = -31 \hat{i} + 38 \hat{j} + 9 \hat{k} \)
This vector is perpendicular to the plane of \( \triangle ABC \). To find the unit vector, divide it by its magnitude:
\( |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{(-31)^2 + (38)^2 + 9^2} \)
\( \implies |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{961 + 1444 + 81} \)
\( \implies |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{2486} \)
Required unit vector \( = \frac{\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|} \)
\( \implies \) Unit vector \( = \frac{-31 \hat{i} + 38 \hat{j} + 9 \hat{k}}{\sqrt{2486}} \)
(ii) Given position vectors of A, B, C:
\( \vec{A} = 2 \hat{i} – \hat{j} + \hat{k} \)
\( \vec{B} = \hat{i} + \hat{j} + 2 \hat{k} \)
\( \vec{C} = \hat{i} + 0 \hat{j} + 3 \hat{k} \)
First, calculate \( \overrightarrow{\mathrm{BA}} \):
\( \overrightarrow{\mathrm{BA}} = \vec{A} - \vec{B} = (2 \hat{i} – \hat{j} + \hat{k}) - (\hat{i} + \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{BA}} = (2 - 1) \hat{i} + (-1 - 1) \hat{j} + (1 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{BA}} = \hat{i} - 2 \hat{j} - \hat{k} \)
Next, calculate \( \overrightarrow{\mathrm{BC}} \):
\( \overrightarrow{\mathrm{BC}} = \vec{C} - \vec{B} = (\hat{i} + 0 \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{BC}} = (1 - 1) \hat{i} + (0 - 1) \hat{j} + (3 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{BC}} = 0 \hat{i} - \hat{j} + \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} \):
\( \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 0 & -1 & 1 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \hat{i}((-2)(1) - (-1)(-1)) - \hat{j}((1)(1) - (-1)(0)) + \hat{k}((1)(-1) - (-2)(0)) \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = \hat{i}(-2 - 1) - \hat{j}(1 - 0) + \hat{k}(-1 - 0) \)
\( \implies \overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}} = -3 \hat{i} - \hat{j} - \hat{k} \)
This vector is perpendicular to the plane of \( \triangle ABC \). To find the unit vector, divide it by its magnitude:
\( |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{(-3)^2 + (-1)^2 + (-1)^2} \)
\( \implies |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{9 + 1 + 1} \)
\( \implies |\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}| = \sqrt{11} \)
Required unit vector \( = \frac{\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|} \)
\( \implies \) Unit vector \( = \frac{-3 \hat{i} - \hat{j} - \hat{k}}{\sqrt{11}} \) or \( = \frac{1}{\sqrt{11}}(-3 \hat{i} - \hat{j} - \hat{k}) \)
(iii) Given points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Let their position vectors be \( \vec{p} = \hat{i} – \hat{j} + 2 \hat{k} \), \( \vec{q} = 2 \hat{i} + 0 \hat{j} – \hat{k} \), and \( \vec{r} = 0 \hat{i} + 2 \hat{j} + \hat{k} \).
To find a vector perpendicular to the plane defined by P, Q, R, we can use the cross product of two vectors formed by these points, e.g., \( \overrightarrow{\mathrm{QP}} \) and \( \overrightarrow{\mathrm{QR}} \).
First, calculate \( \overrightarrow{\mathrm{QR}} \):
\( \overrightarrow{\mathrm{QR}} = \vec{r} - \vec{q} = (0 \hat{i} + 2 \hat{j} + \hat{k}) - (2 \hat{i} + 0 \hat{j} – \hat{k}) \)
\( \implies \overrightarrow{\mathrm{QR}} = (0 - 2) \hat{i} + (2 - 0) \hat{j} + (1 - (-1)) \hat{k} \)
\( \implies \overrightarrow{\mathrm{QR}} = -2 \hat{i} + 2 \hat{j} + 2 \hat{k} \)
Next, calculate \( \overrightarrow{\mathrm{QP}} \):
\( \overrightarrow{\mathrm{QP}} = \vec{p} - \vec{q} = (\hat{i} – \hat{j} + 2 \hat{k}) - (2 \hat{i} + 0 \hat{j} – \hat{k}) \)
\( \implies \overrightarrow{\mathrm{QP}} = (1 - 2) \hat{i} + (-1 - 0) \hat{j} + (2 - (-1)) \hat{k} \)
\( \implies \overrightarrow{\mathrm{QP}} = -\hat{i} - \hat{j} + 3 \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}} \):
\( \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 2 \\ -1 & -1 & 3 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}} = \hat{i}((2)(3) - (2)(-1)) - \hat{j}((-2)(3) - (2)(-1)) + \hat{k}((-2)(-1) - (2)(-1)) \)
\( \implies \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}} = \hat{i}(6 + 2) - \hat{j}(-6 + 2) + \hat{k}(2 + 2) \)
\( \implies \overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}} = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} \)
This vector is perpendicular to the plane of PQR. To find the unit vector, divide it by its magnitude:
\( |\overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}| = \sqrt{8^2 + 4^2 + 4^2} \)
\( \implies |\overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}| = \sqrt{64 + 16 + 16} \)
\( \implies |\overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}| = \sqrt{96} = \sqrt{16 \times 6} = 4 \sqrt{6} \)
Required unit vector \( = \frac{\overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}}{|\overrightarrow{\mathrm{QR}} \times \overrightarrow{\mathrm{QP}}|} \)
\( \implies \) Unit vector \( = \frac{8 \hat{i} + 4 \hat{j} + 4 \hat{k}}{4 \sqrt{6}} \)
\( \implies \) Unit vector \( = \frac{4(2 \hat{i} + \hat{j} + \hat{k})}{4 \sqrt{6}} \)
\( \implies \) Unit vector \( = \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \)
In simple words: To find a unit vector perpendicular to a plane defined by three points, we first create two vectors using these points (like two sides of the triangle). Then we find the cross product of these two vectors. This new vector is perpendicular to the plane. Finally, we divide this vector by its own length (magnitude) to get a unit vector, which means its length is exactly one.

🎯 Exam Tip: When finding a vector perpendicular to a plane containing three points, ensure you form two vectors originating from a common point (e.g., \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)). The cross product of these two vectors will be normal to the plane. Remember to simplify radicals in the final answer.

Question 9. Find the sine of the angle between the vectors.
(i) \( \vec{a} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \), \( \vec{b} = \hat{i} – \hat{j} + \hat{k} \)
(ii) \( \hat{i} + \hat{j} \) and \( \hat{j} + \hat{k} \)

Answer:
(i) Given \( \vec{a} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} + \hat{k} \)
We use the formula \( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ 1 & -1 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-4)(1) - (5)(-1)) - \hat{j}((3)(1) - (5)(1)) + \hat{k}((3)(-1) - (-4)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-4 + 5) - \hat{j}(3 - 5) + \hat{k}(-3 + 4) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \)
Next, find the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + 2^2 + 1^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6} \)
Now, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5 \sqrt{2} \)
\( |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \)
Finally, calculate \( \sin \theta \):
\( \sin \theta = \frac{\sqrt{6}}{(5 \sqrt{2})(\sqrt{3})} \)
\( \implies \sin \theta = \frac{\sqrt{6}}{5 \sqrt{6}} \)
\( \implies \sin \theta = \frac{1}{5} \)
(ii) Let \( \vec{a} = \hat{i} + \hat{j} + 0 \hat{k} \) and \( \vec{b} = 0 \hat{i} + \hat{j} + \hat{k} \)
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((1)(1) - (0)(1)) - \hat{j}((1)(1) - (0)(0)) + \hat{k}((1)(1) - (1)(0)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} - \hat{j} + \hat{k} \)
Next, find the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{1 + 1 + 1} = \sqrt{3} \)
Now, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \)
\( |\vec{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \)
Finally, calculate \( \sin \theta \):
\( \sin \theta = \frac{\sqrt{3}}{(\sqrt{2})(\sqrt{2})} \)
\( \implies \sin \theta = \frac{\sqrt{3}}{2} \)
In simple words: To find the sine of the angle between two vectors, we use a special formula: divide the length of their cross product by the product of their individual lengths. So, first we calculate the cross product, then find its length. After that, we find the lengths of the original two vectors. Finally, we put these values into the formula to get the sine of the angle.

🎯 Exam Tip: Be cautious when simplifying square roots; ensure you correctly factor out perfect squares. Always remember the formula for sine of the angle between two vectors, \( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} \), and correctly apply it.

Question 10. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \), if \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \).

Answer:
Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
We know the formulas for the magnitude of the cross product and the dot product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \) (Equation 1)
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) (Equation 2)
The problem states that \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \).
So, we can set Equation 1 equal to Equation 2:
\( |\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| \cos \theta \)
Since \( |\vec{a}| \neq 0 \) and \( |\vec{b}| \neq 0 \) (otherwise the cross product and dot product would both be zero, and the angle undefined), we can divide both sides by \( |\vec{a}| |\vec{b}| \):
\( \sin \theta = \cos \theta \)
If \( \cos \theta \neq 0 \), we can divide both sides by \( \cos \theta \):
\( \frac{\sin \theta}{\cos \theta} = 1 \)
\( \implies \tan \theta = 1 \)
The angle \( \theta \) for which \( \tan \theta = 1 \) is \( \frac{\pi}{4} \) (or 45 degrees) in the range \( [0, \pi] \).
In simple words: We are given that the length of the cross product is equal to the dot product. We write down the formulas for both using the angle between the vectors. By setting these formulas equal, we find that the sine of the angle must be equal to its cosine. This means the tangent of the angle is 1, which tells us the angle is 45 degrees or \( \frac{\pi}{4} \) radians.

🎯 Exam Tip: This is a classic problem. Remember that if \( |\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b} \), it directly implies \( \tan \theta = 1 \). Conversely, if \( \tan \theta = 1 \), then \( \sin \theta = \cos \theta \) (for positive values of sine/cosine), leading to the equality.

