CBSE Class 12 Mathematics Vector Algebra VBQs Set 03

Short Answer Questions–I:

Question. Find a vector of magnitude 5 units and parallel to resultant of the vectors \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - 2\hat{j} + \hat{k} \).
Answer: Resultant \( \vec{R} = \vec{a} + \vec{b} = 3\hat{i} + \hat{j} \). Unit vector \( \hat{R} = \frac{3\hat{i} + \hat{j}}{\sqrt{10}} \). Required vector \( = \frac{5}{\sqrt{10}}(3\hat{i} + \hat{j}) = \frac{\sqrt{10}}{2}(3\hat{i} + \hat{j}) \)

Question. For any three vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \), find the value of \( \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) \). 
Answer: \( \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a} + \vec{c} \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{a} + \vec{b} \times \vec{c} - \vec{a} \times \vec{b} + \vec{c} \times \vec{a} - \vec{b} \times \vec{c} = \vec{0} \)

Question. Find \( |\vec{x}| \), if for a unit vector \( \vec{a} \), \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15 \).
Answer: \( |\vec{x}|^2 - |\vec{a}|^2 = 15 \Rightarrow |\vec{x}|^2 - 1 = 15 \Rightarrow |\vec{x}|^2 = 16 \Rightarrow |\vec{x}| = 4 \)

Question. If \( \vec{a} \) and \( \vec{b} \) are two unit vectors such that \( \vec{a} + \vec{b} \) is also a unit vector, then find the angle between \( \vec{a} \) and \( \vec{b} \). 
Answer: \( |\vec{a} + \vec{b}|^2 = 1^2 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1 \Rightarrow 1 + 1 + 2 \cos \theta = 1 \Rightarrow \cos \theta = -1/2 \Rightarrow \theta = \frac{2\pi}{3} \text{ or } 120^\circ \)

Question. Find the value of \( a + b \), if the points (2, a, 3)(3, –5, b) and (–1, 11, 9) are collinear. 
Answer: Direction ratios are proportional: \( \frac{3-2}{-1-3} = \frac{-5-a}{11-(-5)} = \frac{b-3}{9-b} \Rightarrow \frac{1}{-4} = \frac{-5-a}{16} = \frac{b-3}{9-b} \). Solving, \( a = -1, b = 1 \). So \( a+b = 0 \)

Question. Find a vector \( \vec{r} \) equally inclined to the three axes and whose magnitude is \( 3\sqrt{3} \) units. 
Answer: \( l=m=n \Rightarrow 3l^2 = 1 \Rightarrow l = \pm 1/\sqrt{3} \). Required vector \( \vec{r} = 3\sqrt{3} \left[ \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \right] = \pm 3(\hat{i} + \hat{j} + \hat{k}) \)

Question. Find the angle between unit vectors \( \vec{a} \) and \( \vec{b} \) so that \( \sqrt{3}\vec{a} - \vec{b} \) is also a unit vector. 
Answer: \( |\sqrt{3}\vec{a} - \vec{b}|^2 = 1 \Rightarrow 3|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3}\vec{a} \cdot \vec{b} = 1 \Rightarrow 3 + 1 - 2\sqrt{3} \cos \theta = 1 \Rightarrow \cos \theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^\circ \text{ or } \frac{\pi}{6} \)

Question. Find \( |\vec{a}| \) and \( |\vec{b}| \), if \( |\vec{a}| = 2|\vec{b}| \) and \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 12 \). 
Answer: \( |\vec{a}|^2 - |\vec{b}|^2 = 12 \Rightarrow 4|\vec{b}|^2 - |\vec{b}|^2 = 12 \Rightarrow 3|\vec{b}|^2 = 12 \Rightarrow |\vec{b}| = 2 \text{ and } |\vec{a}| = 4 \)

