Vector JEE Mathematics Worksheets Set 01

Subjective Questions

Question. The position vector of two points A and B are \( 6\vec{a} + 2\vec{b} \) and \( \vec{a} - 3\vec{b} \). If a point C divides AB in the ratio 3 : 2 then show that the position vector of C is \( 3\vec{a} - \vec{b} \).
Answer: \( \vec{c} = \frac{3(\vec{a} - 3\vec{b}) + 2(6\vec{a} + 2\vec{b})}{5} \)
\( \vec{c} = 3\vec{a} - \vec{b} \)

Question. If \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k}) \) are two lines, then find the equation of acute angle bisector of two lines.
Answer: \( \vec{r} = (1, 2, 3) + \lambda\underbrace{(1, -1, 1)}_{\vec{d}_1} \)
\( \vec{r} = (1, 2, 3) + \mu\underbrace{(1, 1, -1)}_{\vec{d}_2} \)
\( \cos \theta = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|} = \frac{1 - 1 - 1}{\sqrt{3}\sqrt{3}} = \frac{-1}{3} \)
so acute angle bisector
\( \vec{r} = (1, 2, 3) + t(\hat{a} - \hat{b}) \)
\( = (1, 2, 3) + \frac{t}{\sqrt{3}}[(1, -1, 1) - (1, 1, -1)] \)
\( = (1, 2, 3) + \lambda(\hat{j} - \hat{k}) \)

Question. (i) If \( \hat{e}_1 \) and \( \hat{e}_2 \) are two unit vectors such that \( \hat{e}_1 - \hat{e}_2 \) is also a unit vector, then find the angle \( \theta \) between \( \hat{e}_1 \) and \( \hat{e}_2 \).
(ii) Prove that \( \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right)^2 = \left(\frac{\vec{a} - \vec{b}}{|\vec{a}| |\vec{b}|}\right)^2 \)
Answer: (i) \( |\hat{e}_1 - \hat{e}_2|^2 = 1 \)
\( 1 + 1 - 2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \)
(ii) \( \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right)^2 = \frac{a^2}{a^4} + \frac{b^2}{b^4} - \frac{2\vec{a}.\vec{b}}{a^2b^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{2\vec{a}.\vec{b}}{a^2b^2} \)
\( \left(\frac{\vec{a} - \vec{b}}{|\vec{a}| |\vec{b}|}\right)^2 = \frac{1}{a^2b^2}(a^2 + b^2 - 2\vec{a}.\vec{b}) = \frac{1}{b^2} + \frac{1}{a^2} - \frac{2\vec{a}.\vec{b}}{a^2b^2} \)
LHS = RHS.

Question. (i) A vector \( \vec{c} \) is perpendicular to the vectors \( 2\hat{i} + 3\hat{j} - \hat{k}, \hat{i} - 2\hat{j} + 3\hat{k} \) and satisfies the condition \( \vec{c}.(2\hat{i} - \hat{j} + \hat{k}) + 6 = 0 \). Find the vector \( \vec{c} \).
(ii) Given \( |\vec{a}| = 10, |\vec{b}| = 2 \) and \( \vec{a}.\vec{b} = 12 \) then find \( |\vec{a} \times \vec{b}| \).
Answer: (i) \( \vec{c} = \lambda(\vec{a} \times \vec{b}), \vec{a} = (2, 3, -1), \vec{b} = (1, -2, 3) \)
\( \vec{a} \times \vec{b} = 7\hat{i} - 7\hat{j} - 7\hat{k} \)
\( \vec{c} = \lambda(7, -7, -7) = \mu(1, -1, -1) \)
\( \vec{c}.(2, -1, 1) + 6 = 0 \)
\( \mu(2 + 1 - 1) + 6 = 0 \implies \mu = -3 \implies \vec{c} = (-3, 3, 3) \)
(ii) \( |\vec{a}| = 10 ; |\vec{b}| = 2 \)
\( \vec{a}.\vec{b} = 12 \)
\( |\vec{a}| |\vec{b}| \cos\theta = 12 \implies \cos\theta = \frac{12}{20} = \frac{3}{5} \)
\( \sin\theta = \frac{4}{5} \)
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta = 10 \times 2 \times \frac{4}{5} = 16 \)

Question. Find the shortest distance between the lines :
\( \vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) \) and \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}) \)
Answer: Shortest distance between two lines
\( d = \left| \frac{(\vec{a}_2 - \vec{a}_1).(\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \)
\( \vec{a}_1 = (4, -1, 0) \quad \vec{a}_2 = (1, -1, 2) \)
\( \vec{b}_1 = (1, 2, -3) \quad \vec{b}_2 = (2, 4, -5) \)
\( \vec{a}_2 - \vec{a}_1 = (-3, 0, 2) \)
\( \vec{b}_1 \times \vec{b}_2 = (2, -1, 0) \)
\( (\vec{a}_2 - \vec{a}_1).(\vec{b}_1 \times \vec{b}_2) = -6 \)
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{4 + 1 + 0} = \sqrt{5} \)
Shortest distance \( d = \left| \frac{-6}{\sqrt{5}} \right| = \frac{6}{\sqrt{5}} \)

Question. (i) Given units vectors \( \hat{m}, \hat{n} \) and \( \hat{p} \) such that \( (\hat{m}^\wedge \hat{n}) = \hat{p}^\wedge(\hat{m}\times \hat{n}) = \alpha \), then find value of \( [\hat{n} \ \hat{p} \ \hat{m}] \) in terms of \( \alpha \).
(ii) Let \( \vec{a}, \vec{b}, \vec{c} \) be three units vectors and \( \vec{a}.\vec{b} = \vec{a}.\vec{c} = 0 \). If the angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{3} \), then find the value of \( |[\vec{a} \ \vec{b} \ \vec{c}]| \).
Answer: (i) \( (\hat{m}^\wedge \hat{n}) = \hat{p}^\wedge(\hat{m} \times \hat{n}) = \alpha \)
\( [\hat{n} \ \hat{p} \ \hat{m}] = [\hat{p} \ \hat{m} \ \hat{n}] = \hat{p} . (\hat{m} \times \hat{n}) \)
\( = [\hat{n} \ \hat{p} \ \hat{m}] . (\hat{m} \times \hat{n}) = |\hat{p}| \cos \alpha |\hat{m} \times \hat{n}| \)
\( = |\hat{p}| \cos \alpha |\hat{m}| |\hat{n}| \sin \alpha = \cos \alpha \sin \alpha \)
(ii) \( [\vec{a} \ \vec{b} \ \vec{c}] = |\vec{a}.(\vec{b} \times \vec{c})| \)
\( = |\vec{a}| |\vec{b} \times \vec{c}| = |\vec{a}| |\vec{b}| |\vec{c}| \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \)