Question 11. Given \( |\vec{a}| = 2 \); \( |\vec{b}| = 7 \) and \( \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \), find the angle between \( \vec{a} \) and \( \vec{b} \).

Answer:
Given \( |\vec{a}| = 2 \), \( |\vec{b}| = 7 \).
The cross product vector is \( \vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \).
First, find the magnitude of the cross product \( |\vec{a} \times \vec{b}| \):
\( |\vec{a} \times \vec{b}| = \sqrt{3^2 + 2^2 + 6^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{9 + 4 + 36} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{49} = 7 \)
Let \( \theta \) be the required angle between \( \vec{a} \) and \( \vec{b} \).
We use the formula for the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \)
Substitute the known values:
\( 7 = (2)(7) \sin \theta \)
\( \implies 7 = 14 \sin \theta \)
\( \implies \sin \theta = \frac{7}{14} \)
\( \implies \sin \theta = \frac{1}{2} \)
The angle \( \theta \) for which \( \sin \theta = \frac{1}{2} \) is \( \frac{\pi}{6} \) (or 30 degrees) in the range \( [0, \pi] \).
In simple words: We know the lengths of two vectors and the actual cross product vector. We first find the length of this cross product vector. Then, we use the formula that connects the length of the cross product to the lengths of the individual vectors and the sine of the angle between them. By putting in all the known values, we can solve for \( \sin \theta \) and then find the angle \( \theta \).

🎯 Exam Tip: Remember to calculate the magnitude of the cross product vector first if it's given directly. This problem is a direct application of the cross product magnitude formula, so ensure you recall it accurately and perform the calculations carefully.

Question 12. If \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), then prove that \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \).

Answer:
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \).
We know the formulas for the dot product and the magnitude of the cross product:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) (Equation 1)
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \) (Equation 2)
To prove \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \), we can divide Equation 2 by Equation 1:
\( \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} = \frac{|\vec{a}| |\vec{b}| \sin \theta}{|\vec{a}| |\vec{b}| \cos \theta} \)
Assuming \( |\vec{a}| \neq 0 \), \( |\vec{b}| \neq 0 \), and \( \cos \theta \neq 0 \), we can cancel out \( |\vec{a}| \) and \( |\vec{b}| \):
\( \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} = \frac{\sin \theta}{\cos \theta} \)
We know that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \).
Therefore,
\( \implies \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \)
This proves the identity.
In simple words: To show this relationship, we start with the basic definitions of the dot product and the cross product's length, both of which involve the angle between the vectors. If we divide the formula for the cross product's length by the formula for the dot product, the lengths of the individual vectors cancel out. What's left is \( \frac{\sin \theta}{\cos \theta} \), which is exactly equal to \( \tan \theta \).

🎯 Exam Tip: This proof relies on the fundamental definitions of the dot product and the cross product. Make sure you state these definitions clearly and then perform the division to derive the tangent relationship. Remember that the derivation holds when \( \vec{a} \), \( \vec{b} \) are non-zero vectors and \( \cos\theta \neq 0 \).

Question 13. If \( \vec{a} = 4 \hat{i} – \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} – \hat{j} + \hat{k} \), verify that, \( \vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \).

Answer:
Given \( \vec{a} = 4 \hat{i} – \hat{j} + \hat{k} \); \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} – \hat{j} + \hat{k} \)
**Left Hand Side (L.H.S.): \( \vec{a} \times (\vec{b} + \vec{c}) \)**
First, calculate \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} – \hat{j} + \hat{k}) \)
\( \implies \vec{b} + \vec{c} = (1+1)\hat{i} + (1-1)\hat{j} + (1+1)\hat{k} \)
\( \implies \vec{b} + \vec{c} = 2 \hat{i} + 0 \hat{j} + 2 \hat{k} \)
Now, calculate \( \vec{a} \times (\vec{b} + \vec{c}) \):
\( \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 2 & 0 & 2 \end{vmatrix} \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = \hat{i}((-1)(2) - (1)(0)) - \hat{j}((4)(2) - (1)(2)) + \hat{k}((4)(0) - (-1)(2)) \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = \hat{i}(-2 - 0) - \hat{j}(8 - 2) + \hat{k}(0 + 2) \)
\( \implies \vec{a} \times (\vec{b} + \vec{c}) = -2 \hat{i} - 6 \hat{j} + 2 \hat{k} \) (Equation 1)

**Right Hand Side (R.H.S.): \( (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \)**
First, calculate \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1)(1) - (1)(1)) - \hat{j}((4)(1) - (1)(1)) + \hat{k}((4)(1) - (-1)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-1 - 1) - \hat{j}(4 - 1) + \hat{k}(4 + 1) \)
\( \implies \vec{a} \times \vec{b} = -2 \hat{i} - 3 \hat{j} + 5 \hat{k} \)
Next, calculate \( \vec{a} \times \vec{c} \):
\( \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{c} = \hat{i}((-1)(1) - (1)(-1)) - \hat{j}((4)(1) - (1)(1)) + \hat{k}((4)(-1) - (-1)(1)) \)
\( \implies \vec{a} \times \vec{c} = \hat{i}(-1 + 1) - \hat{j}(4 - 1) + \hat{k}(-4 + 1) \)
\( \implies \vec{a} \times \vec{c} = 0 \hat{i} - 3 \hat{j} - 3 \hat{k} \)
Now, sum the two cross products \( (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \):
\( (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = (-2 \hat{i} - 3 \hat{j} + 5 \hat{k}) + (0 \hat{i} - 3 \hat{j} - 3 \hat{k}) \)
\( \implies (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = (-2+0) \hat{i} + (-3-3) \hat{j} + (5-3) \hat{k} \)
\( \implies (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = -2 \hat{i} - 6 \hat{j} + 2 \hat{k} \) (Equation 2)
From Equation 1 and Equation 2, we see that L.H.S. = R.H.S.
Therefore, \( \vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) \) is verified. This property is known as the distributive law of vector cross product over vector addition.
In simple words: We need to check if a rule about cross products works. This rule says that if you cross a vector \( \vec{a} \) with the sum of two other vectors (\( \vec{b} \) and \( \vec{c} \)), it's the same as crossing \( \vec{a} \) with \( \vec{b} \) first, then crossing \( \vec{a} \) with \( \vec{c} \) first, and then adding those two results. We calculated both sides of the equation separately using the given vectors and found that they give the exact same answer, proving the rule.

🎯 Exam Tip: This problem verifies the distributive property of the vector cross product. It is crucial to calculate each part (sum, individual cross products, and final sum/cross product) accurately. Clearly label your Left Hand Side (LHS) and Right Hand Side (RHS) calculations for better presentation and to avoid confusion.

Question 14. Find the area of the parallelogram whose adjacent sides are
(i) \( \hat{i} - 3 \hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} + \hat{k} \)
(ii) \( \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( 3 \hat{i} – 2 \hat{j} + \hat{k} \)
(iii) \( 2 \hat{i} + \hat{j} + 3 \hat{k} \) and \( \hat{i} – \hat{j} \)
(iv) \( 2 \hat{i} \) and \( 3 \hat{j} \).

Answer:
The area of a parallelogram with adjacent sides \( \vec{A} \) and \( \vec{B} \) is given by \( |\vec{A} \times \vec{B}| \).
(i) Let \( \vec{A} = \hat{i} - 3 \hat{j} + \hat{k} \) and \( \vec{B} = \hat{i} + \hat{j} + \hat{k} \)
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((-3)(1) - (1)(1)) - \hat{j}((1)(1) - (1)(1)) + \hat{k}((1)(1) - (-3)(1)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(-3 - 1) - \hat{j}(1 - 1) + \hat{k}(1 + 3) \)
\( \implies \vec{A} \times \vec{B} = -4 \hat{i} + 0 \hat{j} + 4 \hat{k} \)
Now, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{(-4)^2 + 0^2 + 4^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{16 + 0 + 16} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{32} = 4 \sqrt{2} \)
Thus, the required area of the parallelogram is \( 4 \sqrt{2} \) sq. units.
(ii) Let \( \vec{A} = \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( \vec{B} = 3 \hat{i} – 2 \hat{j} + \hat{k} \)
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((2)(1) - (3)(-2)) - \hat{j}((1)(1) - (3)(3)) + \hat{k}((1)(-2) - (2)(3)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(2 + 6) - \hat{j}(1 - 9) + \hat{k}(-2 - 6) \)
\( \implies \vec{A} \times \vec{B} = 8 \hat{i} + 8 \hat{j} - 8 \hat{k} \)
Now, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{8^2 + 8^2 + (-8)^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{64 + 64 + 64} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{192} = 8 \sqrt{3} \)
Thus, the required area of the parallelogram is \( 8 \sqrt{3} \) sq. units.
(iii) Let \( \vec{A} = 2 \hat{i} + \hat{j} + 3 \hat{k} \) and \( \vec{B} = \hat{i} – \hat{j} + 0 \hat{k} \)
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((1)(0) - (3)(-1)) - \hat{j}((2)(0) - (3)(1)) + \hat{k}((2)(-1) - (1)(1)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(0 + 3) - \hat{j}(0 - 3) + \hat{k}(-2 - 1) \)
\( \implies \vec{A} \times \vec{B} = 3 \hat{i} + 3 \hat{j} - 3 \hat{k} \)
Now, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{3^2 + 3^2 + (-3)^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{9 + 9 + 9} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{27} = 3 \sqrt{3} \)
Thus, the required area of the parallelogram is \( 3 \sqrt{3} \) sq. units.
(iv) Let \( \vec{A} = 2 \hat{i} + 0 \hat{j} + 0 \hat{k} \) and \( \vec{B} = 0 \hat{i} + 3 \hat{j} + 0 \hat{k} \)
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((0)(0) - (0)(3)) - \hat{j}((2)(0) - (0)(0)) + \hat{k}((2)(3) - (0)(0)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(0) - \hat{j}(0) + \hat{k}(6) \)
\( \implies \vec{A} \times \vec{B} = 6 \hat{k} \)
Now, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = |6 \hat{k}| = \sqrt{0^2 + 0^2 + 6^2} = \sqrt{36} = 6 \)
Thus, the required area of the parallelogram is \( 6 \) sq. units.
In simple words: To find the area of a parallelogram when you know its two adjacent (next-to-each-other) sides as vectors, you simply calculate the cross product of these two vectors. Then, you find the length (magnitude) of the resulting cross product vector. This length is the area of the parallelogram. Remember that the area is always a positive number.

