JEE Mathematics Vectors MCQs Set E

Question. If \( \vec{u}, \vec{v} \) and \( \vec{w} \) are three non-coplanar vectors, then \( (\vec{u} + \vec{v} - \vec{w}) . [(\vec{u} - \vec{v}) \times (\vec{v} - \vec{w})] \) equals
(a) 0
(b) \( \vec{u} . \vec{v} \times \vec{w} \)
(c) \( \vec{u} . \vec{w} \times \vec{v} \)
(d) \( 3\vec{u} . \vec{v} \times \vec{w} \)
Answer: (b) \( \vec{u} . \vec{v} \times \vec{w} \)
Solution: \( [\vec{u}\ \vec{v}\ \vec{w}] \ne 0 \)
\( (\vec{u} + \vec{v} - \vec{w}) . [(\vec{u} - \vec{v}) \times (\vec{v} - \vec{w})] \)
\( (\vec{u} + \vec{p}) . [(\vec{u} - \vec{v}) \times \vec{p}] \)
\( (\vec{u} + \vec{p}) . [\vec{u} \times \vec{p} - \vec{v} \times \vec{p}] \)
\( = -\vec{u} . (\vec{v} \times \vec{p}) = -\vec{u} . (\vec{v} \times (\vec{v} - \vec{w})) \)
\( = -\vec{u} . (\vec{v} \times \vec{w}) \)

Question. Consider points A, B, C and D with position vectors \( 7\hat{i} + 4\hat{j} + 7\hat{k} \), \( \hat{i} - 6\hat{j} + 10\hat{k} \), \( -\hat{i} - 3\hat{j} + 4\hat{k} \) and \( 5\hat{i} - \hat{j} + 5\hat{k} \) respectively. The ABCD is a
(a) square
(b) rhombus
(c) rectangle
(d) None of the options
Answer: (d) None of the options
Solution: \( \vec{AB} = -6\hat{i} - 10\hat{j} + 3\hat{k} \)
\( \vec{AD} = -2\hat{i} - 5\hat{j} - 2\hat{k} \)
\( \vec{AB} . \vec{AD} \ne 0 \)
so not a square or rectangle \( \vec{AB} \ne \vec{AD} \) so not a rhombus.

Question. The vectors \( \vec{AB} = 3\hat{i} + 4\hat{k} \) and \( \vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k} \) are the sides of a triangle ABC. The length of the median through A is
(a) \( \sqrt{18} \)
(b) \( \sqrt{72} \)
(c) \( \sqrt{33} \)
(d) \( \sqrt{288} \)
Answer: (c) \( \sqrt{33} \)
Solution: \( \vec{AB} = (3, 0, 4) \)
\( \vec{AC} = (5, -2, 4) \)
Let \( \vec{A} \) be origin. D is the mid point of BC
D(4, -1, 4)
\( \vec{AD} = (4, -1, 4) \)
\( |\vec{AD}| = \sqrt{16 + 1 + 16} = \sqrt{33} \)

Question. Let \( \vec{u} = \hat{i} + \hat{j} \), \( \vec{v} = \hat{i} - \hat{j} \) and \( \vec{w} = \hat{i} + 2\hat{j} + 3\hat{k} \). If \( \hat{n} \) is a unit vector such that \( \vec{u} . \hat{n} = 0 \) and \( \vec{v} . \hat{n} = 0 \) then \( |\vec{w} . \hat{n}| \) is equal to
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (d) 3
Solution: \( \vec{u} = (1, 1, 0) \), \( \vec{v} = (1, -1, 0) \), \( \vec{w} = (1, 2, 3) \)
\( \vec{u}.\hat{n} = 0, \vec{v}.\hat{n} = 0 \) then \( \Rightarrow |\vec{w}.\hat{n}| = |\pm 3| = 3 \)
where \( \hat{n} = \lambda(\vec{u} \times \vec{v}) \Rightarrow \hat{n} = -2\lambda\hat{k} \)
\( |\hat{n}| = 1 \Rightarrow \lambda = \pm 1/2 \Rightarrow 2\lambda = \pm 1 \)

Question. If \( \vec{a} = \hat{i} - \hat{j} \), \( \vec{b} = \hat{i} + \hat{j} \), \( \vec{c} = \hat{i} + 3\hat{j} + 5\hat{k} \) and \( \vec{n} \) be a unit vector such that \( \vec{b} . \vec{n} = 0, \vec{a} . \vec{n} = 0 \) then value of \( |\vec{c} . \vec{n}| \) is
(a) 1
(b) 3
(c) 5
(d) 2
Answer: (c) 5
Solution: \( \vec{a} = \hat{i} - \hat{j} \), \( \vec{b} = \hat{i} + \hat{j} \), \( \vec{c} = \hat{i} + 3\hat{j} + 5\hat{k} \)
\( \Rightarrow \vec{n} = \hat{k} \Rightarrow |\vec{c}.\vec{n}| = 5 \)

