GSEB Class 12 Maths Solutions Chapter 10 Vector Algebra Exercise 10.4

Question 1. Find \( |\vec{a} \times \vec{b}| \), if \( \vec{a} = \hat{i} – 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k} \).
Answer:
Given vectors are \( \vec{a} = \hat{i} – 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k} \).
First, we calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = ((-7)(2) - (7)(-2))\hat{i} - ((1)(2) - (7)(3))\hat{j} + ((1)(-2) - (-7)(3))\hat{k} \)
\( \implies \vec{a} \times \vec{b} = (-14 + 14)\hat{i} - (2 - 21)\hat{j} + (-2 + 21)\hat{k} \)
\( \implies \vec{a} \times \vec{b} = 0\hat{i} - (-19)\hat{j} + 19\hat{k} \)
\( \implies \vec{a} \times \vec{b} = 19\hat{j} + 19\hat{k} \)
Next, we determine the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(0)^2 + (19)^2 + (19)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{19^2 + 19^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{2 \times 19^2} \)
\( \implies |\vec{a} \times \vec{b}| = 19\sqrt{2} \).
In simple words: We find the vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \) by doing a cross product. Then, we calculate the length of this new vector, which comes out to be \( 19\sqrt{2} \).

Exam Tip: Remember to correctly apply the determinant rule for cross products, paying close attention to the signs for each component, especially the negative sign for the \( \hat{j} \) component.

Question 2. Find a unit vector perpendicular to each of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} – \vec{b} \), where \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} – 2\hat{k} \).
Answer:
Given vectors are \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} – 2\hat{k} \).
First, calculate \( \vec{a} + \vec{b} \):
\( \vec{a} + \vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} – 2\hat{k}) \)
\( \implies \vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} \)
\( \implies \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k} = 4\hat{i} + 4\hat{j} \).
Next, calculate \( \vec{a} – \vec{b} \):
\( \vec{a} – \vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} – 2\hat{k}) \)
\( \implies \vec{a} – \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} \)
\( \implies \vec{a} – \vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k} = 2\hat{i} + 4\hat{k} \).
Now, find the cross product of these two resultant vectors, \( (\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) \):
\[ (\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \]
\( \implies (\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) = ((4)(4) - (0)(0))\hat{i} - ((4)(4) - (0)(2))\hat{j} + ((4)(0) - (4)(2))\hat{k} \)
\( \implies (\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) = (16-0)\hat{i} - (16-0)\hat{j} + (0-8)\hat{k} \)
\( \implies (\vec{a} + \vec{b}) \times (\vec{a} – \vec{b}) = 16\hat{i} - 16\hat{j} - 8\hat{k} \).
A unit vector perpendicular to both \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \) is found by dividing this cross product by its magnitude:
\( \hat{n} = \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} – \vec{b})}{|(\vec{a} + \vec{b}) \times (\vec{a} – \vec{b})|} \)
First, calculate the magnitude:
\( |16\hat{i} - 16\hat{j} - 8\hat{k}| = \sqrt{(16)^2 + (-16)^2 + (-8)^2} \)
\( \implies \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \).
So, the unit vector is:
\( \hat{n} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} \)
\( \implies \hat{n} = \pm \frac{8(2\hat{i} - 2\hat{j} - \hat{k})}{24} \)
\( \implies \hat{n} = \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{3} \).
In simple words: First, sum and subtract the given vectors. Then, calculate the cross product of these new vectors to find a vector that is perpendicular to both. Finally, divide this perpendicular vector by its own length to get a unit vector.

Exam Tip: Always remember that a unit vector can point in two opposite directions, so include the \( \pm \) sign in your final answer for a unit vector perpendicular to two given vectors.

