Samacheer Kalvi Class 11 Maths Solutions Chapter 8 Vector Algebra I Exercise 8.4

Question 1. Find the magnitude of \( \vec{a} \times \vec{b} \) if \( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \)
Answer: The given vectors are:
\( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \)
\( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \)

First, we calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i} ((1)(-2) - (3)(5)) - \hat{j} ((2)(-2) - (3)(3)) + \hat{k} ((2)(5) - (1)(3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (-2 - 15) - \hat{j} (-4 - 9) + \hat{k} (10 - 3) \)
\( \implies \vec{a} \times \vec{b} = -17\hat{i} - \hat{j} (-13) + 7\hat{k} \)
\( \implies \vec{a} \times \vec{b} = -17\hat{i} + 13\hat{j} + 7\hat{k} \)

Next, we find the magnitude of \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + (13)^2 + (7)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{289 + 169 + 49} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{507} \)
The magnitude of the cross product is \( \sqrt{507} \). This calculation helps in finding the area of a parallelogram formed by these vectors.
In simple words: To find the magnitude, first calculate the cross product of the two vectors using a determinant. Then, take the square root of the sum of the squares of each component from the cross product.

🎯 Exam Tip: Remember that the cross product of two vectors results in a vector, while its magnitude is a scalar value. Pay close attention to the signs when expanding the determinant.

Question 2. Show that \( \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) = \vec{0} \)
Answer: We need to show that the given vector sum equals the zero vector.
Let's expand each term using the distributive property of the vector cross product:
\( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \)
\( \vec{b} \times (\vec{c} + \vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} \)
\( \vec{c} \times (\vec{a} + \vec{b}) = \vec{c} \times \vec{a} + \vec{c} \times \vec{b} \)

Now, sum these expanded terms:
\( (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (\vec{b} \times \vec{c} + \vec{b} \times \vec{a}) + (\vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \)
Group similar terms. We know that \( \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) \).
So, \( \vec{a} \times \vec{c} = -(\vec{c} \times \vec{a}) \) and \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \) and \( \vec{c} \times \vec{b} = -(\vec{b} \times \vec{c}) \).

Substitute these into the sum:
\( \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} - (\vec{a} \times \vec{b}) + \vec{c} \times \vec{a} - (\vec{b} \times \vec{c}) \)
\( \implies \vec{a} \times \vec{b} - \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{c} \times \vec{a} + \vec{b} \times \vec{c} - \vec{b} \times \vec{c} \)
\( \implies \vec{0} + \vec{a} \times \vec{c} - \vec{a} \times \vec{c} + \vec{b} \times \vec{c} - \vec{b} \times \vec{c} \)
\( \implies \vec{0} + \vec{0} + \vec{0} = \vec{0} \)
Thus, the sum of the given vector products is indeed the zero vector. This demonstrates a useful property of vector algebra, showing that such a combination of cyclic cross products always cancels out.
In simple words: Break down each part of the problem into simpler cross products. Then, use the rule that swapping the order of vectors in a cross product changes its sign. When you put all the parts back together, they will cancel each other out, leaving nothing.

🎯 Exam Tip: Remember the property \( \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \) when simplifying vector cross product expressions. This is key to terms cancelling out.

Question 3. Find the vector of magnitude \( 10\sqrt{3} \) that are perpendicular to the plane which contains \( \hat{i} + 2\hat{j} + \hat{k} \) and \( \hat{i} + 3\hat{j} + 4\hat{k} \)
Answer: Let the two given vectors be:
\( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \)
\( \vec{b} = \hat{i} + 3\hat{j} + 4\hat{k} \)

A vector perpendicular to the plane containing \( \vec{a} \) and \( \vec{b} \) is given by their cross product \( \vec{a} \times \vec{b} \).
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 1 & 3 & 4 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i} ((2)(4) - (1)(3)) - \hat{j} ((1)(4) - (1)(1)) + \hat{k} ((1)(3) - (2)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (8 - 3) - \hat{j} (4 - 1) + \hat{k} (3 - 2) \)
\( \implies \vec{a} \times \vec{b} = 5\hat{i} - 3\hat{j} + \hat{k} \)

Now, we find the magnitude of this vector:
\( |\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (-3)^2 + (1)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{25 + 9 + 1} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{35} \)

The unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is:
\( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{5\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{35}} \)

