JEE Mathematics Vectors MCQs Set 04

Question. If the vector \( \vec{b} \) is collinear with the vector \( \vec{a} = (2\sqrt{2}, -1, 4) \) and \( |\vec{b}| = 10 \), then
(a) \( \vec{a} \pm \vec{b} = 0 \)
(b) \( \vec{a} \pm 2\vec{b} = 0 \)
(c) \( 2\vec{a} \pm \vec{b} = 0 \)
(d) None of the options
Answer: (c) \( 2\vec{a} \pm \vec{b} = 0 \)
Solution: \( \vec{a} = (2\sqrt{2}, -1, 4) \)
\( |\vec{b}| = 10 \)
\( \vec{b} = \lambda \vec{a} \)
\( |\vec{b}|^2 = \lambda^2 |\vec{a}|^2 \)
\( 100 = \lambda^2 (8 + 1 + 16) \)
\( \lambda^2 = 4 \Rightarrow \lambda = \pm 2 \)
\( 2\vec{a} \pm \vec{b} = 0 \)

Question. The vertices of a triangle are A(1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the internal bisector of the angle A is
(a) \( \hat{i} + \hat{j} + 2\hat{k} \)
(b) \( 2\hat{i} - 2\hat{j} + \hat{k} \)
(c) \( 2\hat{i} + 2\hat{j} - \hat{k} \)
(d) \( 2\hat{i} + 2\hat{j} + \hat{k} \)
Answer: (d) \( 2\hat{i} + 2\hat{j} + \hat{k} \)
Solution: \( \vec{p} = (3, 2, -1) \)
\( \hat{p} = \frac{(3,2,-1)}{\sqrt{14}} \)
\( \vec{q} = (1, 2, 3) \)
\( \hat{q} = \frac{1}{\sqrt{14}} (1, 2, 3) \)
Angle Bisector = \( \hat{p} + \hat{q} = \frac{1}{\sqrt{14}} (4, 4, 2) \)

Question. Let \( \vec{a} = \hat{i} + \hat{j} \) and \( \vec{b} = 2\hat{i} - \hat{k} \). The point of intersection of the lines \( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \) and \( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \) is
(a) \( -\hat{i} + \hat{j} + 2\hat{k} \)
(b) \( 3\hat{i} - \hat{j} + \hat{k} \)
(c) \( 3\hat{i} + \hat{j} - \hat{k} \)
(d) \( \hat{i} - \hat{j} - \hat{k} \)
Answer: (c) \( 3\hat{i} + \hat{j} - \hat{k} \)
Solution: \( \vec{a} = \hat{i} + \hat{j} \), \( \vec{b} = 2\hat{i} - \hat{k} \)
\( \vec{r} \times \vec{a} = \vec{b} \times \vec{a} \) .....(i)
\( \vec{r} \times \vec{b} = \vec{a} \times \vec{b} \) .....(ii)
Add (i) & (ii)
\( \vec{r} \times (\vec{a} \times \vec{b}) = 0 \Rightarrow \vec{r} = (\vec{a} + \vec{b}) = (3, 1, -1) \)

Question. If \( |\vec{a}| = 5 \), \( |\vec{a} - \vec{b}| = 8 \) and \( |\vec{a} + \vec{b}| = 10 \), then \( |\vec{b}| \) is equal to
(a) \( 1 \)
(b) \( \sqrt{57} \)
(c) \( 3 \)
(d) None of the options
Answer: (b) \( \sqrt{57} \)
Solution: \( |\vec{a} - \vec{b}| = 8 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}.\vec{b} = 64 \) ..(i)
\( |\vec{a} + \vec{b}| = 10 \Rightarrow |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}.\vec{b} = 100 \) ..(ii)
Add (i) and (ii) equation
\( 2|\vec{a}|^2 + 2|\vec{b}|^2 = 164 \)
\( |\vec{b}|^2 = 82 - 25 \)
\( |\vec{b}| = \sqrt{57} \)

Question. Angle between diagonals of a parallelogram whose side are represented by \( \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} - \hat{k} \)
(a) \( \cos^{-1}\left(\frac{1}{3}\right) \)
(b) \( \cos^{-1}\left(\frac{1}{2}\right) \)
(c) \( \cos^{-1}\left(\frac{4}{9}\right) \)
(d) \( \cos^{-1}\left(\frac{5}{9}\right) \)
Answer: (a) \( \cos^{-1}\left(\frac{1}{3}\right) \)
Solution: Diagonals are \( \vec{a} + \vec{b} = (3, 0, 0) \)
and \( \vec{a} - \vec{b} = (1, 2, 2) \)
\( \cos \theta = \frac{(\vec{a} + \vec{b}).(\vec{a} - \vec{b})}{|\vec{a} + \vec{b}|\ |\vec{a} - \vec{b}|} = \frac{3}{3.3} = \frac{1}{3} \)
\( \theta = \cos^{-1}\left(\frac{1}{3}\right) \)

Question. Vector \( \vec{a} \) and \( \vec{b} \) make an angle \( \theta = \frac{2\pi}{3} \). if \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), then \( \{(\vec{a} + 3\vec{b}) \times (3\vec{a} - \vec{b})\}^2 \) is equal to
(a) 225
(b) 250
(c) 275
(d) 300
Answer: (d) 300
Solution: \( \{(\vec{a} + 3\vec{b}) \times (3\vec{a} - \vec{b})\}^2 \)
\( = 100 (\vec{b} \times \vec{a})^2 = 100 |\vec{b}|^2 |\vec{a}|^2 \sin^2 \theta \)
\( = 100 \times 4 \times 1 \times \frac{3}{4} = 300 \)