Question 3. If a unit vector \( \vec{a} \) makes angle \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \) and an acute angle \( \theta \) with \( \hat{k} \), then find \( \theta \) and hence the components of \( \vec{a} \).
Answer:
Let the unit vector be \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \).
Since \( \vec{a} \) is a unit vector, its magnitude is 1:
\( |a_1\hat{i} + a_2\hat{j} + a_3\hat{k}| = 1 \).
The angles made by \( \vec{a} \) with the coordinate axes are given as:
Angle with \( \hat{i} \) (x-axis) is \( \alpha = \frac{\pi}{3} \).
Angle with \( \hat{j} \) (y-axis) is \( \beta = \frac{\pi}{4} \).
Angle with \( \hat{k} \) (z-axis) is \( \theta \) (acute angle).
We know that the direction cosines are \( \cos \alpha = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}| |\hat{i}|} = a_1 \), \( \cos \beta = a_2 \), \( \cos \gamma = a_3 \).
For \( \hat{i} \): \( \vec{a} \cdot \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{i} = a_1 \).
Also, \( \vec{a} \cdot \hat{i} = |\vec{a}| |\hat{i}| \cos(\frac{\pi}{3}) \).
Since \( |\vec{a}| = 1 \) and \( |\hat{i}| = 1 \), we have:
\( a_1 = (1)(1) \cos(\frac{\pi}{3}) \)
\( \implies a_1 = \frac{1}{2} \).
For \( \hat{j} \): \( \vec{a} \cdot \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot \hat{j} = a_2 \).
Also, \( \vec{a} \cdot \hat{j} = |\vec{a}| |\hat{j}| \cos(\frac{\pi}{4}) \).
\( \implies a_2 = (1)(1) \cos(\frac{\pi}{4}) \)
\( \implies a_2 = \frac{1}{\sqrt{2}} \).
Since \( \vec{a} \) is a unit vector, the sum of the squares of its components must be 1:
\( a_1^2 + a_2^2 + a_3^2 = 1 \)
Substitute the values of \( a_1 \) and \( a_2 \):
\( (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + a_3^2 = 1 \)
\( \implies \frac{1}{4} + \frac{1}{2} + a_3^2 = 1 \)
\( \implies \frac{1+2}{4} + a_3^2 = 1 \)
\( \implies \frac{3}{4} + a_3^2 = 1 \)
\( \implies a_3^2 = 1 - \frac{3}{4} \)
\( \implies a_3^2 = \frac{1}{4} \)
\( \implies a_3 = \pm \frac{1}{2} \).
Since \( \theta \) is an acute angle with \( \hat{k} \), \( \cos \theta \) must be positive, so \( a_3 = \frac{1}{2} \).
Now we find \( \theta \). We know \( a_3 = \cos \theta \).
\( \cos \theta = \frac{1}{2} \).
Since \( \theta \) is acute, \( \theta = \frac{\pi}{3} \).
Thus, the components of \( \vec{a} \) are \( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \).
In simple words: We know the angles a unit vector makes with two axes. We use these angles to find the first two parts (components) of the vector. Because it's a unit vector, the sum of the squares of its parts must equal one, which helps us find the third part. The angle with the third axis is then determined from this third part.

Exam Tip: Remember the identity \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) for direction cosines of any vector. This is a quick way to find the third angle or component.

Question 4. Show that \( (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b}) \).
Answer:
We start by expanding the Left Hand Side (L.H.S.) of the given equation:
L.H.S. \( = (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) \)
Using the distributive property of the cross product:
\( \implies \vec{a} \times (\vec{a} + \vec{b}) - \vec{b} \times (\vec{a} + \vec{b}) \)
\( \implies (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b}) \).
We know that the cross product of a vector with itself is the zero vector, i.e., \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \).
Also, we know that \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \).
Substitute these properties into our expanded expression:
\( \implies \vec{0} + (\vec{a} \times \vec{b}) - (-(\vec{a} \times \vec{b})) - \vec{0} \)
\( \implies (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) \)
\( \implies 2(\vec{a} \times \vec{b}) \).
This matches the Right Hand Side (R.H.S.) of the equation.
Therefore, \( (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b}) \) is shown to be true.
In simple words: We expand the cross product on the left side, using the rule that a vector crossed with itself is zero, and that changing the order of vectors in a cross product changes its sign. After simplifying, both sides of the equation become equal, proving the identity.

Exam Tip: Always remember the fundamental properties of the cross product: \( \vec{v} \times \vec{v} = \vec{0} \) and \( \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) \). These are crucial for simplifying vector identities.