Question. Let \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = 2\hat{i} - \hat{j} + \hat{k}, \vec{c} = 3\hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{d} = 3\hat{i} - \hat{j} - 2\hat{k} \), then
(i) If \( \vec{a} \times (\vec{b} \times \vec{c}) = p\vec{a} + q\vec{b} + r\vec{c} \), then find value of p, q are r.
(ii) Find the value of \( (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}).\vec{d} \)
Answer: \( \vec{a} = (1, 2, 3) \quad \vec{b} = (2, -1, 1) \)
\( \vec{c} = (3, 2, 1) \quad \vec{d} = (3, -1, -2) \)
(i) \( \vec{a} \times (\vec{b} \times \vec{c}) = p\vec{a} + q\vec{b} + r\vec{c} \)
\( (\vec{a}.\vec{c})\vec{b} - (\vec{a}.\vec{b})\vec{c} \)
\( 10\vec{b} - 3\vec{c} = p\vec{a} + q\vec{b} + r\vec{c} \)
\( p = 0, q = 10, r = -3 \)
(ii) \( (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}).\vec{d} \quad [(\vec{a} \times \vec{b}) = \vec{p}] \)
\( \vec{p} \times (\vec{a} \times \vec{c}).\vec{d} \)
\( [(\vec{p}.\vec{c})\vec{a} - (\vec{p}.\vec{a})\vec{c}] .\vec{d} \)
\( [\vec{a} \ \vec{b} \ \vec{c}](\vec{a}.\vec{d}) = (20) (-5) = -100 \)

Question. Given that \( \vec{x} + \frac{1}{p^2}(\vec{p}.\vec{x})\vec{p} = \vec{q} \), then show that \( \vec{p}.\vec{x} = \frac{1}{2}(\vec{p}.\vec{q}) \) and find \( \vec{x} \) in terms of \( \vec{p} \) and \( \vec{q} \).
Answer: \( \vec{x} + \frac{1}{p^2}(\vec{p}.\vec{x})\vec{p} = \vec{q} \) and \( \vec{p}.\vec{x} = \frac{1}{2}(\vec{p}.\vec{q}) \)
\( \vec{x} + \frac{(\vec{p}.\vec{q})\vec{p}}{2p^2} = \vec{q} \implies \vec{x} = \vec{q} - \frac{(\vec{p}.\vec{q})\vec{p}}{2p^2} \)

Question. Are the following set of vectors linearly independent?
(i) \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \vec{b} = 3\hat{i} - 6\hat{j} + 9\hat{k} \)
(ii) \( \vec{a} = -2\hat{i} - 4\hat{k}, \vec{b} = \hat{i} - 2\hat{j} - \hat{k}, \vec{c} = \hat{i} - 4\hat{j} + 3\hat{k} \)
Answer: (i) \( \vec{a} = (1, -2, 3) \quad \vec{b} = (3, -6, 9) \)
\( \vec{b} = 3(1, -2, 3) \)
\( \vec{b} = 3\vec{a} \) Linearly dependent
(ii) \( \vec{a} = (-2, 0, -4) \quad \vec{b} = (1, -2, -1) \)
\( \vec{c} = (1, -4, 3) \)
\( x\vec{a} + y\vec{b} + z\vec{c} = 0 \)
\( x(-2, 0, -4) + y(1, -2, -1) + z(1, -4, 3) = 0 \)
\( \implies x = 0, y = 0, z = 0 \) Linear Independent

Question. It is given that \( \vec{x} = \frac{\vec{b} \times \vec{c}}{[\vec{a}\vec{b}\vec{c}]}; \vec{y} = \frac{\vec{c} \times \vec{a}}{[\vec{a}\vec{b}\vec{c}]}, \vec{z} = \frac{\vec{a} \times \vec{b}}{[\vec{a}\vec{b}\vec{c}]} \), where \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar vectors. Show that \( \vec{x}, \vec{y}, \vec{z} \) also forms a non-coplanar system. Find the value of \( \vec{x}.(\vec{a} + \vec{b}) + \vec{y}.(\vec{b} + \vec{c}) + \vec{z}.(\vec{c} + \vec{a}) \).
Answer: \( \vec{x} = \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}; \ \vec{y} = \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}; \ \vec{z} = \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]} \)
\( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar \( \implies [\vec{a} \vec{b} \vec{c}] \neq 0 \)
\( [\vec{x} \vec{y} \vec{z}] = \vec{x}.(\vec{y} \times \vec{z}) \)
\( = \frac{1}{[\vec{a} \vec{b} \vec{c}]^3} [(\vec{b} \times \vec{c}).\{(\vec{c} \times \vec{a}) \times (\vec{a} \times \vec{b})\}] \)
Let \( \vec{p} = \vec{c} \times \vec{a} \)
\( = \frac{1}{[\vec{a} \vec{b} \vec{c}]^3} [(\vec{b} \times \vec{c}).\{(\vec{p}.\vec{b})\vec{a} - (\vec{p}.\vec{a})\vec{b}\}] \)
\( = \frac{1}{[\vec{a} \vec{b} \vec{c}]^3} [(\vec{b} \times \vec{c}).[\vec{a} \vec{b} \vec{c}]\vec{a}] \)
\( = \frac{[\vec{a} \vec{b} \vec{c}]^2}{[\vec{a} \vec{b} \vec{c}]^3} = \frac{1}{[\vec{a} \vec{b} \vec{c}]} \neq 0 \)
\( \vec{x}.(\vec{a} + \vec{b}) + \vec{y}.(\vec{b} + \vec{c}) + \vec{z}.(\vec{c} + \vec{a}) \)
\( \vec{x}.(\vec{a} + \vec{b}) = \frac{(\vec{b} \times \vec{c})}{[\vec{a} \vec{b} \vec{c}]}.(\vec{a} + \vec{b}) = \frac{[\vec{b} \vec{c} \vec{a}]}{[\vec{a} \vec{b} \vec{c}]} = 1 \quad \dots (1) \)
Similarly \( \vec{y}.(\vec{b} + \vec{c}) = 1 \quad \dots (2) \)
\( \vec{z}.(\vec{c} + \vec{a}) = 1 \quad \dots (3) \)
So sum of (1) (2) (3) = 3