Question. Unit vector perpendicular to the plane of the triangle ABC with position vectors \( \vec{a}, \vec{b}, \vec{c} \) of the vertices A, B, C is
(a) \( \frac{(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{\Delta} \)
(b) \( \frac{(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{2\Delta} \)
(c) \( \frac{(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{4\Delta} \)
(d) None of the options
Answer: (b) \( \frac{(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{2\Delta} \)
Solution: Perpendicular to the plane of \( \Delta ABC \) will be area vector
\( \vec{A} = \frac{1}{2} (\vec{AB} \times \vec{AC}) = \frac{1}{2} [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] \)
\( \vec{A} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}] \)
Unit vector \( \hat{A} = \frac{(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{2\Delta} \)

Question. The value of \( [ (\vec{a} + 2\vec{b} - \vec{c}), (\vec{a} - \vec{b}), (\vec{a} - \vec{b} - \vec{c}) ] \) is equal to the box product
(a) \( [\vec{a}\ \vec{b}\ \vec{c}] \)
(b) \( 2[\vec{a}\ \vec{b}\ \vec{c}] \)
(c) \( 3[\vec{a}\ \vec{b}\ \vec{c}] \)
(d) \( 4[\vec{a}\ \vec{b}\ \vec{c}] \)
Answer: (c) \( 3[\vec{a}\ \vec{b}\ \vec{c}] \)
Solution: \( [(\vec{a} + 2\vec{b} - \vec{c}), (\vec{a} - \vec{b}), (\vec{a} - \vec{b} - \vec{c})] \)
\( (\vec{a} + 2\vec{b} - \vec{c}) . [(\vec{a} - \vec{b}) \times (\vec{a} - \vec{b} - \vec{c})] \)
\( (\vec{a} + 2\vec{b} - \vec{c}) . [(\vec{b} - \vec{a}) \times \vec{c}] \)
\( (\vec{a} + 2\vec{b} - \vec{c}) . [\vec{b} \times \vec{a} - \vec{a} \times \vec{c}] \)
\( = [\vec{a}\ \vec{b}\ \vec{c}] - 0 + 0 + 2[\vec{a}\ \vec{b}\ \vec{c}] - 0 + 0 \)
\( = 3[\vec{a}\ \vec{b}\ \vec{c}] \)

Question. If \( \vec{b} \) and \( \vec{c} \) are two non-collinear vectors such that \( \vec{a} \parallel (\vec{b} \times \vec{c}) \), then \( (\vec{a} \times \vec{b}) . (\vec{a} \times \vec{c}) \) is equal to
(a) \( \vec{a}^2 (\vec{b}.\vec{c}) \)
(b) \( \vec{b}^2 (\vec{a}.\vec{c}) \)
(c) \( \vec{c}^2 (\vec{a}.\vec{b}) \)
(d) None of the options
Answer: (a) \( \vec{a}^2 (\vec{b}.\vec{c}) \)
Solution: \( \vec{a} \parallel (\vec{b} \times \vec{c}) \Rightarrow \vec{a}.\vec{b} = 0, \vec{a}.\vec{c} = 0 \)
\( (\vec{a} \times \vec{b}) . (\vec{a} \times \vec{c}) = [(\vec{a} \times \vec{b})\ \vec{a}\ \vec{c}] \)
\( = [\vec{a}\ \vec{c}\ \vec{a} \times \vec{b}] = \vec{a} . [\vec{c} \times (\vec{a} \times \vec{b})] \)
\( = \vec{a} . [(\vec{c}.\vec{b})\vec{a} - (\vec{a}.\vec{c})\vec{b}] = \vec{a}^2 (\vec{b}.\vec{c}) \)

Question. Vector of length 3 unit which is perpendicular to \( \hat{i} + \hat{j} + \hat{k} \) and lies in the plane of \( \hat{i} + \hat{j} + \hat{k} \) and \( 2\hat{i} - 3\hat{j} \)
(a) \( \frac{3}{\sqrt{6}} (\hat{i} - 2\hat{j} + \hat{k}) \)
(b) \( \frac{3}{\sqrt{6}} (2\hat{i} - \hat{j} - \hat{k}) \)
(c) \( \frac{3}{\sqrt{114}} (8\hat{i} - 7\hat{j} - \hat{k}) \)
(d) \( \frac{3}{\sqrt{114}} (-7\hat{i} + 8\hat{j} - \hat{k}) \)
Answer: (d) \( \frac{3}{\sqrt{114}} (-7\hat{i} + 8\hat{j} - \hat{k}) \)
Solution: \( \vec{a} = (1, 1, 1) \), \( \vec{b} = (1, 1, 1) \), \( \vec{c} = (2, -3, 0) \)
\( \vec{v} = \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a}.\vec{c})\vec{b} - (\vec{a}.\vec{b})\vec{c} = (-7, 8, -1) \)
\( \hat{v} = \frac{(-7, 8, -1)}{\sqrt{114}} \)
Reqd. Vector = \( \frac{3}{\sqrt{114}} (-7\hat{i} + 8\hat{j} - \hat{k}) \)