Question. If \( \vec{u} = \vec{a} - \vec{b} \), \( \vec{v} = \vec{a} + \vec{b} \) and \( |\vec{a}| = |\vec{b}| = 2 \), then \( |\vec{u} \times \vec{v}| \) is equal to
(a) \( \sqrt{2(16 - (\vec{a} . \vec{b})^2)} \)
(b) \( 2\sqrt{(16 - (\vec{a} . \vec{b})^2)} \)
(c) \( 2\sqrt{(4 - (\vec{a} . \vec{b})^2)} \)
(d) \( \sqrt{2(4 - (\vec{a} . \vec{b})^2)} \)
Answer: (b) \( 2\sqrt{(16 - (\vec{a} . \vec{b})^2)} \)
Solution: \( \vec{u} = \vec{a} - \vec{b} \), \( \vec{v} = \vec{a} + \vec{b} \), \( |\vec{a}| = |\vec{b}| = 2 \)
\( |\vec{u} \times \vec{v}| = |(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})| = 2|\vec{a} \times \vec{b}| = 2|\vec{a}|\ |\vec{b}| \sin\theta \)
\( = 2|\vec{a}|\ |\vec{b}| \sqrt{\frac{|\vec{a}|^2 |\vec{b}|^2 - (\vec{a}.\vec{b})^2}{|\vec{a}|^2 |\vec{b}|^2}} = 2\sqrt{16 - (\vec{a}.\vec{b})^2} \)

Question. Equation of a line which passes through a point with position vector \( \vec{c} \), parallel to the plane \( \vec{r} . \vec{n} = 1 \) and perpendicular to the line \( \vec{r} = \vec{a} + t\vec{b} \) is
(a) \( \vec{r} = \vec{c} + \lambda(\vec{c} - \vec{a}) \times \vec{n} \)
(b) \( \vec{r} = \vec{c} + \lambda(\vec{a} \times \vec{n}) \)
(c) \( \vec{r} = \vec{c} + \lambda(\vec{b} \times \vec{n}) \)
(d) \( \vec{r} = \vec{c} + \lambda(\vec{b} \times \vec{n}) \times \vec{a} \)
Answer: (c) \( \vec{r} = \vec{c} + \lambda(\vec{b} \times \vec{n}) \)
Solution: \( \vec{r}.\vec{n} = 1 \); \( \vec{r} = \vec{a} + t\vec{b} \)
Direction of line will be = \( (\vec{b} \times \vec{n}) \)
passing through = \( \vec{c} \)
\( \vec{r} = \vec{c} + \lambda(\vec{b} \times \vec{n}) \)

Question. Points L, M and N lie on the sides AB, BC and CA of the triangle ABC such that \( l(AL) : l(LB) = l(BM) : l(MC) = l(CN) : l(NA) = m : n \), then the areas of the triangles LMN and ABC are in the ratio
(a) \( \frac{m^2}{n^2} \)
(b) \( \frac{m^2 - mn + n^2}{(m + n)^2} \)
(c) \( \frac{m^2 - n^2}{m^2 + n^2} \)
(d) \( \frac{m^2 + n^2}{(m + n)^2} \)
Answer: (b) \( \frac{m^2 - mn + n^2}{(m + n)^2} \)
Solution: Assume A(0), B(\( \vec{b} \)), C(\( \vec{c} \))
Position vector of L, M, N
\( L\left(\frac{m\vec{b}}{m+n}\right) \), \( N\left(\frac{n\vec{c}}{m+n}\right) \), \( M\left(\frac{m\vec{c} + n\vec{b}}{m+n}\right) \)
Area of \( \Delta ABC = \frac{1}{2}|\vec{c} \times \vec{b}| \)
Area of \( \Delta LMN = \frac{1}{2}|\vec{LN} \times \vec{LM}| \)
\( = \frac{1}{2} \left( \frac{n\vec{c} - m\vec{b}}{m+n} \right) \times \left( \frac{m\vec{c} + n\vec{b} - m\vec{b}}{m+n} \right) \)
\( = \frac{1}{2(m+n)^2} | (n(n-m) + m^2) (\vec{c} \times \vec{b}) | \)
\( \frac{\text{Area}(\Delta LMN)}{\text{Area}(\Delta ABC)} = \frac{n^2 - mn + m^2}{(m+n)^2} \)

Question. Let \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) and \( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \) be three non-zero vectors such that \( \vec{c} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \). If the angle between \( \vec{a} \) and \( \vec{b} \) is \( \pi/6 \), then \( \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}^2 \) is equal to
(a) 0
(b) 1
(c) \( \frac{1}{4}(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \)
(d) \( \frac{3}{4}(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)(c_1^2 + c_2^2 + c_3^2) \)
Answer: (c) \( \frac{1}{4}(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \)
Solution: \( [\vec{a}\ \vec{b}\ \vec{c}]^2 = \begin{vmatrix} \vec{a}.\vec{a} & \vec{a}.\vec{b} & \vec{a}.\vec{c} \\ \vec{b}.\vec{a} & \vec{b}.\vec{b} & \vec{b}.\vec{c} \\ \vec{c}.\vec{a} & \vec{c}.\vec{b} & \vec{c}.\vec{c} \end{vmatrix} \)
Given \( \vec{c}.\vec{a} = 0 \) & \( \vec{c}.\vec{b} = 0 \)
\( = \begin{vmatrix} |\vec{a}|^2 & |\vec{a}| |\vec{b}| \cos\frac{\pi}{6} & 0 \\ |\vec{a}| |\vec{b}| \cos\frac{\pi}{6} & |\vec{b}|^2 & 0 \\ 0 & 0 & |\vec{c}|^2 \end{vmatrix} \)
\( = |\vec{c}|^2 [ |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \frac{\pi}{6} ] \)
\( = |\vec{c}|^2 |\vec{a}|^2 |\vec{b}|^2 \left[1 - \frac{3}{4}\right] = \frac{1}{4} |\vec{c}|^2 |\vec{a}|^2 |\vec{b}|^2 \)
\( = \frac{1}{4} (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \)