Question. Find the distance between the parallel planes \( \vec{r}.(2\hat{i} - 3\hat{j} + 6\hat{k}) = 5 \) and \( \vec{r}.(6\hat{i} - 9\hat{j} + 18\hat{k}) + 20 = 0 \).
Answer: \( \vec{r}.(2\hat{i} - 3\hat{j} + 6\hat{k}) = 5 \)
\( \vec{r}.(6\hat{i} - 9\hat{j} + 18\hat{k}) = -20 \implies \vec{r}.(2\hat{i} - 3\hat{j} + 6\hat{k}) = \frac{-20}{3} \)
distance = \( \frac{5 + \frac{20}{3}}{\sqrt{4 + 9 + 36}} = \frac{35}{3 \times 7} = \frac{5}{3} \) units.

Question. If \( \vec{a}, \vec{b} \) are two unit vectors and \( \theta \) is the angle between them, then show that
(i) \( \sin\frac{\theta}{2} = \frac{1}{2} |\vec{a} - \vec{b}| \)
(ii) \( \cos\frac{\theta}{2} = \frac{1}{2} |\vec{a} + \vec{b}| \)
Answer: (i) \( \frac{1}{2} |\vec{a} - \vec{b}| = \frac{1}{2} \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta} \)
\( = \frac{1}{2} \sqrt{1 + 1 - 2\cos\theta} = \frac{1}{2} \sqrt{2(1 - \cos\theta)} \)
\( = \frac{1}{2} \sqrt{2 \cdot 2\sin^2\frac{\theta}{2}} = \sin\frac{\theta}{2} \)
(ii) \( \frac{1}{2} |\vec{a} + \vec{b}| = \frac{1}{2} \sqrt{1 + 1 + 2\cos\theta} = \cos\frac{\theta}{2} \)

Question. (i) Let \( \vec{A} = 2\hat{i} + \hat{k}, \vec{B} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{C} = 4\hat{i} - 3\hat{j} + 7\hat{k} \). Determine a vector \( \vec{R} \) satisfying \( \vec{R} \times \vec{B} = \vec{C} \times \vec{B} \) and \( \vec{R}.\vec{A} = 0 \)
(ii) Find vector \( \vec{v} \) which is coplanar with the vectors \( \hat{i} + \hat{j} - 2\hat{k} \) and \( \hat{i} - 2\hat{j} + \hat{k} \) and is orthogonal to the vector \( -2\hat{i} + \hat{j} + \hat{k} \). It is given that the projection of \( \vec{v} \) along the vector \( \hat{i} - \hat{j} + \hat{k} \) is equal to \( 6\sqrt{3} \).
Answer: (i) \( \vec{A} = (2, 0, 1) \quad \vec{B} = (1, 1, 1) \)
\( \vec{C} = (4, -3, 7) \)
\( \vec{R}.\vec{A} = 0 \)
\( \vec{R} \times \vec{B} = \vec{C} \times \vec{B} \implies \vec{A} \times (\vec{R} \times \vec{B}) = \vec{A} \times (\vec{C} \times \vec{B}) \)
\( (\vec{A}.\vec{B})\vec{R} - (\vec{A}.\vec{R})\vec{B} = (\vec{A}.\vec{B})\vec{C} - (\vec{A}.\vec{C})\vec{B} \)
\( (\vec{A}.\vec{B})\vec{R} = (\vec{A}.\vec{B})\vec{C} - (\vec{A}.\vec{C})\vec{B} \)
\( \vec{R} = \vec{C} - \frac{\vec{A}.\vec{C}}{\vec{A}.\vec{B}}\vec{B} = \vec{C} - \frac{(8+7)}{(2+1)}\vec{B} = \vec{C} - \frac{15}{3}\vec{B} \)
\( = \vec{C} - 5\vec{B} = (4, -3, 7) - 5(1, 1, 1) \)
\( \vec{R} = (-1, -8, -2) \)
(ii) Let \( \vec{v} = (\alpha, \beta, \gamma) \)
\( \vec{v}.(\vec{a} \times \vec{b}) = 0 \)
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & -2 & 1 \end{vmatrix} = -3\hat{i} - 3\hat{j} - 3\hat{k} \)
\( -3\alpha - 3\beta - 3\gamma = 0 \quad \dots (1) \)
\( -2\alpha + \beta + \gamma = 0 \quad \dots (2) \)
\( \frac{\vec{v}.\vec{c}}{|\vec{c}|} = 6\sqrt{3} \implies \frac{\alpha - \beta + \gamma}{\sqrt{3}} = 6\sqrt{3} \)
\( \alpha - \beta + \gamma = 18 \quad \dots(3) \)
By solving (1), (2), (3)
\( \alpha = 0, \beta = -9, \gamma = 9 \)
\( \vec{v} = 9(-\hat{j} + \hat{k}) \)