Question 5. Find \( \lambda \) and \( \mu \), if \( (2\hat{i} + 6\hat{j} + 27\hat{k} ) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k} ) = \vec{0} \).
Answer:
Given that the cross product of two vectors is the zero vector, this implies that the two vectors are parallel (collinear).
The cross product is given as:
\[ (2\hat{i} + 6\hat{j} + 27\hat{k} ) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k} ) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} \]
Since this cross product is \( \vec{0} \), each component of the resulting vector must be zero.
\( \implies (6\mu - 27\lambda)\hat{i} - (2\mu - 27)\hat{j} + (2\lambda - 6)\hat{k} = \vec{0} \).
Comparing the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) to zero:
1. \( 6\mu - 27\lambda = 0 \)
2. \( -(2\mu - 27) = 0 \implies 2\mu - 27 = 0 \implies 2\mu = 27 \implies \mu = \frac{27}{2} \).
3. \( 2\lambda - 6 = 0 \implies 2\lambda = 6 \implies \lambda = 3 \).
Now, we check if these values satisfy the first equation:
Substitute \( \lambda = 3 \) and \( \mu = \frac{27}{2} \) into \( 6\mu - 27\lambda \):
\( 6(\frac{27}{2}) - 27(3) = 3(27) - 81 = 81 - 81 = 0 \).
The values satisfy all three equations.
Therefore, \( \lambda = 3 \) and \( \mu = \frac{27}{2} \).
In simple words: When the cross product of two vectors is zero, it means they are parallel. We set up the cross product calculation and make each resulting component equal to zero. Solving these simple equations helps us find the unknown values of \( \lambda \) and \( \mu \).

Exam Tip: If two vectors are collinear (parallel), their cross product is the zero vector. This means their corresponding components are proportional, which can sometimes provide an alternative way to solve for unknowns like \( \lambda \) and \( \mu \).

Question 6. Given that \( \vec{a}.\vec{b} = 0 \) and \( \vec{a} \times \vec{b} = \vec{0} \). What can you conclude about the vectors a and b?
Answer:
We are given two conditions:
1. \( \vec{a}.\vec{b} = 0 \)
This means that either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \) or \( \vec{a} \) is perpendicular to \( \vec{b} \) (i.e., the angle between them is \( \frac{\pi}{2} \)).
2. \( \vec{a} \times \vec{b} = \vec{0} \)
This means that either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \) or \( \vec{a} \) is parallel to \( \vec{b} \) (i.e., the angle between them is 0 or \( \pi \)).
If \( \vec{a} \) and \( \vec{b} \) are non-zero vectors, they cannot be both perpendicular and parallel at the same time. The only way for both conditions to hold simultaneously is if at least one of the vectors is the zero vector.
If \( \vec{a} = \vec{0} \), then \( \vec{a}.\vec{b} = \vec{0}.\vec{b} = 0 \) and \( \vec{a} \times \vec{b} = \vec{0} \times \vec{b} = \vec{0} \). These conditions are satisfied.
If \( \vec{b} = \vec{0} \), then \( \vec{a}.\vec{b} = \vec{a}.\vec{0} = 0 \) and \( \vec{a} \times \vec{b} = \vec{a} \times \vec{0} = \vec{0} \). These conditions are also satisfied.
Therefore, if both \( \vec{a}.\vec{b} = 0 \) and \( \vec{a} \times \vec{b} = \vec{0} \) are true, we can conclude that either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \) (or both).
In simple words: The first condition tells us the vectors are either perpendicular or one of them is zero. The second condition tells us they are either parallel or one of them is zero. Since vectors cannot be both perpendicular and parallel at the same time, the only way both conditions can be true is if one of the vectors is the zero vector.

Exam Tip: Understand the implications of the dot product and cross product being zero. \( \vec{a}.\vec{b} = 0 \) means perpendicular or a zero vector, while \( \vec{a} \times \vec{b} = \vec{0} \) means parallel or a zero vector.