🎯 Exam Tip: The area of a parallelogram formed by adjacent vectors \( \vec{A} \) and \( \vec{B} \) is \( |\vec{A} \times \vec{B}| \). Always set up the cross product determinant carefully and compute its magnitude. Remember that area is a scalar, so it should be a positive value.

Question 15. Find the area of a parallelogram whose diagonals are \( 3 \hat{i} + 4 \hat{j} \) and \( \hat{i} + \hat{j} + \hat{k} \).

Answer:
Let the diagonals of the parallelogram be \( \vec{d_1} \) and \( \vec{d_2} \).
Given \( \vec{d_1} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k} \) and \( \vec{d_2} = \hat{i} + \hat{j} + \hat{k} \).
The area of a parallelogram with diagonals \( \vec{d_1} \) and \( \vec{d_2} \) is given by the formula:
Area \( = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \)
First, find the cross product \( \vec{d_1} \times \vec{d_2} \):
\( \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{vmatrix} \)
\( \implies \vec{d_1} \times \vec{d_2} = \hat{i}((4)(1) - (0)(1)) - \hat{j}((3)(1) - (0)(1)) + \hat{k}((3)(1) - (4)(1)) \)
\( \implies \vec{d_1} \times \vec{d_2} = \hat{i}(4 - 0) - \hat{j}(3 - 0) + \hat{k}(3 - 4) \)
\( \implies \vec{d_1} \times \vec{d_2} = 4 \hat{i} - 3 \hat{j} - \hat{k} \)
Next, find the magnitude of this cross product:
\( |\vec{d_1} \times \vec{d_2}| = \sqrt{4^2 + (-3)^2 + (-1)^2} \)
\( \implies |\vec{d_1} \times \vec{d_2}| = \sqrt{16 + 9 + 1} \)
\( \implies |\vec{d_1} \times \vec{d_2}| = \sqrt{26} \)
Finally, calculate the area of the parallelogram:
Area \( = \frac{1}{2} \sqrt{26} \) sq. units.
In simple words: When you know the two diagonal vectors of a parallelogram, its area is found by first calculating the cross product of these two diagonals. Then, you find the length of that resulting cross product vector. Finally, you divide this length by two. This gives you the area of the parallelogram.

🎯 Exam Tip: Remember the special formula for the area of a parallelogram when given its diagonals: Area \( = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \). This is different from using adjacent sides. Ensure you correctly apply the cross product and magnitude calculations.

Question 16. If \( \vec{a} = 2 \hat{i} – 3 \hat{j} + \hat{k} \), \( \vec{b} = -\hat{i} + \hat{k} \), \( \vec{c} = 2 \hat{j} – \hat{k} \), find the area of the parallelogram having diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \).

Answer:
Given \( \vec{a} = 2 \hat{i} – 3 \hat{j} + \hat{k} \); \( \vec{b} = -\hat{i} + 0 \hat{j} + \hat{k} \) and \( \vec{c} = 0 \hat{i} + 2 \hat{j} – \hat{k} \).
Let the diagonals of the parallelogram be \( \vec{d_1} = \vec{a} + \vec{b} \) and \( \vec{d_2} = \vec{b} + \vec{c} \).
First, calculate \( \vec{d_1} \):
\( \vec{d_1} = \vec{a} + \vec{b} = (2 \hat{i} – 3 \hat{j} + \hat{k}) + (-\hat{i} + 0 \hat{j} + \hat{k}) \)
\( \implies \vec{d_1} = (2 - 1) \hat{i} + (-3 + 0) \hat{j} + (1 + 1) \hat{k} \)
\( \implies \vec{d_1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \)
Next, calculate \( \vec{d_2} \):
\( \vec{d_2} = \vec{b} + \vec{c} = (-\hat{i} + 0 \hat{j} + \hat{k}) + (0 \hat{i} + 2 \hat{j} – \hat{k}) \)
\( \implies \vec{d_2} = (-1 + 0) \hat{i} + (0 + 2) \hat{j} + (1 - 1) \hat{k} \)
\( \implies \vec{d_2} = -\hat{i} + 2 \hat{j} + 0 \hat{k} \)
Now, find the cross product \( \vec{d_1} \times \vec{d_2} \):
\( \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} \)
\( \implies \vec{d_1} \times \vec{d_2} = \hat{i}((-3)(0) - (2)(2)) - \hat{j}((1)(0) - (2)(-1)) + \hat{k}((1)(2) - (-3)(-1)) \)
\( \implies \vec{d_1} \times \vec{d_2} = \hat{i}(0 - 4) - \hat{j}(0 + 2) + \hat{k}(2 - 3) \)
\( \implies \vec{d_1} \times \vec{d_2} = -4 \hat{i} - 2 \hat{j} - \hat{k} \)
Next, find the magnitude of this cross product:
\( |\vec{d_1} \times \vec{d_2}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} \)
\( \implies |\vec{d_1} \times \vec{d_2}| = \sqrt{16 + 4 + 1} \)
\( \implies |\vec{d_1} \times \vec{d_2}| = \sqrt{21} \)
Finally, calculate the area of the parallelogram using the diagonal formula:
Area \( = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \)
\( \implies \) Area \( = \frac{1}{2} \sqrt{21} \) sq. units.
In simple words: This problem asks us to find the area of a parallelogram when its diagonals are given as sums of other vectors. First, we need to calculate the actual diagonal vectors by adding the given component vectors. Once we have the two diagonals, we find their cross product. Then, we calculate the length of this cross product vector. Finally, we divide this length by two to get the parallelogram's area.

🎯 Exam Tip: Pay close attention to the definition of the diagonals. Here, they are sums of other vectors, so compute these sums accurately first. Then, apply the diagonal area formula \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \). Avoid common arithmetic errors in vector addition and cross product calculation.

Question 17. If \( \vec{p} \) and \( \vec{q} \) are unit vectors forming an angle of \( 30^\circ \); find the area of the parallelogram having \( \vec{a} = \vec{p} + 2\vec{q} \) and \( \vec{b} = 2\vec{p} + \vec{q} \) as its diagonals.

Answer:
Given that \( \vec{p} \) and \( \vec{q} \) are unit vectors, so \( |\vec{p}| = 1 \) and \( |\vec{q}| = 1 \).
The angle between \( \vec{p} \) and \( \vec{q} \) is \( \theta = 30^\circ = \frac{\pi}{6} \).
The diagonals of the parallelogram are \( \vec{a} = \vec{p} + 2\vec{q} \) and \( \vec{b} = 2\vec{p} + \vec{q} \).
The area of a parallelogram given its diagonals is \( \frac{1}{2} |\vec{a} \times \vec{b}| \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = (\vec{p} + 2\vec{q}) \times (2\vec{p} + \vec{q}) \)
Using the distributive property of cross product:
\( \implies \vec{a} \times \vec{b} = (\vec{p} \times 2\vec{p}) + (\vec{p} \times \vec{q}) + (2\vec{q} \times 2\vec{p}) + (2\vec{q} \times \vec{q}) \)
We know that \( \vec{X} \times \vec{X} = \vec{0} \) and \( \vec{Y} \times \vec{X} = -\vec{X} \times \vec{Y} \).
\( \implies \vec{a} \times \vec{b} = 2(\vec{p} \times \vec{p}) + (\vec{p} \times \vec{q}) + 4(\vec{q} \times \vec{p}) + 2(\vec{q} \times \vec{q}) \)
\( \implies \vec{a} \times \vec{b} = 2(\vec{0}) + (\vec{p} \times \vec{q}) + 4(-\vec{p} \times \vec{q}) + 2(\vec{0}) \)
\( \implies \vec{a} \times \vec{b} = (\vec{p} \times \vec{q}) - 4(\vec{p} \times \vec{q}) \)
\( \implies \vec{a} \times \vec{b} = -3(\vec{p} \times \vec{q}) \)
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = |-3(\vec{p} \times \vec{q})| \)
\( \implies |\vec{a} \times \vec{b}| = 3 |\vec{p} \times \vec{q}| \)
We know that \( |\vec{p} \times \vec{q}| = |\vec{p}| |\vec{q}| \sin \theta \).
Substitute \( |\vec{p}|=1 \), \( |\vec{q}|=1 \), and \( \theta = 30^\circ \):
\( |\vec{p} \times \vec{q}| = (1)(1) \sin 30^\circ \)
\( \implies |\vec{p} \times \vec{q}| = 1 \times \frac{1}{2} = \frac{1}{2} \)
So, \( |\vec{a} \times \vec{b}| = 3 \left(\frac{1}{2}\right) = \frac{3}{2} \)
Finally, calculate the area of the parallelogram:
Area \( = \frac{1}{2} |\vec{a} \times \vec{b}| \)
\( \implies \) Area \( = \frac{1}{2} \left(\frac{3}{2}\right) = \frac{3}{4} \) sq. units.
In simple words: We are given that two unit vectors, \( \vec{p} \) and \( \vec{q} \), are at a 30-degree angle. We also have two diagonal vectors of a parallelogram, expressed in terms of \( \vec{p} \) and \( \vec{q} \). To find the parallelogram's area, we first find the cross product of these diagonal vectors. Using vector rules, this simplifies to \( -3 \) times the cross product of \( \vec{p} \) and \( \vec{q} \). Then, we use the formula for the magnitude of the cross product of \( \vec{p} \) and \( \vec{q} \) to find its value. Finally, we put all these pieces together to calculate the area using the diagonal formula.

