CBSE Class 12 Mathematics Vector Algebra Class Test Set 01

Class Test Questions with Solutions

Question 1. Find a vector \( \vec{r} \) of magnitude \( 3\sqrt{2} \) units which makes an angle of \( \frac{\pi}{4} \) and \( \frac{\pi}{2} \) with y and z-axes, respectively. [1]
Answer:
Here \( m = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( n = \cos\frac{\pi}{2} = 0 \)
Therefore, \( l^2 + m^2 + n^2 = 1 \) gives
\( l^2 + \frac{1}{2} + 0 = 1 \Rightarrow l = \pm\frac{1}{\sqrt{2}} \)
Hence, the required vector \( \vec{r} = 3\sqrt{2}(l\hat{i} + m\hat{j} + n\hat{k}) \) is given by
\( \vec{r} = 3\sqrt{2}\left(\pm\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}\right) \Rightarrow \vec{r} = \pm 3\hat{i} + 3\hat{j} \)

Question 2. Let the vectors \( \vec{a} \) and \( \vec{b} \) be such that \( |\vec{a}| = 3 \) and \( |\vec{b}| = \frac{\sqrt{2}}{3} \), then \( \vec{a} \times \vec{b} \) is a unit vector. Find the angle between \( \vec{a} \) and \( \vec{b} \). [1]
Answer:
Given \( |\vec{a} \times \vec{b}| = 1 \)
⇒ \( |\vec{a}||\vec{b}|\sin\theta = 1 \)
or \( 3 \times \frac{\sqrt{2}}{3}\sin\theta = 1 \)
⇒ \( \sin\theta = \frac{1}{\sqrt{2}} \) or \( \theta = \frac{\pi}{4} \)

Question 3. Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors and θ is the angle between them. If \( \vec{a} + \vec{b} \) is a unit vector, then find θ. [1]
Answer:
Given \( |\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1 \)
⇒ \( |\vec{a} + \vec{b}| = 1 \)
or \( |\vec{a} + \vec{b}|^2 = 1 \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1 \) \( \{\because |\vec{a}|^2 = \vec{a} \cdot \vec{a}\} \)
⇒ \( |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 1 \) or \( 1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1 \)
⇒ \( 2|\vec{a}||\vec{b}|\cos\theta = -1 \) or \( 2 \times 1 \times 1\cos\theta = -1 \)
⇒ \( \cos\theta = \frac{-1}{2} \Rightarrow \theta = \frac{2\pi}{3} \)

Question 4. If θ is the angle between any two vectors \( \vec{a} \) and \( \vec{b} \), if \( |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \), then θ is? [1]
Answer:
\( |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \) (given)
⇒ \( |\vec{a}||\vec{b}|\cos\theta = |\vec{a}||\vec{b}|\sin\theta \)
⇒ tan θ = 1 ⇒ \( \theta = \frac{\pi}{4} \)

Question 5. Find \( |\vec{a}| \) and \( |\vec{b}| \), if \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8 \) and \( |\vec{a}| = 8|\vec{b}| \). [2]
Answer:
Given \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8 \) and \( |\vec{a}| = 8|\vec{b}| \)
⇒ \( \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8 \)
⇒ \( |\vec{a}|^2 - |\vec{b}|^2 = 8 \) \( \left\{\because \vec{a} \cdot \vec{a} = |\vec{a}|^2,\ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\right\} \)
or \( [8|\vec{b}|]^2 - |\vec{b}|^2 = 8 \)
⇒ \( 63|\vec{b}|^2 = 8 \)
or \( |\vec{b}|^2 = \frac{8}{63} \)
⇒ \( |\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}} \) (magnitude of a vector cannot be negative)
∴ \( |\vec{a}| = 8|\vec{b}| = \frac{16\sqrt{2}}{3\sqrt{7}} \)

Question 6. If a unit vector \( \vec{a} \) makes angles \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \) and an acute angle θ with \( \hat{k} \), then find θ and hence, the components of \( \vec{a} \). [2]
Answer:
The direction cosines of the vector \( \vec{a} \) are \( l = \cos\frac{\pi}{3} = \frac{1}{2} \), \( m = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and n = cos θ
∴ \( l^2 + m^2 + n^2 = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4} \Rightarrow n = \frac{1}{2} \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \)
Since \( \vec{a} \) is a unit vector,
∴ \( \vec{a} = l\hat{i} + m\hat{j} + n\hat{k} \)
⇒ \( \vec{a} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k} \)
Hence, the components of \( \vec{a} \) are \( \frac{1}{2}\hat{i}, \frac{1}{\sqrt{2}}\hat{j}, \frac{1}{2}\hat{k} \)