Question. Vector \( \vec{x} \) satisfying the relation \( \vec{A}.\vec{x} = c \) and \( \vec{A} \times \vec{x} = \vec{B} \) is
(a) \( \frac{c\vec{A} - (\vec{A} \times \vec{B})}{|\vec{A}|} \)
(b) \( \frac{c\vec{A} - (\vec{A} \times \vec{B})}{|\vec{A}|^2} \)
(c) \( \frac{c\vec{A} + (\vec{A} \times \vec{B})}{|\vec{A}|^2} \)
(d) \( \frac{c\vec{A} - 2(\vec{A} \times \vec{B})}{|\vec{A}|^2} \)
Answer: (b) \( \frac{c\vec{A} - (\vec{A} \times \vec{B})}{|\vec{A}|^2} \)
Solution: \( \vec{A}.\vec{x} = c \)
\( \vec{A} \times \vec{x} = \vec{B} \)
take cross with \( \vec{A} \)
\( \vec{A} \times (\vec{A} \times \vec{x}) = \vec{A} \times \vec{B} \)
\( (\vec{A}.\vec{x})\vec{A} - |\vec{A}|^2 \vec{x} = \vec{A} \times \vec{B} \)
\( \vec{x} = \frac{c\vec{A} - \vec{A} \times \vec{B}}{|\vec{A}|^2} \)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are linearly independent vectors, then which one of the following set of vectors is linearly dependent ?
(a) \( \vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a} \)
(b) \( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a} \)
(c) \( \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} \)
(d) None of the options
Answer: (b) \( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a} \)
Solution: \( x\vec{a} + y\vec{b} + z\vec{c} = 0 \)
(For linearly independent vector)
\( x(\vec{a} - \vec{b}) + y(\vec{b} - \vec{c}) + z(\vec{c} - \vec{a}) \)
\( x\vec{a} + y\vec{b} + z\vec{c} - (x\vec{b} + y\vec{c} + z\vec{a}) \)
linear combination \( \Rightarrow 0 - 0 = 0 \)
So \( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a} \) are linearly dependent vector.

Question. If line \( \vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \) is parallel to the plane \( \vec{r} . (3\hat{i} - 2\hat{j} - m\hat{k}) = 14 \), then the value of \( m \) is
(a) 2
(b) -2
(c) 0
(d) can not be predicted with these informations
Answer: (a) 2
Solution: \( \vec{r} = \vec{a} + \lambda\vec{p} \), \( \vec{r} . \vec{n} = 14 \) so \( \vec{p} . \vec{n} = 0 \)
\( (2, 1, 2) . (3, -2, -m) = 0 \)
\( 6 - 2 - 2m = 0 \Rightarrow m = 2 \)

Question. Let \( \vec{a}, \vec{b}, \vec{c} \) be vectors of length 3, 4, 5 respectively. Let \( \vec{a} \) be perpendicular to \( \vec{b} + \vec{c} \), \( \vec{b} \) to \( \vec{c} + \vec{a} \) and \( \vec{c} \) to \( \vec{a} + \vec{b} \). Then \( |\vec{a} + \vec{b} + \vec{c}| \)
(a) \( 2\sqrt{5} \)
(b) \( 2\sqrt{2} \)
(c) \( 10\sqrt{5} \)
(d) \( 5\sqrt{2} \)
Answer: (d) \( 5\sqrt{2} \)
Solution: \( \vec{a}.(\vec{b} + \vec{c}) = 0 \) ...(i)
\( \vec{b}.(\vec{c} + \vec{a}) = 0 \) ...(ii)
\( \vec{c}.(\vec{a} + \vec{b}) = 0 \) ....(iii)
Add all equation \( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} = 0 \)
\( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 0} = \sqrt{9 + 16 + 25} = 5\sqrt{2} \)

Question. Given \( \vec{a} = x\hat{i} + y\hat{j} + 2\hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{c} = \hat{i} + 2\hat{j} \); \( (\vec{a} \wedge \vec{b}) = \pi/2 \), \( \vec{a} . \vec{c} = 4 \), then
(a) \( [\vec{a}\ \vec{b}\ \vec{c}]^2 = |\vec{a}| \)
(b) \( [\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}| \)
(c) \( [\vec{a}\ \vec{b}\ \vec{c}] = 0 \)
(d) \( [\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|^2 \)
Answer: (d) \( [\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|^2 \)
Solution: \( \vec{a} = (x, y, 2) \), \( \vec{b} = (1, -1, 1) \)
\( \vec{c} = (1, 2, 0) \)
\( \vec{a}.\vec{b} = 0 \), \( \vec{a}.\vec{c} = 4 \)
\( x - y + 2 = 0 \) ... (1)
\( x + 2y = 4 \) ... (2)
\( x = 0, y = 2 \)
\( \vec{a} = (0, 2, 2) \)
\( [\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} . (\vec{b} \times \vec{c}) \)
\( \vec{b} \times \vec{c} = (-2, 1, 3) \)
\( = (0, 2, 2) . (-2, 1, 3) = 2 + 6 = 8 \quad |\vec{a}|^2 \)