🎯 Exam Tip: This problem requires combining several vector concepts: distributive property of cross products, properties of \( \vec{X} \times \vec{X} = \vec{0} \), \( \vec{Y} \times \vec{X} = -\vec{X} \times \vec{Y} \), magnitude of cross products, and the area formula for a parallelogram with given diagonals. Carefully substitute values and simplify. A common mistake is forgetting the \( \frac{1}{2} \) factor for diagonal area.

Question 18. Find the area of the triangle whose adjacent sides are determined by the vectors.
(i) \( \overrightarrow{\mathrm{OA}} = 3 \hat{i} + 2 \hat{j} – \hat{k} \), \( \overrightarrow{\mathrm{OB}} = \hat{i} + 3 \hat{j} + \hat{k} \)
(ii) \( 3 \hat{i} + 4 \hat{j} \) and \( -5 \hat{i} + 7 \hat{j} \)

Answer:
The area of a triangle with adjacent sides \( \vec{A} \) and \( \vec{B} \) is given by \( \frac{1}{2} |\vec{A} \times \vec{B}| \).
(i) Let the adjacent sides be \( \vec{A} = \overrightarrow{\mathrm{OA}} = 3 \hat{i} + 2 \hat{j} – \hat{k} \) and \( \vec{B} = \overrightarrow{\mathrm{OB}} = \hat{i} + 3 \hat{j} + \hat{k} \).
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & 3 & 1 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((2)(1) - (-1)(3)) - \hat{j}((3)(1) - (-1)(1)) + \hat{k}((3)(3) - (2)(1)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(2 + 3) - \hat{j}(3 + 1) + \hat{k}(9 - 2) \)
\( \implies \vec{A} \times \vec{B} = 5 \hat{i} - 4 \hat{j} + 7 \hat{k} \)
Next, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{5^2 + (-4)^2 + 7^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{25 + 16 + 49} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{90} = \sqrt{9 \times 10} = 3 \sqrt{10} \)
Finally, calculate the area of the triangle:
Area \( = \frac{1}{2} |\vec{A} \times \vec{B}| \)
\( \implies \) Area \( = \frac{1}{2} (3 \sqrt{10}) = \frac{3 \sqrt{10}}{2} \) sq. units.
(ii) Let the adjacent sides be \( \vec{A} = 3 \hat{i} + 4 \hat{j} + 0 \hat{k} \) and \( \vec{B} = -5 \hat{i} + 7 \hat{j} + 0 \hat{k} \).
First, find the cross product \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ -5 & 7 & 0 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((4)(0) - (0)(7)) - \hat{j}((3)(0) - (0)(-5)) + \hat{k}((3)(7) - (4)(-5)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(21 + 20) \)
\( \implies \vec{A} \times \vec{B} = 0 \hat{i} + 0 \hat{j} + 41 \hat{k} \)
Next, find the magnitude of the cross product:
\( |\vec{A} \times \vec{B}| = \sqrt{0^2 + 0^2 + 41^2} \)
\( \implies |\vec{A} \times \vec{B}| = \sqrt{41^2} = 41 \)
Finally, calculate the area of the triangle:
Area \( = \frac{1}{2} |\vec{A} \times \vec{B}| \)
\( \implies \) Area \( = \frac{1}{2} (41) = \frac{41}{2} \) sq. units.
In simple words: To find the area of a triangle when you are given the vectors that represent two of its sides that meet at a point, you first find the cross product of these two side vectors. Then, you calculate the length of the resulting cross product vector. Finally, you divide this length by two to get the area of the triangle.

🎯 Exam Tip: Remember that the area of a triangle formed by adjacent vectors \( \vec{A} \) and \( \vec{B} \) is half the magnitude of their cross product, i.e., \( \frac{1}{2} |\vec{A} \times \vec{B}| \). Ensure you correctly identify the adjacent sides if points are given, or use the vectors directly if they represent sides. Simplify square roots if possible.

Question 19. Find by vector method, the area of triangle A B C whose vertices are
(i) A(1, 3, 2), B(2, -1, 1) and C(-1, 2, 3)
(ii) A(1, 2, 3), B(2, 5, -1) and C(-1, 1, 2).

Answer:
(i) Given vertices A(1, 3, 2), B(2, -1, 1), C(-1, 2, 3).
First, determine the position vectors of the vertices:
\( \vec{A} = \hat{i} + 3 \hat{j} + 2 \hat{k} \)
\( \vec{B} = 2 \hat{i} - \hat{j} + \hat{k} \)
\( \vec{C} = -\hat{i} + 2 \hat{j} + 3 \hat{k} \)
Next, form two side vectors of the triangle, originating from a common vertex. Let's use \( \overrightarrow{\mathrm{AB}} \) and \( \overrightarrow{\mathrm{AC}} \).
Calculate \( \overrightarrow{\mathrm{AB}} \):
\( \overrightarrow{\mathrm{AB}} = \vec{B} - \vec{A} = (2 \hat{i} - \hat{j} + \hat{k}) - (\hat{i} + 3 \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AB}} = (2 - 1) \hat{i} + (-1 - 3) \hat{j} + (1 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AB}} = \hat{i} - 4 \hat{j} - \hat{k} \)
Calculate \( \overrightarrow{\mathrm{AC}} \):
\( \overrightarrow{\mathrm{AC}} = \vec{C} - \vec{A} = (-\hat{i} + 2 \hat{j} + 3 \hat{k}) - (\hat{i} + 3 \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AC}} = (-1 - 1) \hat{i} + (2 - 3) \hat{j} + (3 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AC}} = -2 \hat{i} - \hat{j} + \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} \):
\( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & -1 \\ -2 & -1 & 1 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}((-4)(1) - (-1)(-1)) - \hat{j}((1)(1) - (-1)(-2)) + \hat{k}((1)(-1) - (-4)(-2)) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}(-4 - 1) - \hat{j}(1 - 2) + \hat{k}(-1 - 8) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = -5 \hat{i} + \hat{j} - 9 \hat{k} \)
Next, find the magnitude of the cross product:
\( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{(-5)^2 + 1^2 + (-9)^2} \)
\( \implies |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{25 + 1 + 81} \)
\( \implies |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{107} \)
Finally, calculate the area of \( \triangle ABC \):
Area \( = \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \)
\( \implies \) Area \( = \frac{1}{2} \sqrt{107} \) sq. units.
(ii) Given vertices A(1, 2, 3), B(2, 5, -1), C(-1, 1, 2).
First, determine the position vectors of the vertices:
\( \vec{A} = \hat{i} + 2 \hat{j} + 3 \hat{k} \)
\( \vec{B} = 2 \hat{i} + 5 \hat{j} – \hat{k} \)
\( \vec{C} = -\hat{i} + \hat{j} + 2 \hat{k} \)
Next, form two side vectors of the triangle, e.g., \( \overrightarrow{\mathrm{AB}} \) and \( \overrightarrow{\mathrm{AC}} \).
Calculate \( \overrightarrow{\mathrm{AB}} \):
\( \overrightarrow{\mathrm{AB}} = \vec{B} - \vec{A} = (2 \hat{i} + 5 \hat{j} – \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AB}} = (2 - 1) \hat{i} + (5 - 2) \hat{j} + (-1 - 3) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AB}} = \hat{i} + 3 \hat{j} - 4 \hat{k} \)
Calculate \( \overrightarrow{\mathrm{AC}} \):
\( \overrightarrow{\mathrm{AC}} = \vec{C} - \vec{A} = (-\hat{i} + \hat{j} + 2 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AC}} = (-1 - 1) \hat{i} + (1 - 2) \hat{j} + (2 - 3) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AC}} = -2 \hat{i} - \hat{j} - \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} \):
\( \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -4 \\ -2 & -1 & -1 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}((3)(-1) - (-4)(-1)) - \hat{j}((1)(-1) - (-4)(-2)) + \hat{k}((1)(-1) - (3)(-2)) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \hat{i}(-3 - 4) - \hat{j}(-1 - 8) + \hat{k}(-1 + 6) \)
\( \implies \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = -7 \hat{i} + 9 \hat{j} + 5 \hat{k} \)
Next, find the magnitude of the cross product:
\( |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{(-7)^2 + 9^2 + 5^2} \)
\( \implies |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{49 + 81 + 25} \)
\( \implies |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| = \sqrt{155} \)
Finally, calculate the area of \( \triangle ABC \):
Area \( = \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \)
\( \implies \) Area \( = \frac{1}{2} \sqrt{155} \) sq. units.
In simple words: To find the area of a triangle when you are given the coordinates of its three corner points (vertices), you first turn these points into position vectors. Then, pick one vertex and create two side vectors starting from that point (e.g., \( \vec{AB} \) and \( \vec{AC} \)). After that, find the cross product of these two side vectors. Calculate the length of this cross product vector, and finally, divide that length by two to get the area of the triangle.

🎯 Exam Tip: When given vertices, always form vectors representing two adjacent sides, such as \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \), from a common vertex. Ensure correct subtraction of position vectors to get the side vectors. The formula for the area of a triangle is \( \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \).