Question 7. Let the vectors \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are given as \( a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), \( b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) and \( c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \). Thus, show that \( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
Answer:
To prove the distributive property of the cross product, we will evaluate both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) separately.
First, calculate \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) + (c_1\hat{i} + c_2\hat{j} + c_3\hat{k}) \)
\( \implies \vec{b} + \vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k} \).
Now, calculate the L.H.S. \( = \vec{a} \times (\vec{b} + \vec{c}) \):
\[ \vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ (b_1+c_1) & (b_2+c_2) & (b_3+c_3) \end{vmatrix} \]
\( \implies \hat{i}[a_2(b_3+c_3) - a_3(b_2+c_2)] - \hat{j}[a_1(b_3+c_3) - a_3(b_1+c_1)] + \hat{k}[a_1(b_2+c_2) - a_2(b_1+c_1)] \)
\( \implies \hat{i}[(a_2b_3-a_3b_2) + (a_2c_3-a_3c_2)] - \hat{j}[(a_1b_3-a_3b_1) + (a_1c_3-a_3c_1)] + \hat{k}[(a_1b_2-a_2b_1) + (a_1c_2-a_2c_1)] \) ..... (1)
Next, calculate the R.H.S. \( = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
First, \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = (a_2b_3-a_3b_2)\hat{i} - (a_1b_3-a_3b_1)\hat{j} + (a_1b_2-a_2b_1)\hat{k} \).
Second, \( \vec{a} \times \vec{c} \):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{c} = (a_2c_3-a_3c_2)\hat{i} - (a_1c_3-a_3c_1)\hat{j} + (a_1c_2-a_2c_1)\hat{k} \).
Now, add \( \vec{a} \times \vec{b} \) and \( \vec{a} \times \vec{c} \):
R.H.S. \( = [(a_2b_3-a_3b_2) + (a_2c_3-a_3c_2)]\hat{i} - [(a_1b_3-a_3b_1) + (a_1c_3-a_3c_1)]\hat{j} + [(a_1b_2-a_2b_1) + (a_1c_2-a_2c_1)]\hat{k} \) ..... (2)
Comparing (1) and (2), we observe that the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) are identical.
Thus, L.H.S. = R.H.S.
Therefore, it is shown that \( \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
In simple words: To show that vector cross product distributes over addition, we calculate both sides of the equation. First, we add vectors \( \vec{b} \) and \( \vec{c} \), then cross the result with \( \vec{a} \). Separately, we cross \( \vec{a} \) with \( \vec{b} \) and \( \vec{a} \) with \( \vec{c} \), and then add those two results. By carefully expanding the determinants for each cross product, we see that both final expressions are exactly the same, proving the property.

Exam Tip: When showing vector identities, a clear, step-by-step expansion of both sides of the equation, comparing the final expressions, is usually the most straightforward and error-proof method.

Question 8. If either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b} = \vec{0} \). Is the converse true? Justify your answer with an example.
Answer:
Part 1: If either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b} = \vec{0} \).
This statement is true.
If \( \vec{a} = \vec{0} \), then \( \vec{a} \times \vec{b} = \vec{0} \times \vec{b} = \vec{0} \).
If \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b} = \vec{a} \times \vec{0} = \vec{0} \).
Part 2: Is the converse true?
The converse statement is: If \( \vec{a} \times \vec{b} = \vec{0} \), then either \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \).
This statement is false.
Justification with an example:
We know that \( \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \hat{n} \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), and \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
If \( \vec{a} \times \vec{b} = \vec{0} \), it means that \( |\vec{a}||\vec{b}|\sin\theta = 0 \).
This can happen in three cases:
1. \( |\vec{a}| = 0 \implies \vec{a} = \vec{0} \)
2. \( |\vec{b}| = 0 \implies \vec{b} = \vec{0} \)
3. \( \sin\theta = 0 \)
If \( \sin\theta = 0 \), it implies that \( \theta = 0 \) or \( \theta = \pi \). In this case, the vectors \( \vec{a} \) and \( \vec{b} \) are parallel or collinear.
Example: Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k} \).
Here, \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \).
However, \( \vec{b} = 2\vec{a} \), meaning \( \vec{a} \) and \( \vec{b} \) are parallel.
Their cross product is:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \]
\( \implies \hat{i}(1 \times 2 - 1 \times 2) - \hat{j}(1 \times 2 - 1 \times 2) + \hat{k}(1 \times 2 - 1 \times 2) \)
\( \implies \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0} \).
In this example, \( \vec{a} \times \vec{b} = \vec{0} \), but neither \( \vec{a} \) nor \( \vec{b} \) is the zero vector. This happens because they are parallel.
Therefore, the converse is not true.
In simple words: The statement "if either vector is zero, their cross product is zero" is correct. But the opposite, "if their cross product is zero, then one of them must be zero", is not always true. This is because if two non-zero vectors are parallel, their cross product is also zero.