Question. \( (\vec{d} + \vec{a}) . (\vec{a} \times (\vec{b} \times (\vec{c} \times \vec{d}))) \) simplifies to
(a) \( (\vec{b} . \vec{d}) [\vec{a}\ \vec{c}\ \vec{d}] \)
(b) \( (\vec{b} . \vec{c}) [\vec{a}\ \vec{b}\ \vec{d}] \)
(c) \( (\vec{b} . \vec{a}) [\vec{a}\ \vec{b}\ \vec{d}] \)
(d) None of the options
Answer: (a) \( (\vec{b} . \vec{d}) [\vec{a}\ \vec{c}\ \vec{d}] \)
Solution: \( (\vec{d} + \vec{a}) . (\vec{a} \times (\vec{b} \times (\vec{c} \times \vec{d}))) \)
\( (\vec{d} + \vec{a}) . (\vec{a} \times \{(\vec{b}.\vec{d})\vec{c} - (\vec{b}.\vec{c})\vec{d}\}) \)
\( (\vec{d} + \vec{a}) . [(\vec{b}.\vec{d})(\vec{a} \times \vec{c}) - (\vec{b}.\vec{c})(\vec{a} \times \vec{d})] \)
\( (\vec{b}.\vec{d}) [\vec{a}\ \vec{c}\ \vec{d}] \)

Question. Let \( \vec{r} \) be a vector perpendicular to \( \vec{a} + \vec{b} + \vec{c} \), where \( [\vec{a}\ \vec{b}\ \vec{c}] = 2 \). If \( \vec{r} = l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b}) \), then \( (l + m + n) \) is equal to
(a) 2
(b) 1
(c) 0
(d) None of the options
Answer: (c) 0
Solution: \( \vec{r} . (\vec{a} + \vec{b} + \vec{c}) = 0 \)
\( \vec{r} = l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b}) \)
\( \vec{r}.\vec{a} = l[\vec{a}\ \vec{b}\ \vec{c}] \) ...(i)
\( \vec{r}.\vec{b} = m[\vec{a}\ \vec{b}\ \vec{c}] \) ...(ii)
\( \vec{r}.\vec{c} = n[\vec{a}\ \vec{b}\ \vec{c}] \) ...(iii)
Add them \( \vec{r} . (\vec{a} + \vec{b} + \vec{c}) = (l + m + n) [\vec{a}\ \vec{b}\ \vec{c}] \Rightarrow l + m + n = 0 \)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are three non-coplanar non-zero vectors and \( \vec{r} \) is any vector in space, then \( (\vec{a} \times \vec{b}) \times (\vec{r} \times \vec{c}) + (\vec{b} \times \vec{c}) \times (\vec{r} \times \vec{a}) + (\vec{c} \times \vec{a}) \times (\vec{r} \times \vec{b}) \) is equal to
(a) \( 2[\vec{a}, \vec{b}, \vec{c}] \vec{r} \)
(b) \( 3[\vec{a}, \vec{b}, \vec{c}] \vec{r} \)
(c) \( [\vec{a}, \vec{b}, \vec{c}] \vec{r} \)
(d) None of the options
Answer: (a) \( 2[\vec{a}, \vec{b}, \vec{c}] \vec{r} \)
Solution: \( \vec{P} \times (\vec{r} \times \vec{c}) + \vec{W} \times (\vec{r} \times \vec{a}) + (\vec{c} \times \vec{a}) \times \vec{v} \)
\( = (\vec{P}.\vec{c})\vec{r} - (\vec{P}.\vec{r})\vec{c} + (\vec{W}.\vec{a})\vec{r} - (\vec{W}.\vec{r})\vec{a} + (\vec{c}.\vec{V})\vec{a} - (\vec{a}.\vec{V})\vec{c} \)
\( = [\vec{a}\ \vec{b}\ \vec{c}]\vec{r} - [\vec{a}\ \vec{b}\ \vec{r}]\vec{c} + [\vec{b}\ \vec{c}\ \vec{a}]\vec{r} - [\vec{b}\ \vec{c}\ \vec{r}]\vec{a} + [\vec{c}\ \vec{r}\ \vec{b}]\vec{a} - [\vec{a}\ \vec{r}\ \vec{b}]\vec{c} \)
\( = 2 [\vec{a}\ \vec{b}\ \vec{c}]\vec{r} \)

Question. Given the vertices A (2, 3, 1), B(4, 1, -2), C(6, 3, 7) & D(-5, -4, 8) of a tetrahedron. The length of the altitude drawn from the vertex D is
(a) 7
(b) 9
(c) 11
(d) None of the options
Answer: (c) 11
Solution: Altitude from D = \( \frac{\text{Volume of Tetrahedron}}{\text{Area of Face ABC}} \)
\( = \frac{\frac{1}{6}|[\vec{AD}\ \vec{AC}\ \vec{AB}]|}{\frac{1}{2}|\vec{AB} \times \vec{AC}|} = 11 \)

Question. If a, b, c are pth, qth, rth terms of an H.P. and \( \vec{u} = (q - r)\hat{i} + (r - p)\hat{j} + (p - q)\hat{k} \), \( \vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c} \), then
(a) \( \vec{u}, \vec{v} \) are parallel vectors
(b) \( \vec{u}, \vec{v} \) are orthogonal vectors
(c) \( \vec{u} . \vec{v} = 1 \)
(d) \( \vec{u} \times \vec{v} = \hat{i} + \hat{j} + \hat{k} \)
Answer: (b) \( \vec{u}, \vec{v} \) are orthogonal vectors
Solution: Let A be the first term and D is common difference
\( \frac{1}{a} = A + (p - 1)D \), \( \frac{1}{b} = A + (q - 1)D \), \( \frac{1}{c} = A + (r - 1)D \)
\( \vec{u} . \vec{v} = \left(\frac{q - r}{a}\right) + \left(\frac{r - p}{b}\right) + \left(\frac{p - q}{c}\right) \)
\( = q\left[\frac{1}{a} - \frac{1}{c}\right] + r\left[\frac{1}{b} - \frac{1}{a}\right] + p\left[\frac{1}{c} - \frac{1}{b}\right] \)
\( = q[A + (p - 1)D - A - (r - 1)D] \)
Similarly \( \vec{u} . \vec{v} = 0 \)