Question. In triangle ABC using vector method show that the distance between the circumcentre and the orthocentre is \( R\sqrt{1 - 8\cos A\cos B\cos C} \), where R is the circumradius of the triangle ABC.
Answer: Let circumcentre \( (\vec{O}) = \vec{O} \)
\( G = \frac{H + 2O}{3} \implies \vec{H} = 3\vec{G} \)
\( \vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \)
\( \vec{H} = \vec{a} + \vec{b} + \vec{c} \)
\( |\vec{OH}| = |\vec{a} + \vec{b} + \vec{c}| \)
\( = \sqrt{a^2 + b^2 + c^2 + 2\sum \vec{a}.\vec{b}} \)
\( = \sqrt{3R^2 + 2R^2\sum \cos 2A} \)
\( = \sqrt{3R^2 + 2R^2(-1 - 4\cos A \cos B \cos C)} \)
\( = R\sqrt{1 - 8\cos A \cos B \cos C} \)

Question. Find the equation of line of intersection of the planes \( \vec{r}.(3\hat{i} - \hat{j} + \hat{k}) = 1 \) and \( \vec{r}.(\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \).
Answer: \( \vec{r}.(3\hat{i} - \hat{j} + \hat{k}) = 1 \) and \( \vec{r}.(\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \)
\( \vec{p} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} = -2\hat{i} + 7\hat{j} + 13\hat{k} \)
Equation of planes
\( 3x - y + z = 1 \) and \( x + 4y - 2z = 2 \)
put z = 0 \( 3x - y = 1, x + 4y = 2 \)
\( x = \frac{6}{13}; \ y = \frac{5}{13} \)
Equation of line will be
\( \frac{x - \frac{6}{13}}{-2} = \frac{y - \frac{5}{13}}{7} = \frac{z - 0}{13} \)
or \( \vec{r} = \left(\frac{6}{13}, \frac{5}{13}, 0\right) + \lambda(-2, 7, 13) \)

Question. Find the point R in which the line AB cuts the plane CDE, where position vectors of points A, B, C, D, E are respectively \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k}, \vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}, \vec{c} = -4\hat{i} + 4\hat{k}, \vec{d} = 2\hat{i} - 2\hat{j} + 2\hat{k} \) and \( \vec{e} = 4\hat{i} + \hat{j} + 2\hat{k} \).
Answer: Let Point of intersection is P.
equation of plane CDE
\( -3(x + 4) + 2(y - 0) - 11(z - 4) = 0 \) ...(i)
Let any point on line AB is
P \( (1 + \lambda, 2 - \lambda, 1 + \lambda) \)
P will lie on plane ....(i)
so \( \lambda = 11/8 \)
so P(19/8, 11/8, 19/8)

Question. Examine for coplanarity of the following sets of points
(i) \( 4\hat{i} + 8\hat{j} + 12\hat{k}, \ 2\hat{i} + 4\hat{j} + 6\hat{k}, \ 3\hat{i} + 5\hat{j} + 4\hat{k}, \ 5\hat{i} + 8\hat{j} + 5\hat{k} \).
(ii) \( 3\vec{a} + 2\vec{b} - 5\vec{c}, \ 3\vec{a} + 8\vec{b} + 5\vec{c}, \ -3\vec{a} + 2\vec{b} + \vec{c}, \ \vec{a} + 4\vec{b} - 3\vec{c} \).
Answer: (i) Let \( \vec{A} = (4, 8, 12) \)
\( \vec{B} = (2, 4, 6); \ \vec{C} = (3, 5, 4) \)
\( \vec{D} = (5, 8, 5) \)
\( \vec{AB} = (-2, -4, -6) ; \ \vec{AC} = (-1, -3, -8) \)
\( \vec{AD} = (1, 0, -7) \)
\( [\vec{AB} \ \vec{AC} \ \vec{AD}] = 0 \)
\( \implies \) Hence points are coplanar.
(ii) Same as part (i)

Question. The position vectors of the angular points of a tetrahedron are \( A(3\hat{i} - 2\hat{j} + \hat{k}), B(3\hat{i} + \hat{j} + 5\hat{k}), C(4\hat{i} + \hat{k}) \) and \( D(\hat{i}) \). Then find the acute angle between the lateral faces ADC and the base ABC.
Answer: Vector normal to plane ABC is \( 8\hat{i} - 4\hat{j} + 3\hat{k} \)
Vector normal to plane ADC is \( -2\hat{i} + \hat{j} + 6\hat{k} \)
So acute angle between plane ABC & ADC
\( \cos \theta = \frac{2}{\sqrt{89}\sqrt{41}} \)

Advanced Subjective Questions

Question. If \(\vec{a}\) & \(\vec{b}\) are non collinear vectors such that, \(\vec{p} = (x+4y)\vec{a} + (2x+y+1)\vec{b}\) & \(\vec{q} = (y-2x+2)\vec{a} + (2x-3y-1)\vec{b}\), find x & y such that \(3\vec{p} = 2\vec{q}\).
Answer: \(3\vec{p} = (3x + 12y)\vec{a} + (6x + 3y + 3)\vec{b}\)
\(2\vec{q} = (2y - 4x + 4)\vec{a} + (4x - 6y - 2)\vec{b}\)
\(3\vec{p} = 2\vec{q} \Rightarrow 3x + 12y = -4x + 2y + 4\)
\(7x + 10 y = 4\) ....(1)
and \(6x + 3y + 3 = 4x - 6y - 2\)
\(2x + 9y = -5\) ....(2)
Solving equation 1 and 2
\(x = 2\), \(y = 1\)

Question. (a) Show that the points \(\vec{a} - 2\vec{b} + 3\vec{c}\) ; \(2\vec{a} + 3\vec{b} - 4\vec{c}\) & \(-7\vec{b} + 10\vec{c}\) are collinear.
(b) Prove that the points A = (1, 2, 3), B (3, 4, 7), C(–3, –2, –5) are collinear & find the ratio in which B divides AC.
Answer: (a) Let \(\vec{P}, \vec{Q}, \& \vec{R}\) be three vectors
\(\vec{PQ} = \vec{a} + 5\vec{b} - 7\vec{c}\)
\(\vec{QR} = -2\vec{a} - 10\vec{b} + 14\vec{c}\)
\(\vec{QR} = -2(\vec{PQ})\)
Here collinear.
(b) \(\vec{AB} = (2, 2, 4)\)
\(\vec{AC} = (-6, -6, -12) \Rightarrow \vec{AC} = -3\vec{AB}\)
Here \(\vec{A}, \vec{B}, \vec{C}\) are collinear
Let the ratio be \(k : 1\)
So, \(\frac{-3k - 1}{k - 1} = 3 \Rightarrow k = \frac{1}{3}\) (externally)