Exam Tip: When evaluating converse statements, remember to consider all conditions under which the original statement's conclusion can be true. For cross products, parallelism is a key alternative to a zero vector for a zero result.

Question 9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 3).
Answer:
The position vectors of the vertices of \( \triangle ABC \) are given as:
\( \vec{A} = \hat{i} + \hat{j} + 2\hat{k} \)
\( \vec{B} = 2\hat{i} + 3\hat{j} + 5\hat{k} \)
\( \vec{C} = \hat{i} + 5\hat{j} + 3\hat{k} \)
We need to find two adjacent side vectors, for example, \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
\( \overrightarrow{AB} = \vec{B} - \vec{A} = (2\hat{i} + 3\hat{j} + 5\hat{k}) - (\hat{i} + \hat{j} + 2\hat{k}) \)
\( \implies \overrightarrow{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} \)
\( \implies \overrightarrow{AB} = \hat{i} + 2\hat{j} + 3\hat{k} \).
\( \overrightarrow{AC} = \vec{C} - \vec{A} = (\hat{i} + 5\hat{j} + 3\hat{k}) - (\hat{i} + \hat{j} + 2\hat{k}) \)
\( \implies \overrightarrow{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (3-2)\hat{k} \)
\( \implies \overrightarrow{AC} = 0\hat{i} + 4\hat{j} + 1\hat{k} = 4\hat{j} + \hat{k} \).
Now, we calculate the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 1 \end{vmatrix} \]
\( \implies \hat{i}(2 \times 1 - 3 \times 4) - \hat{j}(1 \times 1 - 3 \times 0) + \hat{k}(1 \times 4 - 2 \times 0) \)
\( \implies \hat{i}(2 - 12) - \hat{j}(1 - 0) + \hat{k}(4 - 0) \)
\( \implies -10\hat{i} - \hat{j} + 4\hat{k} \).
Next, we find the magnitude of this cross product:
\( |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-10)^2 + (-1)^2 + (4)^2} \)
\( \implies \sqrt{100 + 1 + 16} = \sqrt{117} \).
The area of the triangle ABC is half the magnitude of the cross product of two adjacent sides:
Area of \( \triangle ABC = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| \)
\( \implies \text{Area} = \frac{1}{2} \sqrt{117} \) square units.
In simple words: To find the area of a triangle given its vertices, we first calculate two vectors representing two sides starting from the same vertex. Then, we find the cross product of these two vectors. The area of the triangle is half the length (magnitude) of this cross product vector.

Exam Tip: Remember that the area of a triangle with adjacent sides \( \vec{u} \) and \( \vec{v} \) is given by \( \frac{1}{2}|\vec{u} \times \vec{v}| \). Ensure correct calculation of the determinant for the cross product and the magnitude of the resulting vector.

Question 10. Find the area of parallelogram whose adjacent sides are determined by the vectors \( \vec{a} = \hat{i} – \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} – 7\hat{j} + \hat{k} \).
Answer:
Given the adjacent sides of a parallelogram as vectors \( \vec{a} = \hat{i} – \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} – 7\hat{j} + \hat{k} \).
The area of a parallelogram is given by the magnitude of the cross product of its adjacent side vectors.
First, calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} \]
\( \implies \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2)) \)
\( \implies \hat{i}(-1 - (-21)) - \hat{j}(1 - 6) + \hat{k}(-7 - (-2)) \)
\( \implies \hat{i}(-1 + 21) - \hat{j}(-5) + \hat{k}(-7 + 2) \)
\( \implies 20\hat{i} + 5\hat{j} - 5\hat{k} \).
Next, find the magnitude of this cross product:
Area of parallelogram \( = |\vec{a} \times \vec{b}| = |20\hat{i} + 5\hat{j} - 5\hat{k}| \)
\( \implies \sqrt{(20)^2 + (5)^2 + (-5)^2} \)
\( \implies \sqrt{400 + 25 + 25} \)
\( \implies \sqrt{450} \).
To simplify \( \sqrt{450} \):
\( \sqrt{450} = \sqrt{225 \times 2} = \sqrt{225} \times \sqrt{2} = 15\sqrt{2} \).
Thus, the area of the parallelogram is \( 15\sqrt{2} \) square units.
In simple words: We find the area of a parallelogram by taking the cross product of the two vectors that make up its adjacent sides. Then, we calculate the length of this resulting vector, which gives us the area.