Question 20. Show that the points whose position vectors are given below, are collinear.
(i) \( \vec{a} - 2 \vec{b} + 3 \vec{c} \), \( 2 \vec{a} + 3 \vec{b} - 4 \vec{c} \) and \( -7 \vec{b} + 10 \vec{c} \)
(ii) \( 5 \hat{i} + 6 \hat{j} + 7 \hat{k} \), \( 7 \hat{i} – 8 \hat{j} + 9 \hat{k} \) and \( 3 \hat{i} + 20 \hat{j} + 5 \hat{k} \)

Answer:
(i) Let the position vectors of the three given points be \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \):
\( \vec{\alpha} = \vec{a} - 2 \vec{b} + 3 \vec{c} \)
\( \vec{\beta} = 2 \vec{a} + 3 \vec{b} - 4 \vec{c} \)
\( \vec{\gamma} = 0 \vec{a} - 7 \vec{b} + 10 \vec{c} \)
For points to be collinear, the sum of their pairwise cross products must be the zero vector, i.e., \( \vec{\alpha} \times \vec{\beta} + \vec{\beta} \times \vec{\gamma} + \vec{\gamma} \times \vec{\alpha} = \overrightarrow{0} \).
First, calculate \( \vec{\alpha} \times \vec{\beta} \):
\( \vec{\alpha} \times \vec{\beta} = (\vec{a} - 2 \vec{b} + 3 \vec{c}) \times (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) \)
Using the distributive property and \( \vec{X} \times \vec{X} = \vec{0} \), \( \vec{Y} \times \vec{X} = -\vec{X} \times \vec{Y} \):
\( \implies \vec{\alpha} \times \vec{\beta} = (\vec{a} \times 3\vec{b}) + (\vec{a} \times -4\vec{c}) + (-2\vec{b} \times 2\vec{a}) + (-2\vec{b} \times -4\vec{c}) + (3\vec{c} \times 2\vec{a}) + (3\vec{c} \times 3\vec{b}) \)
\( \implies \vec{\alpha} \times \vec{\beta} = 3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{c}) + 4(\vec{a} \times \vec{b}) + 8(\vec{b} \times \vec{c}) + 6(\vec{c} \times \vec{a}) - 9(\vec{b} \times \vec{c}) \)
\( \implies \vec{\alpha} \times \vec{\beta} = 7(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a}) \) (Equation 1)
Next, calculate \( \vec{\beta} \times \vec{\gamma} \):
\( \vec{\beta} \times \vec{\gamma} = (2 \vec{a} + 3 \vec{b} - 4 \vec{c}) \times (-7 \vec{b} + 10 \vec{c}) \)
\( \implies \vec{\beta} \times \vec{\gamma} = (2\vec{a} \times -7\vec{b}) + (2\vec{a} \times 10\vec{c}) + (3\vec{b} \times 10\vec{c}) + (-4\vec{c} \times -7\vec{b}) \)
\( \implies \vec{\beta} \times \vec{\gamma} = -14(\vec{a} \times \vec{b}) + 20(\vec{a} \times \vec{c}) + 30(\vec{b} \times \vec{c}) + 28(\vec{c} \times \vec{b}) \)
\( \implies \vec{\beta} \times \vec{\gamma} = -14(\vec{a} \times \vec{b}) - 20(\vec{c} \times \vec{a}) + 30(\vec{b} \times \vec{c}) - 28(\vec{b} \times \vec{c}) \)
\( \implies \vec{\beta} \times \vec{\gamma} = -14(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{c}) - 20(\vec{c} \times \vec{a}) \) (Equation 2)
Finally, calculate \( \vec{\gamma} \times \vec{\alpha} \):
\( \vec{\gamma} \times \vec{\alpha} = (-7 \vec{b} + 10 \vec{c}) \times (\vec{a} - 2 \vec{b} + 3 \vec{c}) \)
\( \implies \vec{\gamma} \times \vec{\alpha} = (-7\vec{b} \times \vec{a}) + (-7\vec{b} \times 3\vec{c}) + (10\vec{c} \times \vec{a}) + (10\vec{c} \times -2\vec{b}) \)
\( \implies \vec{\gamma} \times \vec{\alpha} = 7(\vec{a} \times \vec{b}) - 21(\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a}) + 20(\vec{b} \times \vec{c}) \)
\( \implies \vec{\gamma} \times \vec{\alpha} = 7(\vec{a} \times \vec{b}) - (\vec{b} \times \vec{c}) + 10(\vec{c} \times \vec{a}) \) (Equation 3)
Now, add Equation 1, Equation 2, and Equation 3:
\( \vec{\alpha} \times \vec{\beta} + \vec{\beta} \times \vec{\gamma} + \vec{\gamma} \times \vec{\alpha} = (7-14+7)(\vec{a} \times \vec{b}) + (-1+2-1)(\vec{b} \times \vec{c}) + (10-20+10)(\vec{c} \times \vec{a}) \)
\( \implies \vec{\alpha} \times \vec{\beta} + \vec{\beta} \times \vec{\gamma} + \vec{\gamma} \times \vec{\alpha} = 0(\vec{a} \times \vec{b}) + 0(\vec{b} \times \vec{c}) + 0(\vec{c} \times \vec{a}) \)
\( \implies \vec{\alpha} \times \vec{\beta} + \vec{\beta} \times \vec{\gamma} + \vec{\gamma} \times \vec{\alpha} = \overrightarrow{0} \)
Since the sum of the pairwise cross products is the zero vector, the given points are collinear.
(ii) Let the position vectors of the three given points be \( \vec{A}, \vec{B}, \vec{C} \):
\( \vec{A} = 5 \hat{i} + 6 \hat{j} + 7 \hat{k} \)
\( \vec{B} = 7 \hat{i} – 8 \hat{j} + 9 \hat{k} \)
\( \vec{C} = 3 \hat{i} + 20 \hat{j} + 5 \hat{k} \)
For points to be collinear, the area of the triangle formed by them must be zero. This means the cross product of two vectors representing two sides of the triangle must be \( \overrightarrow{0} \). Alternatively, we can show that the sum of pairwise cross products is zero, similar to part (i).
First, calculate \( \vec{A} \times \vec{B} \):
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 6 & 7 \\ 7 & -8 & 9 \end{vmatrix} \)
\( \implies \vec{A} \times \vec{B} = \hat{i}((6)(9) - (7)(-8)) - \hat{j}((5)(9) - (7)(7)) + \hat{k}((5)(-8) - (6)(7)) \)
\( \implies \vec{A} \times \vec{B} = \hat{i}(54 + 56) - \hat{j}(45 - 49) + \hat{k}(-40 - 42) \)
\( \implies \vec{A} \times \vec{B} = 110 \hat{i} + 4 \hat{j} - 82 \hat{k} \) (Equation 4)
Next, calculate \( \vec{B} \times \vec{C} \):
\( \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -8 & 9 \\ 3 & 20 & 5 \end{vmatrix} \)
\( \implies \vec{B} \times \vec{C} = \hat{i}((-8)(5) - (9)(20)) - \hat{j}((7)(5) - (9)(3)) + \hat{k}((7)(20) - (-8)(3)) \)
\( \implies \vec{B} \times \vec{C} = \hat{i}(-40 - 180) - \hat{j}(35 - 27) + \hat{k}(140 + 24) \)
\( \implies \vec{B} \times \vec{C} = -220 \hat{i} - 8 \hat{j} + 164 \hat{k} \) (Equation 5)
Finally, calculate \( \vec{C} \times \vec{A} \):
\( \vec{C} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 20 & 5 \\ 5 & 6 & 7 \end{vmatrix} \)
\( \implies \vec{C} \times \vec{A} = \hat{i}((20)(7) - (5)(6)) - \hat{j}((3)(7) - (5)(5)) + \hat{k}((3)(6) - (20)(5)) \)
\( \implies \vec{C} \times \vec{A} = \hat{i}(140 - 30) - \hat{j}(21 - 25) + \hat{k}(18 - 100) \)
\( \implies \vec{C} \times \vec{A} = 110 \hat{i} + 4 \hat{j} - 82 \hat{k} \) (Equation 6)
Now, sum Equation 4, Equation 5, and Equation 6:
\( \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} = (110 - 220 + 110)\hat{i} + (4 - 8 + 4)\hat{j} + (-82 + 164 - 82)\hat{k} \)
\( \implies \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \overrightarrow{0} \)
Since the sum is the zero vector, the points are collinear.
In simple words: To show that three points are in a straight line (collinear) using vectors, we can calculate the cross products of all possible pairs of vectors formed by these points. If the sum of these three cross products turns out to be the zero vector, then the points must be collinear. This is because the cross product of two vectors is zero if they are parallel, and if two vectors sharing a common point are parallel, all three points lie on the same line.

🎯 Exam Tip: For three points with position vectors \( \vec{A}, \vec{B}, \vec{C} \) to be collinear, a common test is to show that \( \overrightarrow{AB} \times \overrightarrow{AC} = \overrightarrow{0} \), or equivalently, \( \vec{A} \times \vec{B} + \vec{B} \times \vec{C} + \vec{C} \times \vec{A} = \overrightarrow{0} \). Choose the method that feels more straightforward for the given vector expressions, being very careful with vector algebra and determinant calculations.

Question 21. Find \( \lambda \) such that \( \vec{a} = \hat{i} + \lambda \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) are parallel.

Answer:
Given vectors \( \vec{a} = \hat{i} + \lambda \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \).
For two vectors to be parallel, their cross product must be the zero vector, i.e., \( \vec{a} \times \vec{b} = \overrightarrow{0} \).
First, find the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \lambda & 3 \\ 3 & 2 & 9 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((\lambda)(9) - (3)(2)) - \hat{j}((1)(9) - (3)(3)) + \hat{k}((1)(2) - (\lambda)(3)) \)
\( \implies \vec{a} \times \vec{b} = (9\lambda - 6) \hat{i} - (9 - 9) \hat{j} + (2 - 3\lambda) \hat{k} \)
\( \implies \vec{a} \times \vec{b} = (9\lambda - 6) \hat{i} + 0 \hat{j} + (2 - 3\lambda) \hat{k} \)
Since \( \vec{a} \) and \( \vec{b} \) are parallel, \( \vec{a} \times \vec{b} = \overrightarrow{0} \).
\( \implies (9\lambda - 6) \hat{i} + 0 \hat{j} + (2 - 3\lambda) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \)
For two vectors to be equal, their corresponding components must be equal. So, we set the components to zero:
From the \( \hat{i} \) component: \( 9\lambda - 6 = 0 \)
\( \implies 9\lambda = 6 \)
\( \implies \lambda = \frac{6}{9} = \frac{2}{3} \)
From the \( \hat{k} \) component: \( 2 - 3\lambda = 0 \)
\( \implies 2 = 3\lambda \)
\( \implies \lambda = \frac{2}{3} \)
Both components give the same value for \( \lambda \).
Therefore, \( \lambda = \frac{2}{3} \).
In simple words: When two vectors are parallel, it means they point in the same (or opposite) direction, and their cross product is zero. So, we calculate the cross product of the given vectors, which will have a variable \( \lambda \) in its components. We then set each component of this cross product to zero and solve for \( \lambda \). Both equations should give the same value for \( \lambda \).