Question. Points X & Y are taken on the sides QR & RS, respectively of a parallelogram PQRS, so that \(\vec{QX} = 4\vec{XR}\) & \(\vec{RY} = 4\vec{YS}\). The line XY cuts the line PR at Z. Prove that \(\vec{PZ} = \left(\frac{21}{25}\right)\vec{PR}\).
Answer: P.V. of Z are :
\(\frac{4\mu\vec{r} + \vec{r}}{\mu + 1} = \frac{\lambda\vec{r} + \vec{q} - \vec{r}}{\lambda + 1}\)
\(\frac{\mu}{5(\mu + 1)} = \frac{1}{\lambda + 1}\) and \(\frac{4\mu + 1}{5(\mu + 1)} = \frac{\lambda - 1}{\lambda + 1}\)
\(\frac{1}{4(\mu + 1)} = \frac{1}{\lambda - 1} \Rightarrow \lambda = \frac{5\mu + 1}{\mu}\)
\(\mu = 4, \lambda = \frac{21}{4} \Rightarrow \frac{PZ}{PR} = \frac{\lambda}{\lambda + 1} = \frac{21}{25}\)

Question. Find out whether the following pairs of lines are parallel, non parallel; & intersecting, or non-parallel & non-intersecting.
(i) \(\vec{r}_1 = \hat{i} + \hat{j} + 2\hat{k} + \lambda(3\hat{i} - 2\hat{j} + 4\hat{k})\)
\(\vec{r}_2 = 2\hat{i} + \hat{j} + 3\hat{k} + \mu(-6\hat{i} + 4\hat{j} - 8\hat{k})\)
(ii) \(\vec{r}_1 = \hat{i} - \hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k})\)
\(\vec{r}_2 = 2\hat{i} + 4\hat{j} + 6\hat{k} + \mu(2\hat{i} + \hat{j} + 3\hat{k})\)
(iii) \(\vec{r}_1 = \hat{i} + \hat{k} + \lambda(\hat{i} + 3\hat{j} + 4\hat{k})\)
\(\vec{r}_2 = 2\hat{i} + 3\hat{j} + \mu(4\hat{i} - \hat{j} + \hat{k})\)
Answer: (i) \(\vec{r}_2 = (2, 1, 3) - 2\mu (3, -2, 4)\)
Here both lines are parallel.
(ii) \(1 + \lambda = 2 + 2\mu\) ....(1)
\(-1 - \lambda = 4 + \mu\) ....(2)
\(3 + \lambda = 6 + 3\mu\) ....(3)
from (1) and (2)
\(3\mu = -6 \Rightarrow \mu = -2\)
then \(\lambda = -3\)
Since \(\lambda = -3\) and \(\mu = -2\) satisfies (3), hence both lines are intersecting.
(iii) \(1 + \lambda = 2 + 4\mu\) ...(1)
\(3\lambda = 3 - \mu\) ...(2)
\(1 + 4\lambda = \mu\) ...(3)
from 2 and 3 \(7\lambda + 1 = 3 \Rightarrow \lambda = \frac{2}{7}\), \(\mu = \frac{15}{7}\)
since these values of \(\lambda\) and \(\mu\) not satisfy 1 hence non-intersecting.

Question. Let OACB be parallelogram with O at the origin & OC a diagonal. Let D be the mid point of OA. Using vector method prove that BD & CO intersect in the same ratio. Determine this ratio.
Answer: P.V. of P are \(\frac{\left(\frac{\lambda}{2} - 1\right)\vec{a} + \vec{c}}{\lambda + 1} = \frac{\mu\vec{c}}{\mu + 1}\)
\(\lambda = 2\), \(\mu = \frac{1}{2}\)
Here BD and CO intersect in the same ratio.

Question. In \(\Delta ABC\), points E and F divide sides AC and AB respectively so that \(\frac{AE}{EC} = 4\) and \(\frac{AF}{FB} = 1\). Suppose D is a point on side BC. Let G be the intersection of EF and AD and suppose D is situated that \(\frac{AG}{GD} = \frac{3}{2}\). If the ratio \(\frac{BD}{DC} = \frac{a}{b}\), where a and b are in their lowest form, find the value of (a + b).
Answer: P.V. of 'E' = \(\frac{(\vec{a} + 4\vec{c})}{5}\)
Now let \(FG : GE = \lambda : 1\)
So. P.V. of G are:
\(\frac{\frac{3\vec{a}}{a+b}\vec{c} + 2\vec{a}}{5} = \frac{\lambda\left(\frac{\vec{a} + 4\vec{c}}{5}\right) + \frac{\vec{a}}{2}}{\lambda + 1}\)
comparing coefficient of \(\vec{a}, \lambda = \frac{1}{2}\)
comparing coefficient of \(\vec{c}\) , \(\frac{4}{15} = \frac{3a}{5(a+b)}\)
\(\Rightarrow \frac{a}{b} = \frac{4}{5} \Rightarrow a + b = 9\)

Question. The resultant of two vectors \(\vec{a}\) & \(\vec{b}\) is perpendicular to \(\vec{a}\). If \(|\vec{b}| = \sqrt{2} |\vec{a}|\) show that the resultant of \(2\vec{a}\) & \(\vec{b}\) is perpendicular to \(\vec{b}\).
Answer: \((\vec{a} + \vec{b}) \cdot \vec{a} = 0\)
\(a^2 + \vec{a} \cdot \vec{b} = 0 \Rightarrow \vec{a} \cdot \vec{b} = -a^2\) ..... (1)
\((2\vec{a} + \vec{b}) \cdot \vec{b} = 0\)
\(2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = 0\)
\(-2a^2 + 2a^2 = 0\) Hence proved.