Question 7. Find the projection of the vector \( \hat{i} + 3\hat{j} + 7\hat{k} \) on the vector \( 7\hat{i} - \hat{j} + 8\hat{k} \). [2]
Answer:
Let \( \vec{a} = \hat{i} + 3\hat{j} + 7\hat{k} \) and \( \vec{b} = 7\hat{i} - \hat{j} + 8\hat{k} \)
Projection of \( \vec{a} \) on \( \vec{b} = \vec{a} \cdot \hat{b} \)
\( = (\hat{i} + 3\hat{j} + 7\hat{k}) \cdot \frac{7\hat{i} - \hat{j} + 8\hat{k}}{\sqrt{49 + 1 + 64}} = \frac{7 - 3 + 56}{\sqrt{114}} = \frac{60}{\sqrt{114}} \)

Question 8. If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \). [3]
Answer:
We have \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \) and \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \)
Now, \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \)
⇒ \( |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \) \( \left[\because |\vec{a}|^2 = \vec{a} \cdot \vec{a},\ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\right] \)
⇒ \( (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \)
⇒ \( \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0 \)
⇒ \( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \)
⇒ \( 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \) \( [\because |\vec{a}| = |\vec{b}| = |\vec{c}| = 1] \)
⇒ \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \)

Question 9. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{c} = \hat{i} - 2\hat{j} + \hat{k} \), find a unit vector parallel to the vector \( 2\vec{a} - \vec{b} + 3\vec{c} \). [3]
Answer:
Let \( \vec{d} = 2\vec{a} - \vec{b} + 3\vec{c} \)
\( = 2(\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k}) \)
\( = 3\hat{i} - 3\hat{j} + 2\hat{k} \)
⇒ \( |\vec{d}| = \sqrt{(3)^2 + (-3)^2 + (2)^2} = \sqrt{22} \)
Now, \( \hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{3\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{22}} \)

Question 10. Show that the points A, B, C with position vectors \( 2\hat{i} - \hat{j} + \hat{k} \), \( \hat{i} - 3\hat{j} - 5\hat{k} \) and \( 3\hat{i} - 4\hat{j} - 4\hat{k} \) respectively are the vertices of a right angled triangle. Hence find the area of the triangle. [3]
Answer:
\( \overrightarrow{AB} = -\hat{i} - 2\hat{j} - 6\hat{k} \) ⇒ \( |\overrightarrow{AB}|^2 = (1^2 + 2^2 + 6^2) = 41 \)
\( \overrightarrow{BC} = 2\hat{i} - \hat{j} + \hat{k} \) ⇒ \( |\overrightarrow{BC}|^2 = (2^2 + 1^2 + 1^2) = 6 \)
\( \overrightarrow{CA} = -\hat{i} + 3\hat{j} + 5\hat{k} \) ⇒ \( |\overrightarrow{CA}|^2 = (1^2 + 3^2 + 5^2) = 35 \)
∵ \( \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0} \)
and \( |\overrightarrow{AB}|^2 = |\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2 \)
Hence A, B, C are the vertices of a right-angled triangle.
Area of \( \Delta = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| \), where \( \overrightarrow{AC} = \hat{i} - 3\hat{j} - 5\hat{k} \)
\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & -6 \\ 1 & -3 & -5 \end{vmatrix} = (-8\hat{i} - 11\hat{j} + 5\hat{k}) \)
Now \( \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|-8\hat{i} - 11\hat{j} + 5\hat{k}| \)
\( = \frac{1}{2}\sqrt{64 + 121 + 25} = \frac{1}{2}\sqrt{210} \) square units

Question 11. The scalar product of the vector \( \hat{i} + \hat{j} + \hat{k} \) with the unit vector along the sum of vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to one. Find the value of λ. [3]
Answer:
Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \)
⇒ \( \vec{b} + \vec{c} = \hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k} = \vec{d} \) (let)
Now, unit vector along \( \vec{d} = \frac{\hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}} \)
∵ \( \vec{a} \cdot \hat{d} = 1 \)
⇒ \( (\hat{i} + \hat{j} + \hat{k}) \cdot \left(\frac{\hat{i}(2 + \lambda) + 6\hat{j} - 2\hat{k}}{\sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2}}\right) = 1 \)
or \( (2 + \lambda) + 6 - 2 = \sqrt{(2 + \lambda)^2 + 36 + 4} \)
⇒ \( 6 + \lambda = \sqrt{4 + 4\lambda + \lambda^2 + 40} \)
or \( 36 + \lambda^2 + 12\lambda = 44 + 4\lambda + \lambda^2 \)
⇒ 8λ = 8
or λ = 1

Question 12. Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices. [3]
Answer:
Area of triangle ABC = \( \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| \)

Now, \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 0\hat{i} + \hat{j} + 2\hat{k} \)
and \( \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \hat{i} + 2\hat{j} + 0\hat{k} \)
⇒ \( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) \)
\( = -4\hat{i} + 2\hat{j} - \hat{k} \)
∴ \( |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \)
⇒ Area of ΔABC = \( \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{\sqrt{21}}{2} \) square units