Question. For a non zero vector \( \vec{A} \) If the equations \( \vec{A} . \vec{B} = \vec{A} . \vec{C} \) and \( \vec{A} \times \vec{B} = \vec{A} \times \vec{C} \) hold simultaneously, then
(a) \( \vec{A} \) is perpendicular to \( \vec{B} - \vec{C} \)
(b) \( \vec{A} = \vec{B} \)
(c) \( \vec{B} = \vec{C} \)
(d) \( \vec{C} = \vec{A} \)
Answer: (c) \( \vec{B} = \vec{C} \)
Solution: \( \vec{A} . \vec{B} = \vec{A} . \vec{C} \) and \( \vec{A} \times \vec{B} = \vec{A} \times \vec{C} \)
\( \vec{A} \times \vec{B} = \vec{A} \times \vec{C} \)
\( \vec{A} \times (\vec{A} \times \vec{B}) = \vec{A} \times (\vec{A} \times \vec{C}) \)
\( (\vec{A} . \vec{B})\vec{A} - |\vec{A}|^2 \vec{B} = (\vec{A} . \vec{C})\vec{A} - |\vec{A}|^2 \vec{C} \)
\( \vec{B} = \vec{C} \)

Question. If the unit vectors \( \vec{e}_1 \) and \( \vec{e}_2 \) are inclined at an angle \( 2\theta \) and \( |\vec{e}_1 - \vec{e}_2| < 1 \), then for \( \theta \in [0, \pi] \), \( \theta \) may lie in the interval
(a) \( \left[0, \frac{\pi}{6}\right) \)
(b) \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \)
(c) \( \left(\frac{5\pi}{6}, \pi\right] \)
(d) \( \left[\frac{\pi}{2}, \frac{5\pi}{6}\right] \)
Answer: (a) \( \left[0, \frac{\pi}{6}\right) \)
Solution: \( |\vec{e}_1 - \vec{e}_2| < 1 \)
\( |\vec{e}_1|^2 + |\vec{e}_2|^2 - 2|\vec{e}_1||\vec{e}_2| \cos 2\theta < 1 \)
\( 2 - 2 \cos 2\theta < 1 \)
\( \cos 2\theta > 1/2 \Rightarrow 0 \le 2\theta < \frac{\pi}{3} = 0 \le \theta < \frac{\pi}{6} \)

Question. A vector \( \vec{a} \) has components 2p and 1 with respect to a rectangular Cartesian system. The system is rotated through a certain angle about the origin in the counterclockwise sense. If with respect to the new system, \( \vec{a} \) has components p + 1 and 1, then
(a) p = 0
(b) p = 1 or p = -1/3
(c) p = -1 or p = 1/3
(d) p = 1 or p = -1
Answer: (b) p = 1 or p = -1/3
Solution: \( \vec{a} = 2p\hat{i} + \hat{j} \) (old)
\( \vec{a} = (p + 1)\hat{i} + \hat{j} \) (New)
\( |\vec{a}|_{\text{old}} = |\vec{a}|_{\text{new}} \)
\( 4p^2 + 1 = (p + 1)^2 + 1 \)
\( 3p^2 - 2p - 1 = 0 \)
\( (3p + 1)(p - 1) = 0 \)
\( p = -1/3 \) or \( p = 1 \)

Question. Taken on side AC of a triangle ABC, a point M such that \( AM = \frac{1}{3} AC \). A point N is taken on the side CB such that \( BN = CB \), then for the point of intersection X of AB and MN which of the following holds good ?
(a) \( \vec{XB} = \frac{1}{3} \vec{AB} \)
(b) \( \vec{AX} = \frac{1}{3} \vec{AB} \)
(c) \( \vec{XN} = \frac{3}{4} \vec{MN} \)
(d) \( \vec{XM} = 3 \vec{XN} \)
Answer: (c) \( \vec{XN} = \frac{3}{4} \vec{MN} \)
Solution: \( \vec{AN} = 2\vec{b} - \vec{c} \)
Position vector of x
\( \frac{\lambda \frac{\vec{c}}{3} + 2\vec{b} - \vec{c}}{\lambda + 1} = \frac{0 + \mu\vec{b}}{\mu + 1} \)
\( \frac{\lambda/3 - 1}{\lambda + 1} = 0 \Rightarrow \lambda = 3 \)
\( \frac{2}{\lambda + 1} = \frac{\mu}{1 + \mu} \Rightarrow \mu = 1 \), \( \vec{x}(\vec{b}/2) \)
Now check options \( \vec{XN} = 2\vec{b} - \vec{c} - \frac{\vec{b}}{2} = \frac{3\vec{b}}{2} - \vec{c} \)
\( \vec{MN} = 2\vec{b} - \vec{c} - \frac{\vec{c}}{3} \Rightarrow \frac{3}{4}\vec{MN} = \frac{3\vec{b}}{2} - \vec{c} \)
So \( \vec{XN} = \frac{3}{4}\vec{MN} \)