We need a vector with magnitude \( 10\sqrt{3} \) that is perpendicular to the plane. This means we multiply the unit vector by the required magnitude.
The desired vector is \( \pm 10\sqrt{3} \left( \frac{5\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{35}} \right) \).
We use \( \pm \) because a vector perpendicular to a plane can point in two opposite directions. This process helps us find a specific vector that meets both direction and length requirements.
In simple words: First, find a vector that is perpendicular to both given vectors by calculating their cross product. Then, find the length of this new vector. Divide the perpendicular vector by its length to get a unit vector (a vector of length 1). Finally, multiply this unit vector by the desired length (\( 10\sqrt{3} \)) to get the final answer. Remember there are two possible directions, so use a plus/minus sign.

🎯 Exam Tip: A vector perpendicular to a plane formed by two vectors \( \vec{a} \) and \( \vec{b} \) is given by their cross product \( \vec{a} \times \vec{b} \). Always remember to consider both positive and negative directions for the final vector.

Question 4. Find the unit vectors perpendicular to each of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \), where \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \)
Answer: Given vectors are:
\( \vec{a} = \hat{i} + \hat{j} + \hat{k} \)
\( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \)

First, find \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \):
\( \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
\( \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k} \)

Let \( \vec{P} = \vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
Let \( \vec{Q} = \vec{a} - \vec{b} = -\hat{j} - 2\hat{k} \)

A vector perpendicular to both \( \vec{P} \) and \( \vec{Q} \) is their cross product \( \vec{P} \times \vec{Q} \):
\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \]
\( \implies \vec{P} \times \vec{Q} = \hat{i} ((3)(-2) - (4)(-1)) - \hat{j} ((2)(-2) - (4)(0)) + \hat{k} ((2)(-1) - (3)(0)) \)
\( \implies \vec{P} \times \vec{Q} = \hat{i} (-6 + 4) - \hat{j} (-4 - 0) + \hat{k} (-2 - 0) \)
\( \implies \vec{P} \times \vec{Q} = -2\hat{i} + 4\hat{j} - 2\hat{k} \)

Now, find the magnitude of this cross product:
\( |\vec{P} \times \vec{Q}| = \sqrt{(-2)^2 + (4)^2 + (-2)^2} \)
\( \implies |\vec{P} \times \vec{Q}| = \sqrt{4 + 16 + 4} \)
\( \implies |\vec{P} \times \vec{Q}| = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \)

The unit vectors perpendicular to \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are:
\( \pm \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} \)
\( \implies \pm \frac{2(-\hat{i} + 2\hat{j} - \hat{k})}{2\sqrt{6}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \)
These unit vectors point in the two opposite directions perpendicular to the plane formed by \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).
In simple words: First, calculate the sum and difference of the two given vectors. Then, find the cross product of these two new vectors. This cross product will be perpendicular to both of them. Finally, divide this perpendicular vector by its own length to get a unit vector, which is a vector of length one. Remember to include both positive and negative directions.

🎯 Exam Tip: When asked for "unit vectors", remember to include the \( \pm \) sign in your final answer, as there are always two opposite unit vectors perpendicular to a given plane.

Question 5. Find the area of the parallelogram whose two adjacent sides are determined by the vectors \( \hat{i} + 2\hat{j} + 3\hat{k} \) and \( 3\hat{i} - 2\hat{j} + \hat{k} \)
Answer: Let the two adjacent sides of the parallelogram be represented by the vectors:
\( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \)
\( \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \)

The area of the parallelogram is the magnitude of the cross product of these two adjacent side vectors, i.e., \( |\vec{a} \times \vec{b}| \).

First, calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i} ((2)(1) - (3)(-2)) - \hat{j} ((1)(1) - (3)(3)) + \hat{k} ((1)(-2) - (2)(3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (2 - (-6)) - \hat{j} (1 - 9) + \hat{k} (-2 - 6) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (2 + 6) - \hat{j} (-8) + \hat{k} (-8) \)
\( \implies \vec{a} \times \vec{b} = 8\hat{i} + 8\hat{j} - 8\hat{k} \)

Next, calculate the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(8)^2 + (8)^2 + (-8)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{64 + 64 + 64} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{3 \times 64} \)
\( \implies |\vec{a} \times \vec{b}| = 8\sqrt{3} \)
The area of the parallelogram with these adjacent sides is \( 8\sqrt{3} \) square units. This shows how the cross product is directly related to geometric areas in vector calculus.
In simple words: To find the area of a parallelogram when given two side vectors, first calculate the cross product of these vectors. Then, find the length (magnitude) of the resulting vector. This length is the area of the parallelogram.