🎯 Exam Tip: The condition for parallel vectors is that their cross product is the zero vector (or that their components are proportional). Setting each component of the cross product to zero gives equations from which you can find the unknown variable. Always check if all components yield a consistent value for the variable.

Question 22. Find the values of \( a \) for which the vectors \( 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) and \( \hat{i} + a \hat{j} + 3 \hat{k} \) are
(i) perpendicular
(ii) parallel.

Answer:
Let \( \vec{\alpha} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} \) and \( \vec{\beta} = \hat{i} + a \hat{j} + 3 \hat{k} \).
(i) For \( \vec{\alpha} \) and \( \vec{\beta} \) to be perpendicular, their dot product must be zero, i.e., \( \vec{\alpha} \cdot \vec{\beta} = 0 \).
\( (3 \hat{i} + 2 \hat{j} + 9 \hat{k}) \cdot (\hat{i} + a \hat{j} + 3 \hat{k}) = 0 \)
\( \implies (3)(1) + (2)(a) + (9)(3) = 0 \)
\( \implies 3 + 2a + 27 = 0 \)
\( \implies 2a + 30 = 0 \)
\( \implies 2a = -30 \)
\( \implies a = -15 \)
(ii) For \( \vec{\alpha} \) and \( \vec{\beta} \) to be parallel, their cross product must be the zero vector, i.e., \( \vec{\alpha} \times \vec{\beta} = \overrightarrow{0} \).
\( \vec{\alpha} \times \vec{\beta} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 9 \\ 1 & a & 3 \end{vmatrix} \)
\( \implies \vec{\alpha} \times \vec{\beta} = \hat{i}((2)(3) - (9)(a)) - \hat{j}((3)(3) - (9)(1)) + \hat{k}((3)(a) - (2)(1)) \)
\( \implies \vec{\alpha} \times \vec{\beta} = (6 - 9a) \hat{i} - (9 - 9) \hat{j} + (3a - 2) \hat{k} \)
\( \implies \vec{\alpha} \times \vec{\beta} = (6 - 9a) \hat{i} + 0 \hat{j} + (3a - 2) \hat{k} \)
Since \( \vec{\alpha} \times \vec{\beta} = \overrightarrow{0} \):
\( (6 - 9a) \hat{i} + 0 \hat{j} + (3a - 2) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \)
Set the components to zero:
From the \( \hat{i} \) component: \( 6 - 9a = 0 \)
\( \implies 6 = 9a \)
\( \implies a = \frac{6}{9} = \frac{2}{3} \)
From the \( \hat{k} \) component: \( 3a - 2 = 0 \)
\( \implies 3a = 2 \)
\( \implies a = \frac{2}{3} \)
Both components yield \( a = \frac{2}{3} \).
In simple words: For two vectors to be perpendicular, their dot product (a simple multiplication and sum of components) must be zero. For them to be parallel, their cross product (a more complex calculation using determinants) must be the zero vector. In both cases, we set the appropriate result to zero and solve the resulting equation(s) for the unknown variable \( a \).

🎯 Exam Tip: Clearly distinguish between the conditions for perpendicularity (dot product = 0) and parallelism (cross product = \( \overrightarrow{0} \)). For parallel vectors, you can also check if the ratios of corresponding components are equal, e.g., \( \frac{3}{1} = \frac{2}{a} = \frac{9}{3} \). This can sometimes be faster than computing the full cross product.

Question 23. Find a unit vector parallel to the xy plane and perpendicular to the vector \( 4 \hat{i} – 3 \hat{j} + \hat{k} \).

Answer:
A vector parallel to the xy-plane has its \( \hat{k} \) component equal to zero. So, let the required vector be \( \vec{v} = \alpha \hat{i} + \beta \hat{j} + 0 \hat{k} \).
Let the given vector be \( \vec{u} = 4 \hat{i} – 3 \hat{j} + \hat{k} \).
The required vector \( \vec{v} \) must be perpendicular to \( \vec{u} \). This means their dot product must be zero:
\( \vec{v} \cdot \vec{u} = 0 \)
\( (\alpha \hat{i} + \beta \hat{j} + 0 \hat{k}) \cdot (4 \hat{i} – 3 \hat{j} + \hat{k}) = 0 \)
\( \implies 4\alpha - 3\beta + 0(1) = 0 \)
\( \implies 4\alpha - 3\beta = 0 \)
\( \implies 4\alpha = 3\beta \)
From this, we can express \( \beta \) in terms of \( \alpha \): \( \beta = \frac{4\alpha}{3} \).
So, the vector \( \vec{v} \) can be written as \( \vec{v} = \alpha \hat{i} + \frac{4\alpha}{3} \hat{j} \).
We can factor out \( \alpha \): \( \vec{v} = \alpha \left( \hat{i} + \frac{4}{3} \hat{j} \right) \).
To simplify, let \( \alpha = 3k \) for some scalar \( k \). Then \( \vec{v} = 3k \left( \hat{i} + \frac{4}{3} \hat{j} \right) = k(3 \hat{i} + 4 \hat{j}) \).
Now we need to find the unit vector in this direction. First, find the magnitude of \( 3 \hat{i} + 4 \hat{j} \):
\( |3 \hat{i} + 4 \hat{j}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)
The unit vector in the direction of \( 3 \hat{i} + 4 \hat{j} \) is \( \frac{3 \hat{i} + 4 \hat{j}}{5} \).
Since \( \vec{v} \) could be in either direction, the unit vector can be \( \pm \frac{1}{5}(3 \hat{i} + 4 \hat{j}) \).
In simple words: A vector that is parallel to the xy-plane has no 'z' component. We represent this general vector with two unknown parts, \( \alpha \) and \( \beta \). Since this vector must also be perpendicular to another given vector, their dot product must be zero. This gives us a relationship between \( \alpha \) and \( \beta \). Using this relationship, we can write the vector in a simpler form. Finally, to make it a "unit" vector, we divide it by its own length. The plus-minus sign shows that there are two such unit vectors, pointing in opposite directions.

🎯 Exam Tip: Remember that a vector parallel to the xy-plane has a zero \( \hat{k} \) component, and a unit vector has a magnitude of 1. The condition for perpendicularity is that the dot product is zero. Systematically apply these conditions to find the unknown components and then normalize the vector.

Question 24. By vector method obtain the perpendicular distance of the point (6,-4,4) from the line passing through the points (2,1,2) and (3,-1,4).

Answer:
Let P be the point (6,-4,4), and let the line pass through points A(2,1,2) and B(3,-1,4).
The position vectors are: \( \vec{p} = 6 \hat{i} - 4 \hat{j} + 4 \hat{k} \), \( \vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k} \), \( \vec{b} = 3 \hat{i} - \hat{j} + 4 \hat{k} \).
The perpendicular distance (d) from a point P to a line passing through points A and B is given by the formula:
\( d = \frac{|\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{AB}}|} \)
First, calculate the vector \( \overrightarrow{\mathrm{AP}} \):
\( \overrightarrow{\mathrm{AP}} = \vec{p} - \vec{a} = (6 \hat{i} - 4 \hat{j} + 4 \hat{k}) - (2 \hat{i} + \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AP}} = (6 - 2) \hat{i} + (-4 - 1) \hat{j} + (4 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AP}} = 4 \hat{i} - 5 \hat{j} + 2 \hat{k} \)
Next, calculate the vector \( \overrightarrow{\mathrm{AB}} \):
\( \overrightarrow{\mathrm{AB}} = \vec{b} - \vec{a} = (3 \hat{i} - \hat{j} + 4 \hat{k}) - (2 \hat{i} + \hat{j} + 2 \hat{k}) \)
\( \implies \overrightarrow{\mathrm{AB}} = (3 - 2) \hat{i} + (-1 - 1) \hat{j} + (4 - 2) \hat{k} \)
\( \implies \overrightarrow{\mathrm{AB}} = \hat{i} - 2 \hat{j} + 2 \hat{k} \)
Now, find the cross product \( \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} \):
\( \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & 2 \\ 1 & -2 & 2 \end{vmatrix} \)
\( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \hat{i}((-5)(2) - (2)(-2)) - \hat{j}((4)(2) - (2)(1)) + \hat{k}((4)(-2) - (-5)(1)) \)
\( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = \hat{i}(-10 + 4) - \hat{j}(8 - 2) + \hat{k}(-8 + 5) \)
\( \implies \overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}} = -6 \hat{i} - 6 \hat{j} - 3 \hat{k} \)
Next, find the magnitude of the cross product \( |\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}| \):
\( |\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}| = \sqrt{(-6)^2 + (-6)^2 + (-3)^2} \)
\( \implies |\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}| = \sqrt{36 + 36 + 9} = \sqrt{81} = 9 \)
Also, find the magnitude of \( |\overrightarrow{\mathrm{AB}}| \):
\( |\overrightarrow{\mathrm{AB}}| = \sqrt{1^2 + (-2)^2 + 2^2} \)
\( \implies |\overrightarrow{\mathrm{AB}}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
Finally, calculate the perpendicular distance \( d \):
\( d = \frac{9}{3} = 3 \) units.
In simple words: To find the shortest distance from a point to a line using vectors, we pick any point on the line (let's call it A) and the point not on the line (P). We also need another point on the line (B). We create two vectors: one from A to P (\( \overrightarrow{AP} \)) and one along the line from A to B (\( \overrightarrow{AB} \)). We then find the cross product of these two vectors and calculate its length. This length is then divided by the length of the vector along the line (\( |\overrightarrow{AB}| \)). The result is the perpendicular distance.