Question. Use vectors to prove that the diagonals of a trapezium having equal non parallel sides are equal & conversely.
Answer: \(\vec{DA} = \vec{CB}\)
\(\vec{DA} = (\vec{b} - \vec{c})\)
In \(\Delta DAC\), \(\vec{DA} + \vec{AC} = \vec{DC}\)
\(\vec{AC} = 2\vec{c} - \vec{b}\)
In \(\Delta DBC\), \(\vec{DB} + \vec{BC} = \vec{DC}\)
\(\vec{DB} = \vec{c} - \vec{BC}\)
\(\vec{DB} = \vec{c} - \vec{b} + \vec{c} = 2\vec{c} - \vec{b}\) Hence proved.

Question. Given three points on the xy plane on O(0, 0), A(1, 0) and B(–1, 0). Point P is moving on the plane satisfying the condition \((\vec{PA} \cdot \vec{PB}) + 3(\vec{OA} \cdot \vec{OB}) = 0\). If the maximum and minimum values of \(|\vec{PA}| |\vec{PB}|\) are M and m respectively then find the value of \(M^2 + m^2\).
Answer: \((1 - x, -y) \cdot (-1 - x, -y) - 1 + x^2 + y^2 - 3 = 0\)
\(x^2 + y^2 = 4\)
\(x^2 \in [0, 4]\)
\(\sqrt{(1 - x)^2 + y^2} \sqrt{(1 + x)^2 + y^2}\)
\(= \sqrt{5 - 2x} \sqrt{5 + 2x}\)
\(= \sqrt{25 - 4x^2}\)
\(M = \sqrt{25 - 4 \times 4} = \sqrt{9}\)
\(m = \sqrt{25 - 4 \times 0} = \sqrt{25}\)
\(M^2 + m^2 = 25 + 9 = 34\)

Question. In the plane of triangle ABC, squares ACXY, BCWZ are described, in the order given, externally to the triangle on AC & BC respectively. Given that \(\vec{CX} = \vec{b}\), \(\vec{CA} = \vec{a}\), \(\vec{CW} = \vec{x}\), \(\vec{CB} = \vec{y}\). Prove that \(\vec{a} \cdot \vec{y} + \vec{x} \cdot \vec{b} = 0\). Deduce that \(\vec{AW} \cdot \vec{BX} = 0\).
Answer: \(\vec{AW} \cdot \vec{BX} = (\vec{x} - \vec{a}) \cdot (\vec{b} - \vec{y})\)
\(= \vec{x} \cdot \vec{b} - \vec{x} \cdot \vec{y} - \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{y} = 0 = \text{RHS}\).

Question. A \(\Delta OAB\) is right angled at O ; squares OALM & OBPQ are constructed on the sides OA and OB externally. Show that the lines AP & BL intersect on the altitude through 'O'.
Answer: Let O be origin.
AP \(2x + y - a = 0\)
BL \(x + 2y - a = 0\)
point of intersection of AP & BL & let P \(\left(\frac{a}{3}, \frac{a}{3}\right)\)
equation of line \(\perp\) to AB & through origin \(y = x\) .
so it is clear that P lie on line \(y = x\).

Question. Given that \(\vec{u} = \hat{i} - 2\hat{j} + 3\hat{k}\) ; \(\vec{v} = 2\hat{i} + \hat{j} + 4\hat{k}\) ; \(\vec{w} = \hat{i} + 3\hat{j} + 3\hat{k}\) and \((\vec{u} \cdot \vec{R} - 10)\hat{i} + (\vec{v} \cdot \vec{R} - 20)\hat{j} + (\vec{w} \cdot \vec{R} - 20)\hat{k} = 0\). Find the unknown vector \(\vec{R}\).
Answer: Let \(\vec{R} = (x, y, z)\)
\((x - 2y + 3z - 10)\hat{i} + (2x + y + 4z - 20)\hat{j} + (x + 3y + 3z - 20)\hat{k} = 0\)
\(x - 2y + 3z = 10\), \(2x + y + 4z = 20\), \(x + 3y + 3z = 20\)
On solving: \(x = -1\), \(y = 2\), \(z = 5\)
So, \(\vec{R} = -\hat{i} + 2\hat{j} + 5\hat{k}\)

Question. The length of the edge of the regular tetrahedron DABC is 'a'. Point E and F are taken on the edges AD and BD respectively such that E divides DA and F divides BD in the ratio 2 : 1 each. Then find the area of triangle CEF.
Answer: Required area \(= \frac{1}{2} |\vec{EC} \times \vec{EF}|\)
\(= \frac{1}{2} \left|\left(\vec{c} - \frac{2\vec{a}}{3}\right) \times \left(\frac{\vec{b} - 2\vec{a}}{3}\right)\right|\)
\(= \frac{1}{2} \left|\frac{\vec{c} \times \vec{b}}{3} - \frac{2\vec{a} \times \vec{c}}{3} + \frac{\vec{b} \times 2\vec{a}}{9}\right|\)
\(= \frac{1}{2} \left|\left(\frac{a^2 \sin \alpha}{3}\right)\hat{\eta}_1 + \left(\frac{2a^2 \sin \alpha}{3}\right)\hat{\eta}_2 + \left(\frac{2a^2 \sin \alpha}{9}\right)\hat{\eta}_3\right|\)
\(= \frac{a^2}{6} \left|(\sin \alpha)\hat{\eta}_1 + (2 \sin \alpha)\hat{\eta}_2 + \left(\frac{2}{3} \sin \alpha\right)\hat{\eta}_3\right|\)
\(= \frac{a^2}{6} \sqrt{\sin^2 \alpha + 4 \sin^2 \alpha + \frac{4}{9} \sin^2 \alpha}\)
\(= \frac{a^2}{6} \times \sqrt{\frac{49 \sin^2 \alpha}{9}}\), \(\sin^2 \alpha = \frac{3}{4}\)
\(\Rightarrow \frac{7a^2}{18} \sin \alpha \Rightarrow \frac{7a^2}{18} \left(\frac{\sqrt{3}}{2}\right) = \frac{7a^2}{12\sqrt{3}}\) sq. units.

