CBSE Class 12 Mathematics Vector Algebra Important Questions Set B

Question. If \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \), then show that \( (\vec{a} - \vec{d}) \) is parallel to \( (\vec{b} - \vec{c}) \), it is being given that \( \vec{a} \neq \vec{d} \) and \( \vec{b} \neq \vec{c} \). 
Answer: Given, \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \)
\( \Rightarrow \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = \vec{c} \times \vec{d} - \vec{b} \times \vec{d} \)
\( \Rightarrow \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d} = 0 \)
\( \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) + (\vec{b} - \vec{c}) \times \vec{d} = 0 \) [By left and right distributive law]
\( \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = 0 \) [\( \because \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \)]
\( \Rightarrow (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = 0 \) [By right distributive law]
\( \Rightarrow (\vec{a} - \vec{d}) \parallel (\vec{b} - \vec{c}) \)

Question. Prove that : \( |\vec{a} \times \vec{b}|^2 = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix} \)
Answer: Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Then,
LHS \( = |\vec{a} \times \vec{b}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) \)
\( = (ab \sin \theta) \hat{n} \cdot (ab \sin \theta) \hat{n} = (a^2 b^2 \sin^2 \theta)(\hat{n} \cdot \hat{n}) = a^2 b^2 \sin^2 \theta \)
\( = a^2 b^2 (1 - \cos^2 \theta) = a^2 b^2 - (ab \cos \theta)^2 \)
\( = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})^2 \) ... (i)
Also, RHS \( = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} \end{vmatrix} = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{b}) \)
\( = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})^2 \) ... (ii)
From (i) and (ii) RHS = LHS Hence proved.

Question. If \( \vec{a}, \vec{b} \) are unit vectors such that the vector \( \vec{a} + 3\vec{b} \) is perpendicular to \( 7\vec{a} - 5\vec{b} \) and \( \vec{a} - 4\vec{b} \) is perpendicular to \( 7\vec{a} - 2\vec{b} \), then find the angle between \( \vec{a} \) and \( \vec{b} \). 
Answer: Let angle between \( \vec{a} \) and \( \vec{b} \) be \( \theta \)
Given, \( (\vec{a} + 3\vec{b}) \perp (7\vec{a} - 5\vec{b}) \Rightarrow (\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0 \)
\( \Rightarrow 7|\vec{a}|^2 + 16(\vec{a} \cdot \vec{b}) - 15|\vec{b}|^2 = 0 \)
\( \Rightarrow 7 + 16 \cos \theta - 15 = 0 \) [\( \because |\vec{a}|^2 = |\vec{b}|^2 = 1 \)]
\( \Rightarrow \cos \theta = \frac{8}{16} = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \)
Also, given that \( (\vec{a} - 4\vec{b}) \perp (7\vec{a} - 2\vec{b}) \)
\( \Rightarrow (\vec{a} - 4\vec{b}) \cdot (7\vec{a} - 2\vec{b}) = 0 \Rightarrow 7|\vec{a}|^2 + 8|\vec{b}|^2 - 30(\vec{a} \cdot \vec{b}) = 0 \)
\( \Rightarrow 15 - 30 \cos \theta = 0 \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \)

Question. Find the value of \( x \) such that the four points with position vectors, \( A(3\hat{i} + 2\hat{j} + \hat{k}), B(4\hat{i} + x\hat{j} + 5\hat{k}), C(4\hat{i} + 2\hat{j} - 2\hat{k}) \) and \( D(6\hat{i} + 5\hat{j} - \hat{k}) \) are coplanar. 
Answer: Given, \( \vec{AB} = \hat{i} + (x - 2)\hat{j} + 4\hat{k}, \vec{AC} = \hat{i} + 0\hat{j} - 3\hat{k}, \vec{AD} = 3\hat{i} + 3\hat{j} - 2\hat{k} \)
\( \Rightarrow \) As \( A, B, C, D \) are coplanar so \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0 \)
\( \Rightarrow \begin{vmatrix} 1 & x-2 & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{vmatrix} = 0 \Rightarrow 1(9) - (x - 2)7 + 4(3) = 0 \Rightarrow 9 - 7x + 14 + 12 = 0 \)
\( \Rightarrow 35 = 7x \Rightarrow x = 5 \)

Objective Type Questions: 

Choose and write the correct option in each of the following questions.