🎯 Exam Tip: The area of a parallelogram defined by vectors \( \vec{a} \) and \( \vec{b} \) as adjacent sides is \( |\vec{a} \times \vec{b}| \). Make sure to correctly compute the determinant and the final magnitude.

Question 6. Find the area of the triangle whose vertices are A(3, -1, 2), B(1, -1, -3), and C(4, -3, 1)
Answer: The given vertices of the triangle ABC are:
A = (3, -1, 2)
B = (1, -1, -3)
C = (4, -3, 1)

First, we find two vectors representing two sides of the triangle from a common vertex. Let's use \( \vec{AB} \) and \( \vec{AC} \).
The position vectors are: \( \vec{OA} = 3\hat{i} - \hat{j} + 2\hat{k} \), \( \vec{OB} = \hat{i} - \hat{j} - 3\hat{k} \), \( \vec{OC} = 4\hat{i} - 3\hat{j} + \hat{k} \).

\( \vec{AB} = \vec{OB} - \vec{OA} = (\hat{i} - \hat{j} - 3\hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) \)
\( \implies \vec{AB} = (1-3)\hat{i} + (-1-(-1))\hat{j} + (-3-2)\hat{k} \)
\( \implies \vec{AB} = -2\hat{i} + 0\hat{j} - 5\hat{k} = -2\hat{i} - 5\hat{k} \)

\( \vec{AC} = \vec{OC} - \vec{OA} = (4\hat{i} - 3\hat{j} + \hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) \)
\( \implies \vec{AC} = (4-3)\hat{i} + (-3-(-1))\hat{j} + (1-2)\hat{k} \)
\( \implies \vec{AC} = \hat{i} - 2\hat{j} - \hat{k} \)

Next, calculate the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} \]
\( \implies \vec{AB} \times \vec{AC} = \hat{i} ((0)(-1) - (-5)(-2)) - \hat{j} ((-2)(-1) - (-5)(1)) + \hat{k} ((-2)(-2) - (0)(1)) \)
\( \implies \vec{AB} \times \vec{AC} = \hat{i} (0 - 10) - \hat{j} (2 - (-5)) + \hat{k} (4 - 0) \)
\( \implies \vec{AB} \times \vec{AC} = -10\hat{i} - \hat{j} (2 + 5) + 4\hat{k} \)
\( \implies \vec{AB} \times \vec{AC} = -10\hat{i} - 7\hat{j} + 4\hat{k} \)

Now, find the magnitude of this cross product:
\( |\vec{AB} \times \vec{AC}| = \sqrt{(-10)^2 + (-7)^2 + (4)^2} \)
\( \implies |\vec{AB} \times \vec{AC}| = \sqrt{100 + 49 + 16} \)
\( \implies |\vec{AB} \times \vec{AC}| = \sqrt{165} \)

The area of the triangle ABC is half the magnitude of the cross product of two of its side vectors:
Area \( = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{165} \) square units.
This formula is efficient for finding the area of a triangle in three-dimensional space using its vertices.
In simple words: First, find two vectors that form two sides of the triangle, starting from the same corner. Then, calculate the cross product of these two vectors. Finally, take half of the length of this cross product vector. This value is the area of the triangle.

🎯 Exam Tip: To find the area of a triangle from its vertices, always calculate two side vectors from a common vertex (e.g., \( \vec{AB} \) and \( \vec{AC} \)), then find \( \frac{1}{2} |\vec{AB} \times \vec{AC}| \).