Question. The volume of the parallelopiped constructed on the diagonals of the faces of the given rectangular parallelopiped is m times the volume of the given parallelopiped. Then m is equal to
(a) 2
(b) 3
(c) 4
(d) None of the options
Answer: (a) 2
Solution: \( V_{\text{Old}} = [\vec{a}\ \vec{b}\ \vec{c}] \)
\( V_{\text{New}} = [\vec{a}+\vec{b}\ \vec{b}+\vec{c}\ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \)
so m = 2

Question. If \( \vec{a} = \vec{b} + \vec{c} \), \( \vec{b} \times \vec{d} = 0 \) and \( \vec{c} . \vec{d} = 0 \) then \( \frac{\vec{d} \times (\vec{a} \times \vec{d})}{\vec{d}^2} \) is equal to
(a) \( \vec{a} \)
(b) \( \vec{b} \)
(c) \( \vec{c} \)
(d) \( \vec{d} \)
Answer: (c) \( \vec{c} \)
Solution: \( \vec{a} = \vec{b} + \vec{c} \), \( \vec{b} \times \vec{d} = 0 \), \( \vec{c} . \vec{d} = 0 \)
assume \( \vec{b} = \hat{i} \), \( \vec{c} = \hat{j} \), \( \vec{d} = 2\hat{i} \)
\( \frac{\vec{d} \times (\vec{a} \times \vec{d})}{|\vec{d}|^2} = \vec{a} - \frac{(\vec{d} . \vec{a})\vec{d}}{|\vec{d}|^2} \)
\( = \vec{a} - \frac{\vec{d}}{2} = \hat{i} + \hat{j} - \hat{i} = \hat{j} = \vec{c} \)

Question. Consider a tetrahedron with faces \( f_1, f_2, f_3, f_4 \). Let \( \vec{a}_1, \vec{a}_2, \vec{a}_3, \vec{a}_4 \) be the vectors whose magnitudes are respectively equal to the areas of \( f_1, f_2, f_3, f_4 \) and whose directions are perpendicular to these faces in the outward direction. Then
(a) \( \vec{a}_1 + \vec{a}_2 + \vec{a}_3 + \vec{a}_4 = 0 \)
(b) \( \vec{a}_1 + \vec{a}_3 = \vec{a}_2 + \vec{a}_4 \)
(c) \( \vec{a}_1 + \vec{a}_2 = \vec{a}_3 + \vec{a}_4 \)
(d) None of the options
Answer: (a) \( \vec{a}_1 + \vec{a}_2 + \vec{a}_3 + \vec{a}_4 = 0 \)
Solution: Let \( \vec{a}_1 \) be of OAB
\( \vec{a}_2 \) be of OBC
\( \vec{a}_3 \) be of OCA
\( \vec{a}_4 \) be of ABC
\( \vec{a}_1 = \frac{1}{2}(\vec{a} \times \vec{b}) \); \( \vec{a}_2 = \frac{1}{2}(\vec{b} \times \vec{c}) \)
\( \vec{a}_3 = \frac{1}{2}(\vec{c} \times \vec{a}) \); \( \vec{a}_4 = \frac{1}{2}[(\vec{c} - \vec{a}) \times (\vec{b} - \vec{a})] \)
\( \vec{a}_1 + \vec{a}_2 + \vec{a}_3 + \vec{a}_4 = 0 \)

Question. In the isosceles triangle ABC, \( |\vec{AB}| = |\vec{BC}| = 8 \) and a point E divides AB internally in the ratio 1 : 3, then the cosine of angle between \( \vec{CE} \) and \( \vec{CA} \) is (where \( |\vec{CA}| = 12 \))
(a) \( -\frac{3\sqrt{7}}{8} \)
(b) \( \frac{3\sqrt{8}}{17} \)
(c) \( \frac{3\sqrt{7}}{8} \)
(d) \( -\frac{3\sqrt{8}}{17} \)
Answer: (c) \( \frac{3\sqrt{7}}{8} \)
Solution: \( |\vec{a}| = 8 \); \( |\vec{c}| = 8 \)
\( |\vec{CA}| = 12 \Rightarrow |\vec{a} - \vec{c}| = 12 \)
\( \Rightarrow \vec{a} . \vec{c} = -8 \)
Let the angle be \( \theta \)
\( \cos \theta = \frac{\vec{CE} . \vec{CA}}{|\vec{CE}|\ |\vec{CA}|} = \frac{\left(\frac{3\vec{a}}{4} - \vec{c}\right) . (\vec{a} - \vec{c})}{\left|\frac{3\vec{a}}{4} - \vec{c}\right|\ |\vec{CA}|} = \frac{3\sqrt{7}}{8} \)

Question. If the vector product of a constant vector \( \vec{OA} \) with a variable vector \( \vec{OB} \) in a fixed plane OAB be a constant vector, then locus of B is
(a) a straight line perpendicular to \( \vec{OA} \)
(b) a circle with centre O radius equal to \( |\vec{OA}| \)
(c) a straight line parallel to \( \vec{OA} \)
(d) None of the options
Answer: (c) a straight line parallel to \( \vec{OA} \)
Solution: The locus of B will be a straight line parallel to \( \vec{OA} \)