Question. The position vectors of the points A, B, C are respectively (1, 1, 1) ; (1, –1, 2) ; (0, 2, –1). Find a unit vector parallel to the plane determined by ABC perpendicular to the vector (1, 0, 1).
Answer: \(\vec{n}_2 = \vec{AB} \times \vec{AC} = (3, -1, -2)\)
\(\vec{n}_3 = (1, 0, 1)\)
\(\hat{n}_1 = \pm \frac{\vec{n}_2 \times \vec{n}_3}{|\vec{n}_2 \times \vec{n}_3|}\)
\(\hat{n}_1 = \pm \frac{1}{3\sqrt{3}}(-1, -5, 1)\)

Question. Let \(\begin{vmatrix} (a_1 - a)^2 & (a_1 - b)^2 & (a_1 - c)^2 \\ (b_1 - a)^2 & (b_1 - b)^2 & (b_1 - c)^2 \\ (c_1 - a)^2 & (c_1 - b)^2 & (c_1 - c)^2 \end{vmatrix} = 0\) and if the vectors \(\vec{\alpha} = \hat{i} + a\hat{j} + a^2\hat{k}\) ; \(\vec{\beta} = \hat{i} + b\hat{j} + b^2\hat{k}\) ; \(\vec{\gamma} = \hat{i} + c\hat{j} + c^2\hat{k}\) are non coplanar, show that the vectors \(\vec{\alpha}_1 = \hat{i} + a_1\hat{j} + a_1^2\hat{k}\) ; \(\vec{\beta}_1 = \hat{i} + b_1\hat{j} + b_1^2\hat{k}\) and \(\vec{\gamma}_1 = \hat{i} + c_1\hat{j} + c_1^2\hat{k}\) are coplanar.
Answer: \(\begin{vmatrix} (a_1 - a)^2 & (a_1 - b)^2 & (a_1 - c)^2 \\ (b_1 - a)^2 & (b_1 - b)^2 & (b_1 - c)^2 \\ (c_1 - a)^2 & (c_1 - b)^2 & (c_1 - c)^2 \end{vmatrix} = 0\)
\(R_1 \rightarrow R_1 - R_2 \quad R_2 \rightarrow R_2 - R_3\)
\(\begin{vmatrix} (a_1 - b_1)(a_1 + b_1 - 2a) & (a_1 - b_1)(a_1 + b_1 - 2b) & (a_1 - b_1)(a_1 + b_1 - 2c) \\ (b_1 - c_1)(b_1 + c_1 - 2a) & (b_1 - c_1)(b_1 + c_1 - 2b) & (b_1 - c_1)(b_1 + c_1 - 2c) \\ (c_1 - a)^2 & (c_1 - b)^2 & (c_1 - c)^2 \end{vmatrix} = 0\)
\((a_1 - b_1)(b_1 - c_1) \begin{vmatrix} (a_1 + b_1 - 2a) & (a_1 + b_1 - 2b) & (a_1 + b_1 - 2c) \\ (b_1 + c_1 - 2a) & (b_1 + c_1 - 2b) & (b_1 + c_1 - 2c) \\ (c_1 - a)^2 & (c_1 - b)^2 & (c_1 - c)^2 \end{vmatrix} = 0\)
\(C_1 \rightarrow C_1 - C_2 \quad \& \quad C_2 \rightarrow C_2 - C_3\)
\((a_1 - b_1)(b_1 - c_1) \begin{vmatrix} 2(b - a) & 2(c - b) & a_1 + b_1 - 2c \\ 2(b - a) & 2(c - b) & b_1 + c_1 - 2c \\ (b - a)(2c_1 - a - b) & (c - b)(2c_1 - b - c) & (c_1 - c)^2 \end{vmatrix} = 0\)
\((a_1 - b_1)(b_1 - c_1)(b - a)(c - b) \begin{vmatrix} 2 & 2 & a_1 + b_1 - 2c \\ 2 & 2 & b_1 + c_1 - 2c \\ 2c_1 - a - b & 2c_1 - b - c & (c_1 - c)^2 \end{vmatrix} = 0\)
\(C_1 \rightarrow C_1 - C_2\)
\((a_1 - b_1)(b_1 - c_1)(b - a)(c - b)(a - b)(c_1 - a_1) = 0\) ...(i)
Given \(\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \neq 0 \Rightarrow (a - b)(b - c)(c - a) \neq 0\)
So from (i) \((a_1 - b_1)(b_1 - c_1)(c_1 - a_1) = 0\)
\(\Rightarrow \begin{vmatrix} 1 & a_1 & a_1^2 \\ 1 & b_1 & b_1^2 \\ 1 & c_1 & c_1^2 \end{vmatrix} = 0\

Question. The length of an edge of a cube \(ABCDA_1B_1C_1D_1\) is equal to unity. A point E taken on the edge \(AA_1\) is such that \(|AE| = 1/3\). A point F is taken on the edge BC such that \(|BF| = 1/4\). If \(O_1\) is the centre of the cube, find the shortest distance of the vertex \(B_1\) from the plane of the \(\Delta O_1EF\).
Answer: Let coordinate axes be three co-terminus edge of cube. So coordinate of A, B, C, D, \(A_1, B_1, C_1, D_1\) will be as shown in diagram
Now \(\because |AE| = 1/3 \Rightarrow \text{So } \frac{EA}{EA_1} = \frac{1}{2}\)
P.V. of E \(\left(0, \frac{1}{3}, 1\right)\)
again \(|BF| = 1/4 \Rightarrow \frac{FB}{FC} = \frac{1}{3}\)
so P.V. of F \(\left(1, 0, \frac{3}{4}\right)\)
equation of plane OEF will be \(\vec{r} \cdot (5\hat{i} + 9\hat{j} + 8\hat{k})\)
so distance of \(B(1, 1, 1)\) from plane OEF
\(= \frac{|5 + 9 + 8 - 11|}{\sqrt{25 + 81 + 64}} = \frac{11}{\sqrt{170}}\)