Question 7. If \( \vec{a}, \vec{b}, \vec{c} \) are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is \( \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \). Also deduce the condition for collinearity of the points A, B and C.
Answer: Let \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of the vertices A, B, C of triangle ABC.
The vectors representing two sides of the triangle, say \( \vec{AB} \) and \( \vec{AC} \), can be written as:
\( \vec{AB} = \vec{b} - \vec{a} \)
\( \vec{AC} = \vec{c} - \vec{a} \)

The area of triangle ABC is given by \( \frac{1}{2} |\vec{AB} \times \vec{AC}| \).
Let's find the cross product \( \vec{AB} \times \vec{AC} \):
\( \vec{AB} \times \vec{AC} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \)
\( \implies (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} \)
We know that \( \vec{a} \times \vec{a} = \vec{0} \) and \( -\vec{b} \times \vec{a} = \vec{a} \times \vec{b} \) and \( -\vec{a} \times \vec{c} = \vec{c} \times \vec{a} \).
So, \( \vec{AB} \times \vec{AC} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} \)
\( \implies \vec{AB} \times \vec{AC} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \)

Therefore, the area of triangle ABC is:
Area \( = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \)
This formula provides a symmetrical way to calculate the area using the position vectors of all three vertices. It is quite elegant.

**Condition for Collinearity:**
Points A, B, C are collinear if they lie on the same straight line. If they are collinear, they cannot form a triangle, which means the area of the triangle formed by them must be zero.
Thus, for points A, B, C to be collinear, the area of triangle ABC must be 0.
\( \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 0 \)
\( \implies |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| = 0 \)
In simple words: The area of a triangle formed by three points (A, B, C) can be found by taking half the length of a special vector sum: the cross product of A and B, plus B and C, plus C and A. If these three points all lie on a straight line, they cannot make a triangle, so the area of the "triangle" they would form is zero. This zero area is the condition for them to be in a straight line.

🎯 Exam Tip: The condition for three points A, B, C with position vectors \( \vec{a}, \vec{b}, \vec{c} \) to be collinear is that the vector \( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \) must be the zero vector.

Question 8. 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 any vector, which can be written in component form as:
\( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \)

Now, let's calculate each cross product and its squared magnitude:

1. Calculate \( \vec{a} \times \hat{i} \):
\( \vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} \)
\( \implies \vec{a} \times \hat{i} = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) \)
We know \( \hat{i} \times \hat{i} = \vec{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), \( \hat{k} \times \hat{i} = \hat{j} \).
\( \implies \vec{a} \times \hat{i} = a_1(\vec{0}) + a_2(-\hat{k}) + a_3(\hat{j}) \)
\( \implies \vec{a} \times \hat{i} = a_3\hat{j} - a_2\hat{k} \)
Now, find the squared magnitude:
\( |\vec{a} \times \hat{i}|^2 = |a_3\hat{j} - a_2\hat{k}|^2 = (a_3)^2 + (-a_2)^2 = a_3^2 + a_2^2 \) --- (1)

2. Calculate \( \vec{a} \times \hat{j} \):
\( \vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j} \)
\( \implies \vec{a} \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_2(\hat{j} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) \)
We know \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \vec{0} \), \( \hat{k} \times \hat{j} = -\hat{i} \).
\( \implies \vec{a} \times \hat{j} = a_1(\hat{k}) + a_2(\vec{0}) + a_3(-\hat{i}) \)
\( \implies \vec{a} \times \hat{j} = a_1\hat{k} - a_3\hat{i} \)
Now, find the squared magnitude:
\( |\vec{a} \times \hat{j}|^2 = |a_1\hat{k} - a_3\hat{i}|^2 = (-a_3)^2 + (a_1)^2 = a_3^2 + a_1^2 \) --- (2)

3. Calculate \( \vec{a} \times \hat{k} \):
\( \vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k} \)
\( \implies \vec{a} \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) + a_3(\hat{k} \times \hat{k}) \)
We know \( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{k} = \vec{0} \).
\( \implies \vec{a} \times \hat{k} = a_1(-\hat{j}) + a_2(\hat{i}) + a_3(\vec{0}) \)
\( \implies \vec{a} \times \hat{k} = a_2\hat{i} - a_1\hat{j} \)
Now, find the squared magnitude:
\( |\vec{a} \times \hat{k}|^2 = |a_2\hat{i} - a_1\hat{j}|^2 = (a_2)^2 + (-a_1)^2 = a_2^2 + a_1^2 \) --- (3)

Add equations (1), (2), and (3):
\( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_3^2 + a_2^2) + (a_3^2 + a_1^2) + (a_2^2 + a_1^2) \)
\( \implies = 2a_1^2 + 2a_2^2 + 2a_3^2 \)
\( \implies = 2(a_1^2 + a_2^2 + a_3^2) \) --- (4)

We also know that the magnitude squared of vector \( \vec{a} \) is:
\( |\vec{a}|^2 = |a_1\hat{i} + a_2\hat{j} + a_3\hat{k}|^2 = a_1^2 + a_2^2 + a_3^2 \) --- (5)

From equations (4) and (5), we can see that:
\( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2 |\vec{a}|^2 \)
This identity shows a relationship between the magnitude of a vector and its cross products with the standard basis vectors. It highlights a fundamental property in vector algebra.
In simple words: First, write any vector as its parts along the x, y, and z axes. Then, find the cross product of this vector with each basic direction vector (i, j, k). Calculate the squared length of each of these three new vectors. When you add these three squared lengths together, the total will be twice the squared length of the original vector.