Question. Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be non-coplanar unit vectors equally inclined to one another at an acute angle \( \theta \). Then \( [\vec{a}\ \vec{b}\ \vec{c}] \) in terms of \( \theta \) is equal to
(a) \( (1 + \cos\theta) \sqrt{\cos 2\theta} \)
(b) \( (1 + \cos\theta) \sqrt{1 - 2\cos 2\theta} \)
(c) \( (1 - \cos\theta) \sqrt{1 + 2\cos 2\theta} \)
(d) None of the options
Answer: (c) \( (1 - \cos\theta) \sqrt{1 + 2\cos 2\theta} \)
Solution: \( [\vec{a}\ \vec{b}\ \vec{c}]^2 = \begin{vmatrix} \vec{a}.\vec{a} & \vec{a}.\vec{b} & \vec{a}.\vec{c} \\ \vec{b}.\vec{a} & \vec{b}.\vec{b} & \vec{b}.\vec{c} \\ \vec{c}.\vec{a} & \vec{c}.\vec{b} & \vec{c}.\vec{c} \end{vmatrix} = \begin{vmatrix} 1 & \cos\theta & \cos\theta \\ \cos\theta & 1 & \cos\theta \\ \cos\theta & \cos\theta & 1 \end{vmatrix} \)
\( = (1 - \cos\theta)^2 (1 + 2\cos\theta) \)
So \( [\vec{a}\ \vec{b}\ \vec{c}] = (1 - \cos\theta) \sqrt{1 + 2\cos\theta} \)

Question. If \( u \) and \( v \) are unit vectors and \( \theta \) is the acute angle between them, then \( 2u \times 3v \) is a unit vector for
(a) Exactly two values of \( \theta \)
(b) More than two values of \( \theta \)
(c) No value of \( \theta \)
(d) Exactly one value of \( \theta \)
Answer: (d) Exactly one value of \( \theta \)
Solution: \( |u| = 1 \); \( |v| = 1 \)
\( |2u \times 3v| = 1 \)
\( |u \times v| = \frac{1}{6} \)
\( |u|\ |v| \sin\theta = \frac{1}{6} \Rightarrow \sin\theta = \frac{1}{6} \)
As \( \theta \) is acute angle than only one value possible

Question. Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k} \). If the vector \( \vec{c} \) lies in the plane of \( \vec{a} \) and \( \vec{b} \), then \( x \) equals
(a) 0
(b) 1
(c) -4
(d) -2
Answer: (d) -2
Solution: \( \vec{a} = (1, 1, 1) \), \( \vec{b} = (1, -1, 2) \), \( \vec{c} = (x, x - 2, -1) \)
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = (3, -1, -2) \)
\( \vec{c} . (\vec{a} \times \vec{b}) = 0 \)
\( 3x - (x - 2) + 2 = 0 \Rightarrow x = -2 \)

Question. The value of \( a \), for which the points A, B, C with position vectors \( 2\hat{i} - \hat{j} - \hat{k} \), \( \hat{i} - 3\hat{j} - 5\hat{k} \) and \( a\hat{i} - 3\hat{j} - \hat{k} \) respectively are the vertices of a right angled triangle with \( C = \pi/2 \) are
(a) -2 and -1
(b) -2 and 1
(c) 2 and -1
(d) 2 and 1
Answer: (d) 2 and 1
Solution: \( \vec{A}(2, -1, -1) \), \( \vec{B}(1, -3, -5) \), \( \vec{C}(a, -3, -1) \)
\( \vec{AC} . \vec{CB} = 0 \Rightarrow (a - 2, -2, 0) . (a - 1, 0, 4) = 0 \)
\( (a - 1)(a - 2) = 0 \Rightarrow a = 1 \) and 2

Question. The distance between the line \( \vec{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + 4\hat{k}) \) and the plane \( \vec{r} . (\hat{i} + 5\hat{j} + \hat{k}) = 5 \) is
(a) 10/3
(b) 3/10
(c) \( \frac{10}{3\sqrt{3}} \)
(d) 10/9
Answer: (c) \( \frac{10}{3\sqrt{3}} \)
Solution: \( \vec{r} = (2, -2, 3) + \lambda(1, -1, 4) \)
\( \vec{r} . (1, 5, 1) = 5 \)
\( \vec{L}.\vec{n} = 1 - 5 + 4 = 0 \)
So line and plane are parallel.
Let a point on the plane (0, 1, 0)
distance = \( |\vec{b}| \cos\theta = \frac{\vec{a}.\vec{b}}{|\vec{a}|} = \frac{2 - 15 + 3}{\sqrt{27}} = \frac{10}{3\sqrt{3}} \)

Question. Image of the point P with position vector \( 7\hat{i} - \hat{j} + 2\hat{k} \) in the line whose vector equation is \( \vec{r} = 9\hat{i} + 5\hat{j} + 5\hat{k} + \lambda(\hat{i} + 3\hat{j} + 5\hat{k}) \) has the position vector.
(a) (-9, 5, 2)
(b) (9, 5, -2)
(c) (9, -5, -2)
(d) None of the options
Answer: (b) (9, 5, -2)
Solution: \( \vec{PN} = (\lambda + 2, 3\lambda + 6, 5\lambda + 3) \)
\( \vec{b} = (1, 3, 5) \), \( \vec{PN} . \vec{b} = 0 \)
\( (\lambda + 2) + 3(3\lambda + 6) + 5(5\lambda + 3) = 0 \)
\( \Rightarrow 10(\lambda + 2) + 5(5\lambda + 3) = 0 \)
\( \Rightarrow 10\lambda + 20 + 25\lambda + 15 = 0 \)
\( \Rightarrow 35\lambda + 35 = 0 \Rightarrow \lambda = -1 \)
\( \vec{N} = (8, 2, 0) \Rightarrow \vec{N} = \frac{\vec{p} + \vec{p'}}{2} \)
\( \Rightarrow \vec{p'} = 2\vec{N} - \vec{p} = 2(8, 2, 0) - (7, -1, 2) = (16, 4, 0) - (7, -1, 2) = (9, 5, -2) \)