Question. Find the unit vector perpendicular to each of the vectors \( \vec{a} = 4\hat{i} + 3\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \). 
Answer: \( \vec{a} \times \vec{b} = 7\hat{i} - 6\hat{j} - 10\hat{k} \). Unit vector \( = \frac{7\hat{i} - 6\hat{j} - 10\hat{k}}{\sqrt{185}} \)

Short Answer Questions–II:

Question. Find a unit vector perpendicular to both of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) where \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \). 
Answer: \( \vec{a}+\vec{b} = 2\hat{i}+3\hat{j}+4\hat{k} \), \( \vec{a}-\vec{b} = -\hat{j}-2\hat{k} \). Cross product \( = -2\hat{i} + 4\hat{j} - 2\hat{k} \). Unit vector \( = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \)

Question. If \( \vec{p} = 5\hat{i} + \lambda\hat{j} - 3\hat{k} \) and \( \vec{q} = \hat{i} + 3\hat{j} - 5\hat{k} \) then find the value of \( \lambda \), so that \( \vec{p} + \vec{q} \) and \( \vec{p} - \vec{q} \) are perpendicular vectors. 
Answer: \( (\vec{p}+\vec{q}) \cdot (\vec{p}-\vec{q}) = 0 \Rightarrow |\vec{p}|^2 - |\vec{q}|^2 = 0 \Rightarrow (25 + \lambda^2 + 9) - (1 + 9 + 25) = 0 \Rightarrow \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1 \)

Question. Let \( \vec{a} = \hat{i} + 4\hat{j} + 2\hat{k} \), \( \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \). Find a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{c} \cdot \vec{d} = 27 \). 
Answer: \( \vec{d} = k(\vec{a} \times \vec{b}) = k(32\hat{i} - \hat{j} - 14\hat{k}) \). Given \( \vec{c} \cdot \vec{d} = 27 \Rightarrow k(64 + 1 - 56) = 27 \Rightarrow 9k = 27 \Rightarrow k=3 \). So \( \vec{d} = 96\hat{i} - 3\hat{j} - 42\hat{k} \)

Question. Show that the four points with position vectors \( 4\hat{i} + 8\hat{j} + 12\hat{k}, 2\hat{i} + 4\hat{j} + 6\hat{k}, 3\hat{i} + 5\hat{j} + 4\hat{k} \) and \( 5\hat{i} + 8\hat{j} + 5\hat{k} \) are coplanar. 
Answer: Let points be A, B, C, D. \( \vec{AB} = -2\hat{i}-4\hat{j}-6\hat{k} \), \( \vec{AC} = -\hat{i}-3\hat{j}-8\hat{k} \), \( \vec{AD} = \hat{i}-7\hat{k} \). Scalar triple product \( [\vec{AB} \quad \vec{AC} \quad \vec{AD}] = 0 \), hence coplanar.

Question. Find \( x \) such that four points A(4, 1, 2), B(5, x, 6), C(5, 1, –1) and D(7, 4, 0) are coplanar. 
Answer: \( \vec{AB}=(1, x-1, 4) \), \( \vec{AC}=(1, 0, -3) \), \( \vec{AD}=(3, 3, -2) \). For coplanarity, determinant \( = 0 \Rightarrow 1(9) - (x-1)(7) + 4(3) = 0 \Rightarrow 9 - 7x + 7 + 12 = 0 \Rightarrow 7x = 28 \Rightarrow x=4 \)

Question. For three vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \) if \( \vec{a} \times \vec{b} = \vec{c} \) and \( \vec{a} \times \vec{c} = \vec{b} \), then prove that \( \vec{a}, \vec{b} \) and \( \vec{c} \) are mutually perpendicular vectors, \( |\vec{b}| = |\vec{c}| \) and \( |\vec{a}| = 1 \). 
Answer: \( \vec{a} \times \vec{b} = \vec{c} \Rightarrow \vec{c} \perp \vec{a}, \vec{c} \perp \vec{b} \). \( \vec{a} \times \vec{c} = \vec{b} \Rightarrow \vec{b} \perp \vec{a}, \vec{b} \perp \vec{c} \). Since \( \vec{a} \perp \vec{b} \) and \( \vec{a} \perp \vec{c} \), vectors are mutually perpendicular. Also \( |\vec{c}| = |\vec{a}||\vec{b}| \sin 90^\circ \) and \( |\vec{b}| = |\vec{a}||\vec{c}| \sin 90^\circ \). Solving, \( |\vec{a}|=1 \) and \( |\vec{b}|=|\vec{c}| \).