Question. \( [(\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}), (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}), (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})] \) is equal to
(a) \( [\vec{a}\ \vec{b}\ \vec{c}]^2 \)
(b) \( [\vec{a}\ \vec{b}\ \vec{c}]^3 \)
(c) \( [\vec{a}\ \vec{b}\ \vec{c}]^4 \)
(d) None of the options
Answer: (c) \( [\vec{a}\ \vec{b}\ \vec{c}]^4 \)
Solution: \( (\vec{a} \times \vec{b}) \times (\vec{b} \times \vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}]\vec{b} \)
\( (\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [\vec{b}\ \vec{c}\ \vec{a}]\vec{c} \)
\( (\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b}) = [\vec{c}\ \vec{a}\ \vec{b}]\vec{a} \)
So box product = \( [\vec{a}\ \vec{b}\ \vec{c}]^3 [\vec{a}\ \vec{b}\ \vec{c}] = [\vec{a}\ \vec{b}\ \vec{c}]^4 \)

Question. If the vectors \( a\hat{i} + \hat{j} + \hat{k} \), \( \hat{i} + b\hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} + c\hat{k} \) (\( a \ne b \ne c \ne 1 \)) are coplanar, then the value of \( \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} \) is equal to
(a) 1
(b) -1
(c) 0
(d) None of the options
Answer: (a) 1
Solution: \( [\vec{A}\ \vec{B}\ \vec{C}] = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \)
\( = \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1 \)

Question. The vectors \( \vec{a} = -4\hat{i} + 3\hat{k}, \vec{b} = 14\hat{i} + 2\hat{j} - 5\hat{k} \) are coinitial. The vector \( \vec{d} \) which is bisecting the angle between the vectors \( \vec{a} \) and \( \vec{b} \) and is having the magnitude \( \sqrt{6} \), is
(a) \( \hat{i} + \hat{j} + 2\hat{k} \)
(b) \( \hat{i} - \hat{j} + 2\hat{k} \)
(c) \( \hat{i} + \hat{j} - 2\hat{k} \)
(d) None of the options
Answer: (a) \( \hat{i} + \hat{j} + 2\hat{k} \)
Solution: \( \vec{d} = \hat{a} + \hat{b} = \frac{-4\hat{i} + 3\hat{k}}{5} + \frac{14\hat{i} + 2\hat{j} - 5\hat{k}}{15} = \frac{-12\hat{i} + 9\hat{k} + 14\hat{i} + 2\hat{j} - 5\hat{k}}{15} \)
\( = \frac{2\hat{i} + 2\hat{j} + 4\hat{k}}{15} = \frac{2}{15}(\hat{i} + \hat{j} + 2\hat{k}) \)

Question. A point taken on each median of a triangle divides the median in the ratio 1 : 3, reckoning from the vertex. Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is
(a) 5 : 13
(b) 25 : 64
(c) 13 : 32
(d) None of the options
Answer: (b) 25 : 64
Solution: \( A_1\left(\frac{\vec{b} + \vec{c}}{4}\right) \), \( B_1\left(\frac{\vec{c} + 3\vec{b}}{4}\right) \), \( C_1\left(\frac{\vec{b} + 3\vec{c}}{4}\right) \)
Area of \( \Delta A_1B_1C_1 = \frac{1}{2} |\vec{A_1B_1} \times \vec{A_1C_1}| \)
Area of \( \Delta ABC = \frac{1}{2} |\vec{b} \times \vec{c}| \)
Ratio = \( \frac{\text{Area of } \Delta A_1B_1C_1}{\text{Area of } \Delta ABC} = \frac{25}{64} \)

Question. If \( \vec{r} . (2\hat{i} + 3\hat{j} - 2\hat{k}) + 3/2 = 0 \) is the equation of a plane and \( \hat{i} - 2\hat{j} + 2\hat{k} \) is a point, then a point equidistant from the plane on the opposite side is
(a) \( 2\hat{i} + 3\hat{j} + \hat{k} \)
(b) \( \hat{i} + \hat{j} + 3\hat{k} \)
(c) \( 2\hat{i} + 3\hat{j} + 3\hat{k} \)
(d) \( 3(\hat{i} + \hat{j} + \hat{k}) \)
Answer: (a) \( 2\hat{i} + 3\hat{j} + \hat{k} \)

Question. If A(1, 1, 1), C(2, -1, 2), the vector equation of the line AB is \( \vec{r} = (\hat{i} + \hat{j} + \hat{k}) + t(6\hat{i} - 3\hat{j} + 2\hat{k}) \) and d is the shortest distance of the point C from AB, then
(a) B(6, -3, 2)
(b) B(5, -4, 1)
(c) d = \( \sqrt{2} \)
(d) d = \( \sqrt{6} \)
Answer: (c) d = \( \sqrt{2} \)
Solution: Let D is of c on line
\( AC = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{6} \)
AD = proj. of AC on AD = \( \frac{1(6) + (-2)(-3) + 1(2)}{7} = 2 \)
So shortest distance \( (CD)^2 = (AC)^2 - (AD)^2 = 6 - 4 = 2 \)
\( CD = \sqrt{2} \)

Question. If \( \vec{b} \) and \( \vec{c} \) are any two perpendicular unit vectors and \( \vec{a} \) is any vector, then \( (\vec{a} . \vec{b})\vec{b} + (\vec{a} . \vec{c})\vec{c} + \frac{\vec{a} . (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|^2} (\vec{b} \times \vec{c}) \) is equal to
(a) \( \vec{a} \)
(b) \( \vec{b} \)
(c) \( \vec{c} \)
(d) None of the options
Answer: (a) \( \vec{a} \)
Solution: Assume \( \vec{b} = \hat{i} \), \( \vec{c} = \hat{j} \) and \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \)
\( \vec{a}.\vec{b} = 1 \), \( \vec{a}.\vec{c} = 1 \)
\( \vec{b} + \vec{c} + \hat{k} = \hat{i} + \hat{j} + \hat{k} = \vec{a} \)