Exam Tip: For the area of a parallelogram, remember that it's simply the magnitude of the cross product of its adjacent sides, unlike a triangle which requires half of this value.

Question 11. Let the vectors \( \vec{a} \) and \( \vec{b} \) are such that \( |\vec{a}| = 3 \) and \( |\vec{b}| = \frac{\sqrt{2}}{3} \), then \( \vec{a} \times \vec{b} \) is a unit vector, if the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{2} \)
Answer: (b) \( \frac{\pi}{4} \)
We are given:
\( |\vec{a}| = 3 \)
\( |\vec{b}| = \frac{\sqrt{2}}{3} \)
\( \vec{a} \times \vec{b} \) is a unit vector, which means \( |\vec{a} \times \vec{b}| = 1 \).
The formula for the magnitude of the cross product is:
\( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \).
Substitute the given values into the formula:
\( 1 = (3)(\frac{\sqrt{2}}{3})\sin\theta \)
\( \implies 1 = \sqrt{2}\sin\theta \)
\( \implies \sin\theta = \frac{1}{\sqrt{2}} \).
For \( \sin\theta = \frac{1}{\sqrt{2}} \), the angle \( \theta \) is \( \frac{\pi}{4} \) (or 45 degrees) within the usual range for angles between vectors (0 to \( \pi \)).
Thus, the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \).
In simple words: We use the formula that connects the length of the cross product, the lengths of the individual vectors, and the sine of the angle between them. By plugging in the given lengths and knowing the cross product has a length of 1 (because it's a unit vector), we can easily find the sine of the angle, and then the angle itself.

Exam Tip: Remember that a "unit vector" simply means its magnitude is 1. Always be careful with trigonometric values for standard angles.

Question 12. Area of rectangle having vertices A, B, C and D with position vectors \( (-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} ) \), \( (\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} ) \), \( (\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}) \) and \( (-\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}) \) is
(a) 1
(b) 3
(c) 2
(d) 4
Answer: (c) 2
Let the position vectors of the vertices be:
\( \vec{A} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \vec{B} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \vec{C} = \hat{i} – \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \vec{D} = -\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k} \)
To find the area of the rectangle ABCD, we can find the lengths of two adjacent sides, for example, \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \), and multiply them.
First, calculate \( \overrightarrow{AB} = \vec{B} - \vec{A} \):
\( \overrightarrow{AB} = (\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) - (-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) \)
\( \implies \overrightarrow{AB} = (1 - (-1))\hat{i} + (\frac{1}{2} - \frac{1}{2})\hat{j} + (4-4)\hat{k} \)
\( \implies \overrightarrow{AB} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \).
The magnitude of \( \overrightarrow{AB} \) is \( |\overrightarrow{AB}| = |2\hat{i}| = 2 \).
Next, calculate \( \overrightarrow{AD} = \vec{D} - \vec{A} \):
\( \overrightarrow{AD} = (-\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}) - (-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) \)
\( \implies \overrightarrow{AD} = (-1 - (-1))\hat{i} + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (4-4)\hat{k} \)
\( \implies \overrightarrow{AD} = ( -1 + 1)\hat{i} + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (0)\hat{k} \)
\( \implies \overrightarrow{AD} = 0\hat{i} - 1\hat{j} + 0\hat{k} = -\hat{j} \).
The magnitude of \( \overrightarrow{AD} \) is \( |\overrightarrow{AD}| = |-\hat{j}| = 1 \).
The area of the rectangle ABCD is the product of the lengths of its adjacent sides:
Area \( = |\overrightarrow{AB}| \times |\overrightarrow{AD}| \)
Area \( = 2 \times 1 = 2 \).
So, the area of the rectangle is 2 square units.
In simple words: To find the area of a rectangle from its corner points, we first find the vectors for two sides that meet at a corner. Then, we calculate the length of each of these side vectors. Finally, we multiply these two lengths together to get the total area of the rectangle.

Exam Tip: For rectangles, adjacent sides are perpendicular, meaning their dot product is zero. The area is simply the product of the magnitudes of these adjacent side vectors. This is usually simpler than using the cross product magnitude for a parallelogram.