🎯 Exam Tip: The formula for the perpendicular distance from a point P to a line passing through A and B is \( d = \frac{|\overrightarrow{\mathrm{AP}} \times \overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{AB}}|} \). Ensure you correctly form the vectors \( \overrightarrow{\mathrm{AP}} \) and \( \overrightarrow{\mathrm{AB}} \) (both starting from a point on the line) and calculate their cross product and magnitudes accurately.

Question 25. Given, \( \vec{a} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \), \( \vec{b} = \frac{1}{7}(3 \hat{i} – 6 \hat{j} + 2 \hat{k}) \), \( \vec{c} = \frac{1}{7}(6 \hat{i} + 2 \hat{j} – 3 \hat{k}) \), \( \hat{i}, \hat{j}, \hat{k} \) being a right handed orthogonal system of unit vectors in space, show that \( \vec{a}, \vec{b}, \vec{c} \) is also another system of unit vectors mutually perpendicular.

Answer:
Given vectors:
\( \vec{a} = \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \)
\( \vec{b} = \frac{1}{7}(3 \hat{i} – 6 \hat{j} + 2 \hat{k}) \)
\( \vec{c} = \frac{1}{7}(6 \hat{i} + 2 \hat{j} – 3 \hat{k}) \)
To show that \( \vec{a}, \vec{b}, \vec{c} \) form a system of unit vectors, we must prove that their magnitudes are 1.
\( |\vec{a}| = \left| \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \right| = \frac{1}{7} \sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{1}{7} \sqrt{49} = \frac{1}{7}(7) = 1 \)
\( |\vec{b}| = \left| \frac{1}{7}(3 \hat{i} – 6 \hat{j} + 2 \hat{k}) \right| = \frac{1}{7} \sqrt{3^2 + (-6)^2 + 2^2} = \frac{1}{7} \sqrt{9 + 36 + 4} = \frac{1}{7} \sqrt{49} = \frac{1}{7}(7) = 1 \)
\( |\vec{c}| = \left| \frac{1}{7}(6 \hat{i} + 2 \hat{j} – 3 \hat{k}) \right| = \frac{1}{7} \sqrt{6^2 + 2^2 + (-3)^2} = \frac{1}{7} \sqrt{36 + 4 + 9} = \frac{1}{7} \sqrt{49} = \frac{1}{7}(7) = 1 \)
Since \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \), they are unit vectors.
To show that they are mutually perpendicular, we must prove that their dot products are zero.
\( \vec{a} \cdot \vec{b} = \left( \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \right) \cdot \left( \frac{1}{7}(3 \hat{i} – 6 \hat{j} + 2 \hat{k}) \right) \)
\( \implies \vec{a} \cdot \vec{b} = \frac{1}{49} ((2)(3) + (3)(-6) + (6)(2)) \)
\( \implies \vec{a} \cdot \vec{b} = \frac{1}{49} (6 - 18 + 12) = \frac{1}{49} (0) = 0 \)
\( \vec{b} \cdot \vec{c} = \left( \frac{1}{7}(3 \hat{i} – 6 \hat{j} + 2 \hat{k}) \right) \cdot \left( \frac{1}{7}(6 \hat{i} + 2 \hat{j} – 3 \hat{k}) \right) \)
\( \implies \vec{b} \cdot \vec{c} = \frac{1}{49} ((3)(6) + (-6)(2) + (2)(-3)) \)
\( \implies \vec{b} \cdot \vec{c} = \frac{1}{49} (18 - 12 - 6) = \frac{1}{49} (0) = 0 \)
\( \vec{c} \cdot \vec{a} = \left( \frac{1}{7}(6 \hat{i} + 2 \hat{j} – 3 \hat{k}) \right) \cdot \left( \frac{1}{7}(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) \right) \)
\( \implies \vec{c} \cdot \vec{a} = \frac{1}{49} ((6)(2) + (2)(3) + (-3)(6)) \)
\( \implies \vec{c} \cdot \vec{a} = \frac{1}{49} (12 + 6 - 18) = \frac{1}{49} (0) = 0 \)
Since \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{b} \cdot \vec{c} = 0 \), and \( \vec{c} \cdot \vec{a} = 0 \), the vectors are mutually perpendicular.
To show that they form a right-handed system, we can verify that \( \vec{a} \times \vec{b} = \vec{c} \).
\( \vec{a} \times \vec{b} = \frac{1}{49} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \frac{1}{49} [\hat{i}((3)(2) - (6)(-6)) - \hat{j}((2)(2) - (6)(3)) + \hat{k}((2)(-6) - (3)(3))] \)
\( \implies \vec{a} \times \vec{b} = \frac{1}{49} [\hat{i}(6 + 36) - \hat{j}(4 - 18) + \hat{k}(-12 - 9)] \)
\( \implies \vec{a} \times \vec{b} = \frac{1}{49} [42 \hat{i} + 14 \hat{j} - 21 \hat{k}] \)
\( \implies \vec{a} \times \vec{b} = \frac{7}{49} [6 \hat{i} + 2 \hat{j} - 3 \hat{k}] \)
\( \implies \vec{a} \times \vec{b} = \frac{1}{7} [6 \hat{i} + 2 \hat{j} - 3 \hat{k}] = \vec{c} \)
Similarly, \( \vec{b} \times \vec{c} = \vec{a} \) and \( \vec{c} \times \vec{a} = \vec{b} \). This confirms they form a right-handed system. All the conditions are met.
In simple words: We are given three new vectors and need to prove they act like the standard \( \hat{i}, \hat{j}, \hat{k} \) vectors. First, we calculate the length of each new vector and show that all of them are exactly 1. This means they are "unit vectors". Next, we take the dot product of each pair of vectors and show that all these dot products are zero. This means they are "mutually perpendicular". Lastly, to confirm they form a right-handed system, we can check if \( \vec{a} \times \vec{b} \) equals \( \vec{c} \), and so on. Since all these conditions are met, they form a new, valid right-handed orthogonal system of unit vectors.

🎯 Exam Tip: To prove a set of vectors forms a right-handed orthogonal system of unit vectors, you must show three things: (1) each vector has magnitude 1 (is a unit vector), (2) each pair of vectors is mutually perpendicular (their dot product is 0), and (3) they satisfy the right-handed rule (e.g., \( \vec{a} \times \vec{b} = \vec{c} \), \( \vec{b} \times \vec{c} = \vec{a} \), \( \vec{c} \times \vec{a} = \vec{b} \)). Be thorough in checking all conditions.

Question 26. \( \vec{a}, \vec{b}, \vec{c} \) represents three sides \( \overrightarrow{\mathrm{BC}} \), \( \overrightarrow{\mathrm{CA}} \), \( \overrightarrow{\mathrm{AB}} \) of a \( \triangle \mathrm{ABC} \) such that \( |\vec{a}| = 13 \), \( |\vec{b}|= 14 \), \( |\vec{c}|= 15 \) Find \( \angle C \).

Answer:
In a triangle ABC, let the lengths of the sides opposite to vertices A, B, C be \( a, b, c \) respectively.
Given: \( |\vec{a}| = 13 \), which means \( BC = 13 \).
Given: \( |\vec{b}| = 14 \), which means \( CA = 14 \).
Given: \( |\vec{c}| = 15 \), which means \( AB = 15 \).
So, in \( \triangle ABC \): \( c = AB = 15 \), \( a = BC = 13 \), \( b = CA = 14 \).
We need to find \( \angle C \). We can use the Cosine Rule:
\( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)
First, calculate the squares of the side lengths:
\( a^2 = 13^2 = 169 \)
\( b^2 = 14^2 = 196 \)
\( c^2 = 15^2 = 225 \)
Now, substitute these values into the Cosine Rule formula:
\( \cos C = \frac{169 + 196 - 225}{2 \times 13 \times 14} \)
\( \implies \cos C = \frac{365 - 225}{364} \)
\( \implies \cos C = \frac{140}{364} \)
Simplify the fraction:
\( \implies \cos C = \frac{140 \div 28}{364 \div 28} \)
\( \implies \cos C = \frac{5}{13} \)
Therefore, \( C = \cos^{-1}\left(\frac{5}{13}\right) \).
In simple words: We are given the lengths of all three sides of a triangle. To find one of the angles, for example, angle C, we use a rule called the Cosine Rule. This rule relates the lengths of the sides to the cosine of one of the angles. We plug in the lengths of the sides into the formula, do the calculations, and then use the inverse cosine function to find the actual angle.

🎯 Exam Tip: When using the Cosine Rule, ensure you correctly identify which side length corresponds to which variable (e.g., \( c \) is opposite \( \angle C \)). Double-check arithmetic, especially squaring numbers and simplifying the fraction, to avoid common errors.

Question 27. \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \); \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \), \( \vec{a} \neq \overrightarrow{0} \) prove that \( \vec{b} = \vec{c} \).