Question. If \( A_1, A_2, A_3, \dots, A_n \) are the vertices of a regular plane polygon with n sides and O is its centre then \( \sum_{i=1}^{n-1} (\vec{OA_i} \times \vec{OA_{i+1}}) \) equals
(a) \( (1 - n) (\vec{OA_2} \times \vec{OA_1}) \)
(b) \( (n - 1) (\vec{OA_2} \times \vec{OA_1}) \)
(c) \( n (\vec{OA_2} \times \vec{OA_1}) \)
(d) None of the options
Answer: (a) \( (1 - n) (\vec{OA_2} \times \vec{OA_1}) \)
Solution: If the radius of circum centre = r
\( |\vec{OA_i}| = r \) where i = 1, 2, 3, ...... n
\( \therefore \sum \vec{OA_i} \times \vec{OA_{i+1}} = \sum |\vec{OA_i}|\ |\vec{OA_{i+1}}| \sin \frac{2\pi}{n} \hat{n} \)
\( = \sum r^2 \sin \frac{2\pi}{n} \hat{n} = (n - 1) r^2 \sin \frac{2\pi}{n} \hat{n} \)
\( = (n - 1) (\vec{OA_1} \times \vec{OA_2}) = (1 - n) (\vec{OA_2} \times \vec{OA_1}) \)

Question. The set of values of 'm' for which the vectors \( \vec{a} = m\hat{i} + (m + 1)\hat{j} + (m + 8)\hat{k} \), \( \vec{b} = (m + 3)\hat{i} + (m + 4)\hat{j} + (m + 5)\hat{k} \) and \( \vec{c} = (m + 6)\hat{i} + (m + 7)\hat{j} + (m + 8)\hat{k} \) are non-coplanar is
(a) R
(b) R - {1}
(c) R - {1, 2}
(d) \( \phi \)
Answer: (a) R
Solution: \( \begin{vmatrix} m & m + 1 & m + 8 \\ m + 3 & m + 4 & m + 5 \\ m + 6 & m + 7 & m + 8 \end{vmatrix} \)
\( R_1 \to R_2 - R_1 \)
\( R_2 \to R_2 - R_3 \)
\( = \begin{vmatrix} 3 & 3 & -3 \\ -3 & -3 & -3 \\ m + 6 & m + 7 & m + 8 \end{vmatrix} = -9 \begin{vmatrix} 1 & 1 & -1 \\ 1 & 1 & 1 \\ m + 6 & m + 7 & m + 8 \end{vmatrix} \)
\( = -9[m + 8 - m - 7] - 1[m + 8 - m - 6] - 1[m + 7 - m - 6] \)
\( = -9 - 2 - 1 = -12 \)

Question. For any four points P, Q, R, S, \( |\vec{PQ} \times \vec{RS} - \vec{QR} \times \vec{PS} + \vec{RP} \times \vec{QS}| \) is equal to 4 times the area of the triangle
(a) PQR
(b) QRS
(c) PRS
(d) PQS
Answer: (b) QRS
Solution: Let S(\( \vec{o} \)), P(\( \vec{a} \)), Q(\( \vec{b} \)), R(\( \vec{c} \))
\( = |\vec{PQ} \times \vec{RS} - \vec{QR} \times \vec{PS} + \vec{RP} \times \vec{QS}| \)
\( = |(\vec{b} - \vec{a}) \times (-\vec{c}) - (\vec{c} - \vec{b}) \times (-\vec{a}) + (\vec{a} - \vec{c}) \times (-\vec{b})| \)
\( = 2 |(\vec{c} \times \vec{b})| = 2 (\vec{b} \times \vec{c}) = 4 \text{ Area of RS} \)

Question. The vector \( \hat{i} + x\hat{j} + 3\hat{k} \) is rotated through an angle of \( \cos^{-1} \frac{11}{14} \) and doubled in magnitude, then it becomes \( 4\hat{i} + (4x - 2)\hat{j} + 2\hat{k} \). The value of 'x' is
(a) -2/3
(b) 2/3
(c) 1/3
(d) 2
Answer: (d) 2
Solution: \( \vec{a} = (1, x, 3) \) \( \cos\theta = 11/14 \)
\( \vec{b} = (4, 4x - 2, 2) \)
\( \cos\theta = \frac{\vec{a}.\vec{b}}{|\vec{a}|\ |\vec{b}|} \), \( |\vec{b}| = 2 |\vec{a}| \)
\( \frac{11}{14} = \frac{4 + x(4x - 2) + 6}{2 |\vec{a}|^2} \Rightarrow x = 2 \) and \( x = -20/17 \)

Question. Given the three vectors \( \vec{a} = -2\hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 5\hat{j} \) and \( \vec{c} = 4\hat{i} + 4\hat{j} - 2\hat{k} \). The projection of the vector \( 3\vec{a} - 2\vec{b} \) on the vector \( \vec{c} \) is
(a) 11
(b) -11
(c) 13
(d) None of the options
Answer: (b) -11
Solution: \( \vec{a} = (-2, 1, 1) \), \( \vec{b} = (1, 5, 0) \), \( \vec{c} = (4, 4, -2) \)
\( \vec{d} = 3\vec{a} - 2\vec{b} = 3(-2, 1, 1) - 2(1, 5, 0) = (-6, 3, 3) - (2, 10, 0) = (-8, -7, 3) \)
Projection = \( |\vec{d}| \cos\theta = \frac{\vec{d}.\vec{c}}{|\vec{c}|} = \frac{-32 - 28 - 6}{\sqrt{16 + 16 + 4}} = \frac{-66}{6} = -11 \)