Question. A particle is acted upon by constant forces \( 4\hat{i} + \hat{j} - 3\hat{k} \) and \( 3\hat{i} + \hat{j} - \hat{k} \) which displace it from a point \( \hat{i} + 2\hat{j} + 3\hat{k} \) to the point \( 5\hat{i} + 4\hat{j} + \hat{k} \). The workdone in standard units by the force is given by
(a) 40
(b) 30
(c) 25
(d) 15
Answer: (a) 40
Solution: \( \vec{F}_1 = (4, 1, -3) \), \( \vec{F}_2 = (3, 1, -1) \)
\( \vec{ds} = (5, 4, 1) - (1, 2, 3) = (4, 2, -2) \)
work done = \( \vec{F}_1 . \vec{ds} + \vec{F}_2 . \vec{ds} = 24 + 16 = 40 \)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar vectors and \( \lambda \) is a real number, then the vectors \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( \lambda\vec{b} + 4\vec{c} \) and \( (2\lambda - 1)\vec{c} \) are non-coplanar for
(a) all values of \( \lambda \)
(b) all except one value of \( \lambda \)
(c) all except two values of \( \lambda \)
(d) no value of \( \lambda \)
Answer: (c) all except two values of \( \lambda \)
Solution: \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar \( \Rightarrow [\vec{a}\ \vec{b}\ \vec{c}] \ne 0 \)
\( [\vec{a} + 2\vec{b} + 3\vec{c} \quad \lambda\vec{b} + 4\vec{c} \quad (2\lambda - 1)\vec{c}] \)
\( = (\vec{a} + 2\vec{b} + 3\vec{c}) . [(\lambda\vec{b} + 4\vec{c}) \times (2\lambda - 1)\vec{c}] \)
\( = \lambda(2\lambda - 1)(\vec{a} + 2\vec{b} + 3\vec{c}) . (\vec{b} \times \vec{c}) \)
\( = \lambda(2\lambda - 1)[\vec{a}\ \vec{b}\ \vec{c}] \)

Question. Let \( \vec{u}, \vec{v}, \vec{w} \) be such that \( |\vec{u}| = 1 \), \( |\vec{v}| = 2 \), \( |\vec{w}| = 3 \). If the projection \( \vec{v} \) along \( \vec{u} \) is equal to that of \( \vec{w} \) along \( \vec{u} \) and \( \vec{v}, \vec{w} \) are perpendicular to each other, then \( |\vec{u} - \vec{v} + \vec{w}| \) equals
(a) 2
(b) \( \sqrt{7} \)
(c) \( \sqrt{14} \)
(d) 14
Answer: (c) \( \sqrt{14} \)
Solution: \( |\vec{u} - \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u}.\vec{v} + \vec{v}.\vec{w} - \vec{w}.\vec{u}) \)
\( = 14 - 2(\vec{u}.\vec{v} + \vec{v}.\vec{w} - \vec{w}.\vec{u}) \)
Given \( \frac{\vec{u}.\vec{v}}{|\vec{u}|} = \frac{\vec{w}.\vec{u}}{|\vec{u}|} \) and \( \vec{w}.\vec{v} = 0 \) & \( \vec{w}.\vec{u} = 0 \)
\( \vec{u}.\vec{v} = \vec{w}.\vec{u} = \frac{\vec{w}.\vec{v}}{|\vec{v}|} \)
\( |\vec{u} - \vec{v} + \vec{w}|^2 = 14 - 2(0) \)
\( |\vec{u} - \vec{v} + \vec{w}|^2 = 14 \Rightarrow |\vec{u} - \vec{v} + \vec{w}| = \sqrt{14} \)

Question. Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be non-zero vectors such that \( (\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} \). If \( \theta \) is the acute angle between the vectors \( \vec{b} \) and \( \vec{c} \), then \( \sin\theta \) equals is
(a) \( 1/3 \)
(b) \( \sqrt{2}/3 \)
(c) \( 2/3 \)
(d) \( 2\sqrt{2}/3 \)
Answer: (d) \( 2\sqrt{2}/3 \)
Solution: \( (\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} \)
\( (\vec{a}.\vec{c})\vec{b} - (\vec{b}.\vec{c})\vec{a} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a} \)
\( \cos\theta = -\frac{1}{3} \Rightarrow \sin\theta = \frac{2\sqrt{2}}{3} \)

Question. \( \vec{a}, \vec{b}, \vec{c} \) are three vectors, such that \( \vec{a} + \vec{b} + \vec{c} = 0 \), \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), \( |\vec{c}| = 3 \) then \( \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} \) is equal to
(a) 0
(b) -7
(c) 7
(d) 1
Answer: (b) -7
Solution: \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), \( |\vec{c}| = 3 \)
\( \vec{a} + \vec{b} + \vec{c} = 0 \)
\( |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \)
\( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}) = 0 \)
\( 1 + 4 + 9 + 2(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}) = 0 \)
\( \Rightarrow \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} = -7 \)