Answer:
Given two conditions:
1. \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \)
2. \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \)
Also given that \( \vec{a} \neq \overrightarrow{0} \). We need to prove \( \vec{b} = \vec{c} \).
**From condition 1: \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \)**
\( \implies \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0 \)
Using the distributive property of dot product:
\( \implies \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \)
This means either \( \vec{a} = \overrightarrow{0} \) or \( \vec{b} - \vec{c} = \overrightarrow{0} \) or \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).
Since we are given \( \vec{a} \neq \overrightarrow{0} \), we have two possibilities:
(i) \( \vec{b} - \vec{c} = \overrightarrow{0} \implies \vec{b} = \vec{c} \)
(ii) \( \vec{a} \perp (\vec{b} - \vec{c}) \) (meaning the angle between \( \vec{a} \) and \( \vec{b} - \vec{c} \) is \( 90^\circ \)) (Equation A)

**From condition 2: \( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \)**
\( \implies \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = \overrightarrow{0} \)
Using the distributive property of cross product:
\( \implies \vec{a} \times (\vec{b} - \vec{c}) = \overrightarrow{0} \)
This means either \( \vec{a} = \overrightarrow{0} \) or \( \vec{b} - \vec{c} = \overrightarrow{0} \) or \( \vec{a} \) is parallel to \( (\vec{b} - \vec{c}) \).
Since we are given \( \vec{a} \neq \overrightarrow{0} \), we have two possibilities:
(iii) \( \vec{b} - \vec{c} = \overrightarrow{0} \implies \vec{b} = \vec{c} \)
(iv) \( \vec{a} || (\vec{b} - \vec{c}) \) (meaning the angle between \( \vec{a} \) and \( \vec{b} - \vec{c} \) is \( 0^\circ \) or \( 180^\circ \)) (Equation B)

Now, consider the possibilities from Equation A and Equation B simultaneously.
If \( \vec{b} - \vec{c} \neq \overrightarrow{0} \), then from (A), \( \vec{a} \) must be perpendicular to \( (\vec{b} - \vec{c}) \), and from (B), \( \vec{a} \) must be parallel to \( (\vec{b} - \vec{c}) \).
A non-zero vector \( \vec{a} \) cannot be both parallel and perpendicular to another non-zero vector \( (\vec{b} - \vec{c}) \) at the same time.
This implies that the only possibility for \( \vec{b} - \vec{c} \) is that it must be the zero vector.
Therefore, \( \vec{b} - \vec{c} = \overrightarrow{0} \)
\( \implies \vec{b} = \vec{c} \)
This proves the statement.
In simple words: We are given two pieces of information: the dot product of \( \vec{a} \) with \( \vec{b} \) is the same as with \( \vec{c} \), and similarly for the cross product. Also, \( \vec{a} \) is not a zero vector. From the dot product rule, we know \( \vec{a} \) must be perpendicular to \( (\vec{b} - \vec{c}) \). From the cross product rule, we know \( \vec{a} \) must be parallel to \( (\vec{b} - \vec{c}) \). The only way for a non-zero vector \( \vec{a} \) to be both parallel and perpendicular to another vector \( (\vec{b} - \vec{c}) \) is if that other vector, \( (\vec{b} - \vec{c}) \), is itself a zero vector. If \( (\vec{b} - \vec{c}) \) is zero, then \( \vec{b} \) must be equal to \( \vec{c} \).

🎯 Exam Tip: This proof elegantly uses the properties of dot and cross products. The key is to manipulate both given equations into forms involving \( (\vec{b} - \vec{c}) \) and then deduce that \( (\vec{b} - \vec{c}) \) must be the zero vector, as it cannot be both parallel and perpendicular to \( \vec{a} \) (since \( \vec{a} \neq \overrightarrow{0} \)).

Question 28. If \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) then show that
(i) \( (\vec{r} \times \hat{i})^2 = y^2 + z^2 \)
(ii) \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + x y = 0 \).

Answer:
Given \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \).
(i) First, calculate the cross product \( \vec{r} \times \hat{i} \):
\( \vec{r} \times \hat{i} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{i} \)
Using the distributive property and standard cross product rules (\( \hat{i} \times \hat{i} = \overrightarrow{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), \( \hat{k} \times \hat{i} = \hat{j} \)):
\( \implies \vec{r} \times \hat{i} = x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i}) \)
\( \implies \vec{r} \times \hat{i} = x(\overrightarrow{0}) + y(-\hat{k}) + z(\hat{j}) \)
\( \implies \vec{r} \times \hat{i} = z \hat{j} - y \hat{k} \)
Now, we need to find \( (\vec{r} \times \hat{i})^2 \), which is the square of its magnitude \( |\vec{r} \times \hat{i}|^2 \):
\( (\vec{r} \times \hat{i})^2 = |z \hat{j} - y \hat{k}|^2 \)
\( \implies (\vec{r} \times \hat{i})^2 = \sqrt{0^2 + z^2 + (-y)^2}^2 \)
\( \implies (\vec{r} \times \hat{i})^2 = z^2 + y^2 \)
Thus, \( (\vec{r} \times \hat{i})^2 = y^2 + z^2 \) is shown.
(ii) We already have \( \vec{r} \times \hat{i} = z \hat{j} - y \hat{k} \).
Next, calculate the cross product \( \vec{r} \times \hat{j} \):
\( \vec{r} \times \hat{j} = (x \hat{i} + y \hat{j} + z \hat{k}) \times \hat{j} \)
Using standard cross product rules (\( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \overrightarrow{0} \), \( \hat{k} \times \hat{j} = -\hat{i} \)):
\( \implies \vec{r} \times \hat{j} = x(\hat{i} \times \hat{j}) + y(\hat{j} \times \hat{j}) + z(\hat{k} \times \hat{j}) \)
\( \implies \vec{r} \times \hat{j} = x(\hat{k}) + y(\overrightarrow{0}) + z(-\hat{i}) \)
\( \implies \vec{r} \times \hat{j} = x \hat{k} - z \hat{i} \)
Now, find the dot product \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) \):
\( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = (0 \hat{i} + z \hat{j} - y \hat{k}) \cdot (-z \hat{i} + 0 \hat{j} + x \hat{k}) \)
\( \implies (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = (0)(-z) + (z)(0) + (-y)(x) \)
\( \implies (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = 0 + 0 - xy \)
\( \implies (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) = -xy \)
We need to show \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = 0 \).
Substitute the result: \( -xy + xy = 0 \)
\( \implies 0 = 0 \)
Thus, \( (\vec{r} \times \hat{i}) \cdot (\vec{r} \times \hat{j}) + xy = 0 \) is shown.
In simple words: This problem asks us to prove two vector identities using a general vector \( \vec{r} \). For part (i), we first calculate the cross product of \( \vec{r} \) with \( \hat{i} \), which gives a new vector. Then we find the square of the length of this new vector. For part (ii), we calculate another cross product, \( \vec{r} \) with \( \hat{j} \). After getting both cross product vectors, we find their dot product and then add \( xy \) to see if the sum is zero. Both parts rely on careful application of vector algebra rules.

🎯 Exam Tip: Remember the basic cross product relations for unit vectors (\( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{i} = -\hat{k} \), etc., and \( \hat{i} \times \hat{i} = \overrightarrow{0} \)). When calculating \( (\vec{A})^2 \), it means \( |\vec{A}|^2 \), the square of the magnitude. Pay attention to signs in cross products and ensure correct component-wise multiplication for dot products.

Question 29. For any vector \( \vec{a} \), prove that \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \).

Answer:
Let the vector \( \vec{a} \) be \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \).
Then, the magnitude of \( \vec{a} \) is \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \), so \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \).
We need to prove \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \).
First, calculate \( \vec{a} \times \hat{i} \):
\( \vec{a} \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} \)
\( \implies \vec{a} \times \hat{i} = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) \)
\( \implies \vec{a} \times \hat{i} = a_1(\overrightarrow{0}) + a_2(-\hat{k}) + a_3(\hat{j}) \)
\( \implies \vec{a} \times \hat{i} = a_3 \hat{j} - a_2 \hat{k} \)
Now, find its magnitude squared \( |\vec{a} \times \hat{i}|^2 \):
\( |\vec{a} \times \hat{i}|^2 = a_3^2 + (-a_2)^2 = a_3^2 + a_2^2 \) (Equation 1)

Next, calculate \( \vec{a} \times \hat{j} \):
\( \vec{a} \times \hat{j} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{j} \)
\( \implies \vec{a} \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_2(\hat{j} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) \)
\( \implies \vec{a} \times \hat{j} = a_1(\hat{k}) + a_2(\overrightarrow{0}) + a_3(-\hat{i}) \)
\( \implies \vec{a} \times \hat{j} = -a_3 \hat{i} + a_1 \hat{k} \)
Now, find its magnitude squared \( |\vec{a} \times \hat{j}|^2 \):
\( |\vec{a} \times \hat{j}|^2 = (-a_3)^2 + a_1^2 = a_3^2 + a_1^2 \) (Equation 2)

Finally, calculate \( \vec{a} \times \hat{k} \):
\( \vec{a} \times \hat{k} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{k} \)
\( \implies \vec{a} \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) + a_3(\hat{k} \times \hat{k}) \)
\( \implies \vec{a} \times \hat{k} = a_1(-\hat{j}) + a_2(\hat{i}) + a_3(\overrightarrow{0}) \)
\( \implies \vec{a} \times \hat{k} = a_2 \hat{i} - a_1 \hat{j} \)
Now, find its magnitude squared \( |\vec{a} \times \hat{k}|^2 \):
\( |\vec{a} \times \hat{k}|^2 = a_2^2 + (-a_1)^2 = a_2^2 + a_1^2 \) (Equation 3)

Add Equation 1, Equation 2, and Equation 3:
\( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \)
\( \implies |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2a_1^2 + 2a_2^2 + 2a_3^2 \)
\( \implies |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2(a_1^2 + a_2^2 + a_3^2) \)
Since \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \):
\( \implies |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \)
This proves the identity.
In simple words: This problem asks us to prove a specific vector identity. We start by assuming a general vector \( \vec{a} \) with components \( a_1, a_2, a_3 \). Then, we individually calculate the cross product of \( \vec{a} \) with each of the standard unit vectors \( \hat{i}, \hat{j}, \hat{k} \). For each of these results, we find the square of its length (magnitude). When we add these three squared lengths together, we find that the sum simplifies to twice the square of the length of the original vector \( \vec{a} \), thus proving the identity.

🎯 Exam Tip: This proof is a direct application of the component form of vectors and cross product rules. Ensure you correctly apply the cross product rules for \( \hat{i}, \hat{j}, \hat{k} \) (e.g., \( \hat{i} \times \hat{k} = -\hat{j} \)) and calculate the magnitude squared for each resulting vector. The summation then becomes straightforward.