Question. If the acute angle that the vector, \( \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} \) makes with the plane of the two vectors \( 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \hat{i} - \hat{j} + 2\hat{k} \) is \( \cot^{-1} \sqrt{2} \) then
(a) \( \alpha (\beta + \gamma) = \beta\gamma \)
(b) \( \beta (\gamma + \alpha) = \gamma\alpha \)
(c) \( \gamma(\alpha + \beta) = \alpha\beta \)
(d) \( \alpha\beta + \beta\gamma + \gamma\alpha = 0 \)
Answer: (a) \( \alpha (\beta + \gamma) = \beta\gamma \)
Solution: Normal Vector \( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} = 5(\hat{i} - \hat{j} - \hat{k}) \)
Let \( \vec{A} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} \). If \( \theta \) is the angle between vector \( \vec{A} \) and plane then \( 90 - \theta \) will be the angle between normal and plane
\( \cos (90 - \theta) = \frac{5\alpha - 5\beta - 5\gamma}{5\sqrt{3} \sqrt{\alpha^2 + \beta^2 + \gamma^2}} \)
\( \sin^2 \theta = \frac{(\alpha - \beta - \gamma)^2}{3(\alpha^2 + \beta^2 + \gamma^2)} \Rightarrow \beta\gamma = \alpha(\beta + \gamma) \)

Question. If a line has a vector equation \( \vec{r} = 2\hat{i} + 6\hat{j} + \lambda(\hat{i} - 3\hat{j}) \), then which of the following statements hold goods ?
(a) the line is parallel to \( 2\hat{i} + 6\hat{j} \)
(b) the line passes through the point \( 2\hat{i} + 3\hat{j} \)
(c) the line passes through the point \( \hat{i} + 9\hat{j} \)
(d) the line is parallel to XY-plane
Answer: (c), (d)
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \)
So, \( x - 2 = \lambda \), \( y - 6 = - 3\lambda \)
for [B], \( x = 2 \), \( y = 3 \)
Since value of \( \lambda \) is not same hence [B] is rejected.
For [C], \( x = 1 \), \( y = 9 \)
Both equation gives \( \lambda = - 1 \)
hence [C] is correct.
Since in given equation \( \hat{k} = 0 \) here [D] is correct.

Question. The vector \( \frac{1}{3} (2\hat{i} - 2\hat{j} + \hat{k}) \) is
(a) a unit vector
(b) makes an angle \( \frac{\pi}{3} \) with the vector \( 2\hat{i} - 4\hat{j} - 3\hat{k} \)
(c) parallel to the vector \( -\hat{i} + \hat{j} - \frac{1}{2}\hat{k} \)
(d) Perpendicular to the vector \( 3\hat{i} + 2\hat{j} - 2\hat{k} \)
Answer: (a), (c), (d)
Let \( \vec{a} = \left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right) \)
\( | \vec{a} | = 1 \Rightarrow \) [A]
\( \cos \alpha = \frac{\left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right) \cdot (2, -4, -3)}{1 \cdot \sqrt{29}} \)
\( = \frac{\frac{4}{3} + \frac{8}{3} - 1}{\sqrt{29}} = \frac{3}{\sqrt{29}} \)
\( \vec{a} \cdot (3, 2, -2) = 0 \Rightarrow \) [D] is correct Answer.
\( \vec{a} \times (-1, 1, -1/2) = \left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right) \times \left(-1, 1, -\frac{1}{2}\right) \)
\( = \frac{2}{3}\hat{k} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} + \frac{1}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{1}{3}\hat{i} = 0 \)
\( \Rightarrow \) [C] is correct options.

Question. The vector \( \vec{c} \), directed along the external bisector of the angle between the vectors \( \vec{a} = 7\hat{i} - 4\hat{j} + 4\hat{k} \) and \( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \) with \( |\vec{c}| = 5\sqrt{6} \), is
(a) \( \frac{5}{3} (\hat{i} - 7\hat{j} + 2\hat{k}) \)
(b) \( \frac{5}{3} (\hat{i} + 7\hat{j} - 2\hat{k}) \)
(c) \( \frac{5}{3} (-\hat{i} + 7\hat{j} - 2\hat{k}) \)
(d) \( \frac{5}{3} (-\hat{i} + 7\hat{j} + 2\hat{k}) \)
Answer: (a), (c)
\( \vec{c} = \lambda(\hat{a} - \hat{b}) = \frac{\lambda}{9}(1, -7, -2) \)
\( |\vec{c}| = 5\sqrt{6} \Rightarrow |\vec{c}| = \sqrt{\frac{\lambda^2}{81}(1 + 49 + 4)} = 5\sqrt{6} \)
\( |\lambda| = 15 \)
\( \vec{c} = \frac{5}{3}(1, -7, -2) \) or \( \vec{c} = \frac{5}{3}(-1, 7, 2) \)
So [A] and [C] are correct Answer.