Question. If \( \vec{a} \cdot \vec{b} = \frac{1}{2} |\vec{a}| |\vec{b}| \), then the angle between \( \vec{a} \) and \( \vec{b} \) is 
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer: (c)

Question. Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors and \( \theta \) is the angle between them. Then \( \vec{a} + \vec{b} \) is unit vector if \( \theta \) is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{2\pi}{3} \)
Answer: (d)

Question. The magnitude of the vector \( 6\hat{i} + 2\hat{j} + 3\hat{k} \) is 
(a) 5
(b) 7
(c) 12
(d) 1
Answer: (b)

Question. Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \). If \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b} = |\vec{b}|^2 \) and \( |\vec{a} - \vec{b}| = \sqrt{7} \) then \( |\vec{b}| \) equals
(a) 7
(b) 14
(c) \( \sqrt{7} \)
(d) 21
Answer: (c)

Question. If \( |\vec{a} \times \vec{b}| = 4 \) and \( \vec{a} \cdot \vec{b} = 2 \) then \( |\vec{a}|^2 |\vec{b}|^2 \) is equal to
(a) 2
(b) 6
(c) 8
(d) 20
Answer: (d)

Question. The value of \( p \) for which \( p(\hat{i} + \hat{j} + \hat{k}) \) is a unit vector is 
(a) 0
(b) \( \frac{1}{\sqrt{3}} \)
(c) 1
(d) \( \sqrt{3} \)
Answer: (b)

Fill in the blanks. 

Question. The area of the parallelogram whose diagonals are \( 2\hat{i} \) and \( -3\hat{k} \) is ___________ square units. 
Answer: 3 sq. units

Question. The sine of the angle between vectors \( \vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k} \) and \( \vec{b} = 4\hat{i} + 3\hat{j} - \hat{k} \) is equal to _________.
Answer: \( \frac{5}{\sqrt{26}} \)

Question. The value of \( \lambda \) for which the vectors \( 2\hat{i} - \lambda\hat{j} + \hat{k} \) and \( \hat{i} + 2\hat{j} - \hat{k} \) are orthogonal is ____________. 
Answer: \( \lambda = \frac{1}{2} \)

Question. If \( \vec{a} = 3\hat{i} - 2\hat{j} + 2\hat{k}, \vec{b} = 6\hat{i} + 4\hat{j} - 2\hat{k} \) and \( \vec{c} = -3\hat{i} - 2\hat{j} + 4\hat{k} \). Then \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is equal to _________.
Answer: 72

Question. The vectors \( \vec{a} = 3\hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{b} = -\hat{i} - \hat{k} \) are the adjacent sides of a parallelogram. The acute angle between its diagonals is _____________ .
Answer: \( \frac{\pi}{4} \)

Very Short Answer Questions:

Question. If \( |\vec{a}| = \sqrt{3} \), \( |\vec{b}| = 2 \) and angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \), then find \( \vec{a} \cdot \vec{b} \). 
Answer: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta = \sqrt{3} \cdot 2 \cdot \cos 60^\circ = \sqrt{3} \cdot 2 \cdot \frac{1}{2} = \sqrt{3}\)

Question. Find the sum of the vectors \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k} \), \( \vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k} \) and \( \vec{c} = \hat{i} - 6\hat{j} - 7\hat{k} \). 
Answer: \( \vec{a} + \vec{b} + \vec{c} = (1 - 2 + 1)\hat{i} + (-2 + 4 - 6)\hat{j} + (1 + 5 - 7)\hat{k} = 0\hat{i} - 4\hat{j} - \hat{k} = -4\hat{j} - \hat{k} \)

Question. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes 1 and 2 respectively and when \( |\vec{a} \times \vec{b}| = \sqrt{3} \). 
Answer: \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta \Rightarrow \sqrt{3} = 1 \cdot 2 \cdot \sin \theta \Rightarrow \sin \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ \)

Question. Find a vector of magnitude \( \sqrt{171} \), which is perpendicular to both of the vectors \( \vec{a} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \).
Answer: \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 3 & -1 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(2+9) + \hat{k}(-1-6) = \hat{i} - 11\hat{j} - 7\hat{k} \). Magnitude \( = \sqrt{1^2 + (-11)^2 + (-7)^2} = \sqrt{171} \). Required vector is \( \hat{i} - 11\hat{j} - 7\hat{k} \).