Question. The vector \(\vec{OP} = \hat{i} + 2\hat{j} + 2\hat{k}\) turns through a right angle, passing through the positive x-axis on the way. Find the vector in its new position.
Answer: \(\cos \theta = \frac{1}{3}\)
\(\cos (90 - \theta) = \frac{a}{3}\)
\(\sin \theta = \frac{a}{3}\)
\(\frac{2\sqrt{2}}{3} = \frac{a}{3} \Rightarrow a = 2\sqrt{2}\)
\(a^2 + b^2 + c^2 = 3\)
\(a + 2b + 2c = 0\)
Solve & get b & c

Question. Find the point R in which the line AB cuts the plane CDE where \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}\), \(\vec{c} = -4\hat{j} + 4\hat{k}\), \(\vec{d} = 2\hat{i} - 2\hat{j} + 2\hat{k}\) & \(\vec{e} = 4\hat{i} + \hat{j} + 2\hat{k}\).
Answer: Equation of plane CDE : -
\(\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}\)
\(\vec{n} = \vec{CD} \times \vec{CE} = 6\hat{i} - 4\hat{j} + 2\hat{k}\)
\(\vec{r} \cdot (6, -4, 2) = 24 \Rightarrow 3x - 2y + z = 12\)
Equation of line AB :
\(\vec{r} = (1, 2, 1) + \lambda(1, -1, 1)\)
P.V. of 'R' : \(1 + \lambda\), \(2 - \lambda\), \(1 + \lambda\)
lies on plane CDE
\(3(1 + \lambda) - 2(2 - \lambda) + 1 + \lambda = 12 \Rightarrow \lambda = 2\)
P.V. of R : \((3\hat{i} + 3\hat{k})\)

Question. (A) Prove that \(|\vec{a} \times \vec{b}| = \sqrt{-\vec{b} \cdot [\vec{a} \times (\vec{a} \times \vec{b})]}\)
(B) Given that \(\vec{a}, \vec{b}, \vec{p}, \vec{q}\) are four vectors such that \(\vec{a} + \vec{b} = \mu\vec{p}\), \(\vec{b} \cdot \vec{q} = 0\) & \((\vec{b})^2 = 1\), where \(\mu\) is a scalar then prove that \(|(\vec{a} \cdot \vec{q})\vec{p} - (\vec{p} \cdot \vec{q})\vec{a}| = |\vec{p} \cdot \vec{q}|\)
Answer: (A) R.H.S. \(\sqrt{-\vec{b} \cdot \{(\vec{a} \cdot \vec{b})\vec{a} - a^2\vec{b}\}}\)
\(\Rightarrow \sqrt{-\vec{b} \cdot \{(ab \cos \theta)\vec{a} - a^2\vec{b}\}}\)
\(\Rightarrow \sqrt{a^2 b^2 (1 - \cos^2 \theta)} = ab \sin \theta\)
\(= |\vec{a} \times \vec{b}| = \text{L.H.S.}\)
(B) LHS. \(|(\vec{a} \cdot \vec{q})\vec{p} - (\vec{p} \cdot \vec{q})\vec{a}|\)
from \(\vec{a} + \vec{b} = \mu\vec{p}\) ....(1)
\(\vec{a} \cdot \vec{q} = \mu \vec{p} \cdot \vec{q}\) ....(2)
\(\Rightarrow |(\mu \vec{p} \cdot \vec{q})\vec{p} - (\vec{p} \cdot \vec{q})\vec{a}| \Rightarrow |(\vec{p} \cdot \vec{q})(\mu \vec{p} - \vec{a})|\)
\(\Rightarrow |(\vec{p} \cdot \vec{q})\vec{b}| = |\vec{p} \cdot \vec{q}| = \text{R.H.S.} \quad (\because |\vec{b}|^2 = 1)\)

Question. Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} - \hat{j} - \hat{k} \) be three vectors. A vector \( \vec{v} \) in the plane of \( \vec{a} \) and \( \vec{b} \), whose projection on \( \vec{c} \) is \( 1/\sqrt{3} \), is given by 
(a) \( \hat{i} - 3\hat{j} + 3\hat{k} \)
(b) \( -3\hat{i} - 3\hat{j} - \hat{k} \)
(c) \( 3\hat{i} - \hat{j} + 3\hat{k} \)
(d) \( \hat{i} + 3\hat{j} - 3\hat{k} \)
Answer: (c) \( 3\hat{i} - \hat{j} + 3\hat{k} \)

Question. The vector(s) which is/are coplanar with vectors \( \hat{i} + \hat{j} + 2\hat{k} \) and \( \hat{i} + 2\hat{j} + \hat{k} \), and perpendicular to the vector \( \hat{i} + \hat{j} + \hat{k} \) is/are
(a) \( \hat{j} - \hat{k} \)
(b) \( -\hat{i} + \hat{j} \)
(c) \( \hat{i} - \hat{j} \)
(d) \( -\hat{j} + \hat{k} \)
Answer: (a) \( \hat{j} - \hat{k} \) and (d) \( -\hat{j} + \hat{k} \)

Question. Let \( \vec{a} = -\hat{i} - \hat{k}, \vec{b} = -\hat{i} + \hat{j} \) and \( \vec{c} = \hat{i} + 2\hat{j} + 3\hat{k} \) be three given vectors. If \( \vec{r} \) is a vector such that \( \vec{r} \times \vec{b} = \vec{c} \times \vec{b} \) and \( \vec{r} \cdot \vec{a} = 0 \), then the value of \( \vec{r} \cdot \vec{b} \) is 
Answer: \( (\vec{r} - \vec{c}) \times \vec{b} = 0 \)
\( \Rightarrow \vec{r} = \vec{c} + \lambda\vec{b} = (1 - \lambda, 2 + \lambda, 3) \)
\( \vec{r} \cdot \vec{a} = 0 \Rightarrow \lambda = 4 \) so \( \vec{r} \cdot \vec{b} = 9 \)