Question. If \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \), then the vectors \( \vec{a} - \vec{d} \) and \( \vec{b} - \vec{c} \) are
(a) collinear
(b) linearly independent
(c) perpendicular
(d) parallel
Answer: (a), (d)
\( \vec{a} \times (\vec{b} - \vec{c}) = \vec{d} \times (\vec{b} - \vec{c}) \)
\( (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = 0 \Rightarrow (\vec{a} - \vec{d}) \parallel (\vec{b} - \vec{c}) \)

Question. \( \hat{a} \) and \( \hat{b} \) are two given unit vectors at right angle. The unit vector equally inclined with \( \hat{a}, \hat{b} \) and \( \hat{a} \times \hat{b} \) will be
(a) \( -\frac{1}{\sqrt{3}} (\hat{a} + \hat{b} + \hat{a} \times \hat{b}) \)
(b) \( \frac{1}{\sqrt{3}} (\hat{a} + \hat{b} + \hat{a} \times \hat{b}) \)
(c) \( \frac{1}{\sqrt{3}} (\hat{a} + \hat{b} - \hat{a} \times \hat{b}) \)
(d) \( -\frac{1}{\sqrt{3}} (\hat{a} + \hat{b} - \hat{a} \times \hat{b}) \)
Answer: (a), (b)
Let \( \hat{a} = \hat{i} \), \( \hat{b} = \hat{j} \) then \( \hat{a} \times \hat{b} = \hat{i} \times \hat{j} = \hat{k} \)
Let required vector be \( \vec{r} \) and it makes \( \alpha, \beta, \gamma \) with x, y, z axis then given \( \alpha = \beta = \gamma \)
\( \Rightarrow \cos \alpha = \cos \beta = \cos \gamma \)
\( l^2 + m^2 + n^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{3}} \)
New \( \vec{r} = (l\hat{i} + m\hat{j} + n\hat{k}) \)
\( \vec{r} = \pm \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \Rightarrow \vec{r} = \pm \frac{1}{\sqrt{3}} (\hat{a} + \hat{b} + \hat{a} \times \hat{b}) \)

Question. If a, b, c are different real numbers and \( a\hat{i} + b\hat{j} + c\hat{k}, b\hat{i} + c\hat{j} + a\hat{k} \) and \( c\hat{i} + a\hat{j} + b\hat{k} \) are position vectors of three non-collinear points A, B, and C, then
(a) centroid of triangle ABC is \( \frac{a + b + c}{3} (\hat{i} + \hat{j} + \hat{k}) \)
(b) \( \hat{i} + \hat{j} + \hat{k} \) is equally inclined to the three vectors
(c) perpendicular from the origin to the plane of triangle ABC meet at centroid
(d) triangle ABC is an equilateral triangle.
Answer: (a), (b), (c), (d)
\( G = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{(a + b + c)}{3} (\hat{i} + \hat{j} + \hat{k}) \)
then [A] and [B] are correct.
\( |\vec{AB}| = |\vec{BC}| = |\vec{CA}| \)
then [D] is also correct.
[C] \( \Rightarrow \) (By observation)

Question. If \( \vec{z}_1 = a\hat{i} + b\hat{j} \) and \( \vec{z}_2 = c\hat{i} + d\hat{j} \) are two vectors in \( \hat{i} \) and \( \hat{j} \) system, where \( |\vec{z}_1| = |\vec{z}_2| = r \) and \( \vec{z}_1 \cdot \vec{z}_2 = 0 \), then \( \vec{w}_1 = a\hat{i} + c\hat{j} \) and \( \vec{w}_2 = b\hat{i} + d\hat{j} \) satisfy
(a) \( |\vec{w}_1| = r \)
(b) \( |\vec{w}_2| = r \)
(c) \( \vec{w}_1 \cdot \vec{w}_2 = 0 \)
(d) None of the options
Answer: (a), (b), (c)
\( |\vec{z}_1| = \sqrt{a^2 + b^2} = a = r \)
\( |\vec{z}_2| = \sqrt{c^2 + d^2} = d = r \)
\( |\vec{w}_1| = \sqrt{a^2 + c^2} = a = r \)
\( |\vec{w}_2| = \sqrt{b^2 + d^2} = d = r \)
\( \vec{w}_1 \cdot \vec{w}_2 = ab + cd \), then \( c = 0 \), \( b = 0 \)
so \( \vec{w}_1 \cdot \vec{w}_2 = 0 \). Hence A, B, C

Question. A line passes through a point A with position vector \( 3\hat{i} + \hat{j} - \hat{k} \) and parallel to the vector \( 2\hat{i} - \hat{j} + 2\hat{k} \). If P is a point on this line such that AP = 15 units, then the position vector of the point P is/are
(a) \( 13\hat{i} + 4\hat{j} - 9\hat{k} \)
(b) \( 13\hat{i} - 4\hat{j} + 9\hat{k} \)
(c) \( 7\hat{i} - 6\hat{j} + 11\hat{k} \)
(d) \( -7\hat{i} + 6\hat{j} - 11\hat{k} \)
Answer: (b), (d)
\( \vec{r} = (3, 1, -1) + \lambda(2, -1, 2) \)
P.V. of 'P' \( = 3 + 2\lambda, 1 - \lambda, -1 + 2\lambda \)
\( | \vec{AP} | = \sqrt{(2\lambda)^2 + (-\lambda)^2 + (2\lambda)^2} = 15 \)
\( \lambda^2 = 25 \Rightarrow \lambda = \pm 5 \)
for \( \lambda = 5 \Rightarrow \) [B] and for \( \lambda = -5 \Rightarrow \) [D]