Question. Write the distance of the point (3, –5, 12) from X-axis. 
Answer: Distance from X-axis \( = \sqrt{y^2 + z^2} = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = 13 \)

Question. If \( \vec{a} \cdot \vec{a} = 0 \) and \( \vec{a} \cdot \vec{b} = 0 \), then what can be concluded about the vector \( \vec{b} \)? 
Answer: Since \( \vec{a} \cdot \vec{a} = 0 \Rightarrow \vec{a} = \vec{0} \). Thus, \( \vec{a} \cdot \vec{b} = \vec{0} \cdot \vec{b} = 0 \) is true for any vector \( \vec{b} \). Hence, \( \vec{b} \) can be any vector.

Question. If \( \vec{a} = 4\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \), then find a unit vector parallel to the vector \( \vec{a} + \vec{b} \). 
Answer: \( \vec{a} + \vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k} \). Unit vector \( = \frac{6\hat{i} - 3\hat{j} + 2\hat{k}}{\sqrt{6^2 + (-3)^2 + 2^2}} = \frac{6\hat{i} - 3\hat{j} + 2\hat{k}}{7} \)

Question. If \( \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \), then find \( |\vec{a} \times \vec{b}| \).
Answer: \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} = -17\hat{i} + 13\hat{j} + 7\hat{k} \). \( |\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + 13^2 + 7^2} = \sqrt{507} \)

Question. If a unit vector \( \vec{a} \) makes angles \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \) and an acute angle \( \theta \) with \( \hat{k} \), then find the value of \( \theta \). Answer: \( \cos^2 \frac{\pi}{3} + \cos^2 \frac{\pi}{4} + \cos^2 \theta = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1 \Rightarrow \cos^2 \theta = \frac{1}{4} \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \text{ or } 60^\circ \)

Question. In a triangle OAC, if B is the mid-point of side AC and \( \vec{OA} = \vec{a}, \vec{OB} = \vec{b} \), then what is \( \vec{OC} \)? 
Answer: Since B is the midpoint of AC, \( \vec{OB} = \frac{\vec{OA} + \vec{OC}}{2} \Rightarrow \vec{b} = \frac{\vec{a} + \vec{OC}}{2} \Rightarrow \vec{OC} = 2\vec{b} - \vec{a} \)

Question. If \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 400 \) and \( |\vec{a}| = 5 \), then write the value of \( |\vec{b}| \). 
Answer: \( |\vec{a}|^2 |\vec{b}|^2 = 400 \Rightarrow 5^2 |\vec{b}|^2 = 400 \Rightarrow 25 |\vec{b}|^2 = 400 \Rightarrow |\vec{b}|^2 = 16 \Rightarrow |\vec{b}| = 4 \)

Question. If \( \hat{a}, \hat{b} \) and \( \hat{c} \) are mutually perpendicular unit vectors, then find the value of \( |2\hat{a} + \hat{b} + \hat{c}| \). 
Answer: \( |2\hat{a} + \hat{b} + \hat{c}|^2 = |2\hat{a}|^2 + |\hat{b}|^2 + |\hat{c}|^2 = 4(1) + 1 + 1 = 6 \). Thus, value is \( \sqrt{6} \)

Question. Find a unit vector in the direction of \( \vec{a} = 3\hat{i} - 2\hat{j} + 6\hat{k} \). 
Answer: \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{\sqrt{3^2 + (-2)^2 + 6^2}} = \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} \)

Question. Write a vector of magnitude 9 units in the direction of vector \( -2\hat{i} + \hat{j} + 2\hat{k} \). 
Answer: Unit vector \( = \frac{-2\hat{i} + \hat{j} + 2\hat{k}}{3} \). Vector of magnitude 9 \( = 9 \left( \frac{-2\hat{i} + \hat{j} + 2\hat{k}}{3} \right) = -6\hat{i} + 3\hat{j} + 6\hat{k} \)