🎯 Exam Tip: This proof relies on the cyclic properties of unit vectors (\( \hat{i} \times \hat{j} = \hat{k} \), etc.) and the definition of magnitude. Double-check your signs and squares carefully.

Question 9. Let \( \vec{a}, \vec{b}, \vec{c} \) be unit vectors such that \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{a} \cdot \vec{c} = 0 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{3} \). Prove that \( \vec{a} = \pm \frac{2}{\sqrt{3}} (\vec{b} \times \vec{c}) \).
Answer: Given that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, which means:
\( |\vec{a}| = 1 \), \( |\vec{b}| = 1 \), \( |\vec{c}| = 1 \)

Also given:
\( \vec{a} \cdot \vec{b} = 0 \)
\( \vec{a} \cdot \vec{c} = 0 \)

Since \( \vec{a} \cdot \vec{b} = 0 \), vector \( \vec{a} \) is perpendicular to vector \( \vec{b} \).
Since \( \vec{a} \cdot \vec{c} = 0 \), vector \( \vec{a} \) is perpendicular to vector \( \vec{c} \).
This implies that \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \).
We know that the cross product \( \vec{b} \times \vec{c} \) is a vector perpendicular to both \( \vec{b} \) and \( \vec{c} \).
Therefore, \( \vec{a} \) must be parallel to \( \vec{b} \times \vec{c} \). This means \( \vec{a} \) can be written as a scalar multiple of \( \vec{b} \times \vec{c} \).
\( \vec{a} = k (\vec{b} \times \vec{c}) \) for some scalar \( k \).

Now, let's find the magnitude of both sides:
\( |\vec{a}| = |k (\vec{b} \times \vec{c})| \)
\( |\vec{a}| = |k| |\vec{b} \times \vec{c}| \)

We know \( |\vec{a}| = 1 \).
Also, the magnitude of the cross product \( |\vec{b} \times \vec{c}| \) is given by \( |\vec{b}| |\vec{c}| \sin \theta \), where \( \theta \) is the angle between \( \vec{b} \) and \( \vec{c} \).
Given \( |\vec{b}| = 1 \), \( |\vec{c}| = 1 \), and \( \theta = \frac{\pi}{3} \).
So, \( |\vec{b} \times \vec{c}| = (1)(1) \sin(\frac{\pi}{3}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \).

Substitute these values back into the magnitude equation:
\( 1 = |k| \left( \frac{\sqrt{3}}{2} \right) \)
\( \implies |k| = \frac{2}{\sqrt{3}} \)
So, \( k = \pm \frac{2}{\sqrt{3}} \).

Substitute the value of \( k \) back into the equation for \( \vec{a} \):
\( \vec{a} = \pm \frac{2}{\sqrt{3}} (\vec{b} \times \vec{c}) \)
Hence, the statement is proven. This demonstrates how vector properties like perpendicularity and cross products are used in proofs involving unit vectors.
In simple words: We are told that vector 'a' is a unit vector and is perpendicular to both 'b' and 'c' (which are also unit vectors). This means 'a' must point in the same direction as the cross product of 'b' and 'c'. We then use the fact that the angle between 'b' and 'c' is 60 degrees to find the length of the cross product of 'b' and 'c'. By comparing the lengths, we find the scaling factor that relates 'a' to the cross product, proving the given statement.

🎯 Exam Tip: When \( \vec{A} \cdot \vec{B} = 0 \), it means \( \vec{A} \) is perpendicular to \( \vec{B} \). If a vector is perpendicular to two non-parallel vectors, it must be parallel to their cross product. This is a crucial concept for such proofs.