Question. If \( \vec{a}, \vec{b} \) are two non-collinear unit vectors and \( \vec{a}, \vec{b}, x\vec{a} - y\vec{b} \) form a triangle, then
(a) \( x = -1; y = 1 \) and \( |\vec{a} + \vec{b}| = 2 \cos \left(\frac{\vec{a} \wedge \vec{b}}{2}\right) \)
(b) \( x = -1; y = 1 \) and \( \cos(\vec{a} \wedge \vec{b}) + |\vec{a} + \vec{b}| \cos(\vec{a} \wedge -(\vec{a} + \vec{b})) = -1 \)
(c) \( |\vec{a} + \vec{b}| = -2 \cot \left(\frac{\vec{a} \wedge \vec{b}}{2}\right) \cos \left(\frac{\vec{a} \wedge \vec{b}}{2}\right) \) and \( x = -1, y = 1 \)
(d) None of the options
Answer: (a), (b)
For a triangle
\( \vec{a} + \vec{b} + x\vec{a} - y\vec{b} = 0 \)
\( \Rightarrow x = -1, y = 1 \)
\( |\vec{a} + \vec{b}| = \sqrt{1 + 1 + 2 \cos \theta} = 2 \cos \frac{\theta}{2} \)

Question. The volume of a right triangular prism \( \text{ABCA}_1\text{B}_1\text{C}_1 \) is equal to 3. If the position vectors of the vertices of the base ABC are A(1, 0, 1), B(2, 0, 0) and C(0, 1, 0), then position vectors of the vertex \( \text{A}_1 \) can be
(a) (2, 2, 2)
(b) (0, 2, 0)
(c) (0, -2, 2)
(d) (0, -2, 0)
Answer: (a), (d)
\( \vec{n} = \vec{AB} \times \vec{AC} = (1, 2, 1) \)
\( \vec{A} = (1, 0, 1) \)
\( \vec{A}_1 = (0, -2, 0) \)
similarly \( \vec{A}_1 = (2, 2, 2) \)

Question. Which of the following statement(s) is/are true ?
(a) If \( \vec{n} \cdot \vec{a} = 0, \vec{n} \cdot \vec{b} = 0 \) and \( \vec{n} \cdot \vec{c} = 0 \) for some non-zero \( \vec{n} \) then \( [\vec{a} \vec{b} \vec{c}] = 0 \)
(b) there exist a vector having direction angles \( \alpha = 30^\circ \) and \( \beta = 45^\circ \)
(c) locus of point for which \( x = 3 \) and \( y = 4 \) is a line parallel to the Z-axis whose distance from the Z-axis is 5
(d) the vertices of a regular tetrahedron are OABC where 'O' is the origin. Then vector \( \vec{OA} + \vec{OB} + \vec{OC} \) is perpendicular to the plane ABC
Answer: (a), (c), (d)
[A]
\( \vec{n} \) is perpendicular to \( \vec{a}, \vec{b}, \vec{c} \)
\( \Rightarrow \vec{a}, \vec{b}, \vec{c} \) are coplanar hence \( [\vec{a} \vec{b} \vec{c}] = 0 \)
[B] \( l^2 + m^2 + n^2 = 1 \)
\( n^2 = 1 - \frac{3}{4} - \frac{1}{2} \)
\( n^2 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \) (not possible)
Hence false, [C] by observation
[D] same as in options [C] of Q.6

Question. If \( \vec{r} = \hat{i} + 5\hat{j} + 5\hat{k} + \lambda(2\hat{i} + \hat{j} + 4\hat{k}) \) and \( \vec{r} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 3 \) are the equations of a line and plane respectively, then which of the following is false ?
(a) line is perpendicular to the plane
(b) line lies in the plane
(c) line is parallel to the plane but does not lie in the plane
(d) line cuts the plane is in one point only
Answer: (a), (c), (d)
\( \vec{a} \cdot \vec{n} = d \)
\( 1 + 10 - 5 = 6 = d \)
\( 2 + 2 - 4 = 0 \)
Hence line lies in plane \( \Rightarrow \) [B] is true.

Question. Unit vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \) are coplanar. A unit vector \( \vec{d} \) is perpendicular to them. If \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \frac{1}{6}\hat{i} - \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k} \), and the angle between \( \vec{a} \) and \( \vec{b} \) is \( 30^\circ \), then \( \vec{c} \) is
(a) \( (\hat{i} - 2\hat{j} + 2\hat{k})/3 \)
(b) \( (\hat{i} - 2\hat{j} + 2\hat{k})/3 \)
(c) \( (-2\hat{i} - 2\hat{j} - \hat{k})/3 \)
(d) \( (-\hat{i} + 2\hat{j} - 2\hat{k})/3 \)
Answer: (a), (d)
\( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \left(\frac{1}{6}, -\frac{1}{3}, \frac{1}{3}\right) \)
\( (|\vec{a}| |\vec{b}| \sin 30^\circ \hat{n}) \times (\vec{c} \times \vec{d}) = \left(\frac{1}{6}, -\frac{1}{3}, \frac{1}{3}\right) \)
\( (\hat{n} \cdot \vec{d})\vec{c} - (\hat{n} \cdot \vec{c})\vec{d} = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \)
\( \vec{c} - 0 = \frac{(1, -2, 2)}{3} \)
\( \vec{c} = \frac{(\hat{i} - 2\hat{j} + 2\hat{k})}{3} = \) [A] and obviously [D] will also be the answer as there can be two unit vectors perpendicular to a plane.