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors such that \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \), then prove that (i) \( \vec{a} = \pm 2(\vec{b} \times \vec{c}) \) (ii) \( [\vec{a} + \vec{b} \quad \vec{b} + \vec{c} \quad \vec{c} + \vec{a}] = \pm 1 \).
Answer: \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \), so \( \vec{a} \parallel \vec{b} \times \vec{c} \). \( |\vec{b} \times \vec{c}| = 1 \cdot 1 \cdot \sin(\pi/6) = 1/2 \). Thus \( \vec{a} = \pm \frac{\vec{b} \times \vec{c}}{1/2} = \pm 2(\vec{b} \times \vec{c}) \). STP part results from \( 2[\vec{a} \quad \vec{b} \quad \vec{c}] = 2(\vec{a} \cdot (1/2 \hat{a})) = \pm 1 \).

Question. The two adjacent sides of a parallelogram are \( 2\hat{i} - 4\hat{j} - 5\hat{k} \) and \( 2\hat{i} + 2\hat{j} + 3\hat{k} \). Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. 
Answer: Diagonals \( \vec{d_1} = 4\hat{i}-2\hat{j}-2\hat{k} \), \( \vec{d_2} = -6\hat{j}-8\hat{k} \). Unit vectors \( = \frac{2\hat{i}-\hat{j}-\hat{k}}{\sqrt{6}} \), \( \frac{-3\hat{j}-4\hat{k}}{5} \). Area \( = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| = \sqrt{404} \) sq units.

Question. Find the angle between the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) if \( \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + \hat{j} - 2\hat{k} \), and hence find a vector perpendicular to both \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \). 
Answer: \( \vec{a}+\vec{b}=5\hat{i}+\hat{k} \), \( \vec{a}-\vec{b}=-\hat{i}-2\hat{j}+5\hat{k} \). \( (\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = -5 + 0 + 5 = 0 \), so angle is \( 90^\circ \). Perpendicular vector \( = 2\hat{i} - 26\hat{j} - 10\hat{k} \).

Question. If \( \vec{a} = 2\hat{i} + \hat{j} - \hat{k} \), \( \vec{b} = 4\hat{i} - 7\hat{j} + \hat{k} \), find a vector \( \vec{c} \) such that \( \vec{a} \times \vec{c} = \vec{b} \) and \( \vec{a} \cdot \vec{c} = 6 \). 
Answer: \( \vec{c} = (3, 1, 1) \) or \( 3\hat{i} + \hat{j} + \hat{k} \)

Question. Using vectors find the area of triangle ABC with vertices A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1). 
Answer: \( \vec{AB} = \hat{i}-3\hat{j}+\hat{k} \), \( \vec{AC} = 3\hat{i}+3\hat{j}-4\hat{k} \). Area \( = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{81+49+144} = \frac{\sqrt{274}}{2} \) sq. units.

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: \( x = 5 \)

Question. If \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) represent two adjacent sides of a parallelogram, find unit vectors parallel to the diagonals of the parallelogram. 
Answer: Diagonals \( \vec{d_1} = 3\hat{i}+6\hat{j}-2\hat{k} \), \( \vec{d_2} = \hat{i}+2\hat{j}-8\hat{k} \). Unit vectors \( = \frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k}) \), \( \frac{1}{\sqrt{69}}(\hat{i}+2\hat{j}-8\hat{k}) \)

Question. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1). 
Answer: Area \( = \frac{\sqrt{274}}{2} \) sq. units.