Question 10. Find the angle between the vector \( 2\hat{i} + \hat{j} - \hat{k} \) and \( \hat{i} + 2\hat{j} + \hat{k} \) using vector product.
Answer: Let the two given vectors be:
\( \vec{a} = 2\hat{i} + \hat{j} - \hat{k} \)
\( \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \)

We use the formula involving the vector product: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \), where \( \theta \) is the angle between the vectors.

First, calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i} ((1)(1) - (-1)(2)) - \hat{j} ((2)(1) - (-1)(1)) + \hat{k} ((2)(2) - (1)(1)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (1 - (-2)) - \hat{j} (2 - (-1)) + \hat{k} (4 - 1) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (1 + 2) - \hat{j} (2 + 1) + \hat{k} (3) \)
\( \implies \vec{a} \times \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} \)

Next, find the magnitude of \( \vec{a} \times \vec{b} \):
\( |\vec{a} \times \vec{b}| = \sqrt{(3)^2 + (-3)^2 + (3)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{9 + 9 + 9} = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \)

Now, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}| = \sqrt{(2)^2 + (1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)
\( |\vec{b}| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \)

Using the formula \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \):
\( 3\sqrt{3} = (\sqrt{6})(\sqrt{6}) \sin \theta \)
\( \implies 3\sqrt{3} = 6 \sin \theta \)
\( \implies \sin \theta = \frac{3\sqrt{3}}{6} \)
\( \implies \sin \theta = \frac{\sqrt{3}}{2} \)

For \( \sin \theta = \frac{\sqrt{3}}{2} \), the angle \( \theta \) is \( \frac{\pi}{3} \) (or 60 degrees). This method effectively uses the area of the parallelogram formed by the vectors to determine the angle between them.
In simple words: First, calculate the cross product of the two vectors. Then, find the length of this cross product vector. Also, find the length of each of the original two vectors. Put these lengths into the formula for the sine of the angle (length of cross product = length of vector A times length of vector B times sine of the angle). Solve for the sine of the angle, then find the angle itself.

🎯 Exam Tip: Remember the vector product formula: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \). This is crucial for finding the angle between two vectors when using their cross product.

Question 10. Find the angle between the vector \( 2\hat{i} + \hat{j} - \hat{k} \) and \( \hat{i} + 2\hat{j} + \hat{k} \) using vector product.
Answer:
Let the given vectors be \( \vec{a} = 2\hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \).
First, we find the cross product of these two vectors:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} \)
\( \implies \hat{i}((1)(1) - (-1)(2)) - \hat{j}((2)(1) - (-1)(1)) + \hat{k}((2)(2) - (1)(1)) \)
\( \implies \hat{i}(1 + 2) - \hat{j}(2 + 1) + \hat{k}(4 - 1) \)
\( \implies 3\hat{i} - 3\hat{j} + 3\hat{k} \)
Next, we calculate the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = |3\hat{i} - 3\hat{j} + 3\hat{k}| \)
\( = \sqrt{3^2 + (-3)^2 + 3^2} \)
\( = \sqrt{9 + 9 + 9} \)
\( = \sqrt{27} \)
\( = 3\sqrt{3} \)
Now, we find the magnitudes of the individual vectors:
\( |\vec{a}| = |2\hat{i} + \hat{j} - \hat{k}| \)
\( = \sqrt{2^2 + 1^2 + (-1)^2} \)
\( = \sqrt{4 + 1 + 1} \)
\( = \sqrt{6} \)
\( |\vec{b}| = |\hat{i} + 2\hat{j} + \hat{k}| \)
\( = \sqrt{1^2 + 2^2 + 1^2} \)
\( = \sqrt{1 + 4 + 1} \)
\( = \sqrt{6} \)
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). We use the formula for the sine of the angle:
\( \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} \)
\( = \frac{3\sqrt{3}}{\sqrt{6} \cdot \sqrt{6}} \)
\( = \frac{3\sqrt{3}}{6} \)
\( = \frac{\sqrt{3}}{2} \)
\( \implies \theta = \frac{\pi}{3} \)
This method helps find the angle even when the dot product method (which uses cosine) might be less direct for certain calculations.
In simple words: We find the cross product of the two given vectors and then calculate its length. We also find the lengths of the original vectors. Using these lengths in a special formula involving sine, we can figure out the angle between the two vectors.

🎯 Exam Tip: When using the vector product to find the angle, remember that the sine of the angle will be positive, as \( \theta \) is typically taken as the acute angle between vectors (\(0 \le \theta \le \pi\)).