Question. Let \( \vec{p} = 2\hat{i} + 3\hat{j} - a\hat{k} \), \( \vec{q} = b\hat{i} + 5\hat{j} - \hat{k} \) and \( \vec{r} = \hat{i} + \hat{j} + 3\hat{k} \). If \( \vec{p}, \vec{q}, \vec{r} \) are coplanar and \( \vec{p} \cdot \vec{q} = 20 \), then a and b have the values
(a) 1, 3
(b) 9, 7
(c) 5, 5
(d) 13, 9
Answer: (a), (d)
\( [\vec{p} \, \vec{q} \, \vec{r}] = \begin{vmatrix} 2 & 3 & -a \\ b & 5 & -1 \\ 1 & 1 & 3 \end{vmatrix} = 0 \quad \dots(1) \)
and \( \vec{p} \cdot \vec{q} = 20 \Rightarrow a = 5 - 2b \quad \dots(2) \)
from equation (1) : \( 5a - 9b - ab + 2q = 0 \)
put \( a = 5 - 2b \), \( b = 3, 9 \)
from equation (2) : (\( a = - 1 \) and \( - 13 \))

Question. The value(s) of \( \alpha \in [0, 2\pi] \) for which vector \( \vec{a} = \hat{i} + 3\hat{j} + (\sin 2\alpha)\hat{k} \) makes an obtuse angle with the z-axis and the vectors \( \vec{b} = (\tan \alpha)\hat{i} - \hat{j} + 2\sqrt{\sin \frac{\alpha}{2}}\hat{k} \) and \( \vec{c} = (\tan \alpha)\hat{i} + (\tan \alpha)\hat{j} + 3\sqrt{\csc \frac{\alpha}{2}}\hat{k} \) are orthogonal, is/are
(a) \( \tan^{-1} 3 \)
(b) \( \pi - \tan^{-1} 2 \)
(c) \( \pi + \tan^{-1} 3 \)
(d) \( 2\pi - \tan^{-1} 2 \)
Answer: (b), (d)
\( \vec{b} \cdot \vec{c} = 0 \)
\( \tan^2 \alpha + \tan \alpha - 6 = 0 \)
\( \tan \alpha = 3 \) and \( \tan \alpha = -2 \)
Also, \( \vec{a} \cdot (0, 0, 1) < 0 \)
\( \sin 2\alpha < 0 \Rightarrow x < 0 \)
\( \sin \alpha \cos \alpha < 0 \)
\( \Rightarrow \alpha \) must lie in second and fourth quadrant.
Now, \( \tan \alpha = 3 \) is rejected. Hence
\( \tan \alpha = - 2 \)
\( -\tan \alpha = 2 \Rightarrow \tan (\pi - \alpha) = 2 \Rightarrow \tan (2\pi - \alpha) = 2 \)
so [B] & [D] is correct options.

Question. If \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k}, \vec{b} = y\hat{i} + z\hat{j} + x\hat{k} \) and \( \vec{c} = z\hat{i} + x\hat{j} + y\hat{k} \), then \( \vec{a} \times (\vec{b} \times \vec{c}) \) is
(a) parallel to \( (y - z)\hat{i} + (z - x)\hat{j} + (x - y)\hat{k} \)
(b) orthogonal to \( \hat{i} + \hat{j} + \hat{k} \)
(c) orthogonal to \( (y + z)\hat{i} + (z + x)\hat{j} + (x + y)\hat{k} \)
(d) orthogonal to \( x\hat{i} + y\hat{j} + z\hat{k} \)
Answer: (a), (b), (c), (d)
\( \vec{m} = \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \)
\( \vec{m} = (xy + yz + zx) ((y - z)\hat{i} + (z - x)\hat{j} + (x - y)\hat{k}) \)
Now, \( \vec{m} \times ((y - z), (z - x), (x - y)) = 0 \Rightarrow \) [A]
\( \vec{m} \cdot (1, 1, 1) = 0 \Rightarrow \) [B]
\( \vec{m} \cdot ((y + z), (z + x), (x + y)) = 0 \Rightarrow \) [C]
and \( \vec{m} \cdot (x, y, z) = 0 \Rightarrow \) [D]
Hence A, B, C, D

Question. If \( \vec{a}, \vec{b}, \vec{c} \) be three non-zero vectors satisfying the condition \( \vec{a} \times \vec{b} = \vec{c} \) and \( \vec{b} \times \vec{c} = \vec{a} \), then
(a) \( \vec{a}, \vec{b}, \vec{c} \) are orthogonal in pairs
(b) \( [\vec{a} \vec{b} \vec{c}] = |\vec{a}|^2 \)
(c) \( [\vec{a} \vec{b} \vec{c}] = |\vec{c}|^2 \)
(d) \( |\vec{b}| = |\vec{c}| \)
Answer: (a), (b), (c)
\( \vec{a} \times \vec{b} = \vec{c} \Rightarrow \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \vec{c} \)
\( \Rightarrow \vec{a} \cdot \vec{c} = 0 \), \( \vec{b} \cdot \vec{c} = 0 \)
Now, \( \vec{b} \times \vec{c} = \vec{a} \)
\( \vec{b} \cdot \vec{a} = 0 \), \( \vec{c} \cdot \vec{a} = 0 \), hence [A],
\( \vec{a} \times \vec{b} = \vec{c} \Rightarrow \vec{c} \cdot (\vec{a} \times \vec{b}) = |\vec{c}|^2 \)
\( [\vec{c} \, \vec{a} \, \vec{b}] = [\vec{a} \, \vec{b} \, \vec{c}] = |\vec{c}|^2 \Rightarrow \) [C]
\( \vec{b} \times \vec{c} = \vec{a} \Rightarrow \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}|^2 \)
\( [\vec{a} \, \vec{b} \, \vec{c}] = |\vec{a}|^2 \Rightarrow \) [B]