Vector Algebra

$\mathrm{AB}=\mathrm{a}, \mathrm{BC}=\mathbf{b}, \mathrm{DA}=\mathbf{a}-\mathbf{b}$

In $\triangle A B C$, using vector concept, we have

$\begin{aligned} & & \mathbf{A B}+\mathbf{B C} & =\mathbf{A C} \\ \Rightarrow & & \mathbf{A C} & =\mathbf{a}+\mathbf{b} \quad \text{.... (i)} \end{aligned}$

In $,$\triangle A D C$,

$\begin{aligned} & A D+D C=A C \\ \Rightarrow \quad & D C=A C-A D \\ = & a+b-(b-a)=2 a \end{aligned}$

In $\triangle M D C$,

$\begin{aligned} & M D+D C=M C \\ \Rightarrow & M D=\frac{b}{2}-2 a \end{aligned}$

[$\because M$ is mid-point of $BC$]

In $\triangle A D X$,

$\begin{aligned} & \mathbf{A D}+\mathbf{D X}=\mathbf{A X} \\ & \mathbf{A D}+\frac{4}{5} \mathbf{D M}=\mathbf{A X} \\ \Rightarrow \quad \mathbf{A X} & =\mathbf{b}-\mathbf{a}+\frac{4}{5}\left(2 \mathbf{a}-\frac{\mathbf{b}}{2}\right) \\ & =\mathbf{b}-\mathbf{a}+\frac{8}{5} \mathbf{a}-\frac{2}{5} \mathbf{b}=\frac{3}{5}(\mathbf{a}+\mathbf{b}) \\ \Rightarrow \quad \mathbf{A X} & =\frac{3}{5} \mathbf{A C} \quad \text{[from Eq. (i)]} \end{aligned}$

which says that $\mathbf{A X} \| \mathbf{A C}$ and one point $A$ is common there.

Therefore, $A, X$ and $C$ are collinear.

$\begin{aligned} &\begin{array}{rrr} \text { } & (3 a-5 b) \perp(2 a+b) \\ \Rightarrow & \quad(3 a-5 b)(2 a+b)=0 \\ \Rightarrow & 6 a \cdot a+3 a \cdot b-10 b \cdot a-5 b \cdot b=0 \\ \Rightarrow & 6|a|^2+3 a \cdot b-10 a \cdot b-5|b|^2=0 \\ \Rightarrow & 6|a|^2-5|b|^2=7 a \cdot b \quad \text{.... (i)} \end{array}\\ &\text { Also, }(a+4 b) \perp(-a+b) \end{aligned}$

$\begin{array}{rr} \Rightarrow & (a+4 b) \cdot(-\mathbf{a}+\mathbf{b})=0 \\ \Rightarrow & -\mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}-4 \mathbf{b} \cdot \mathbf{a}+4 \mathbf{b} \cdot \mathbf{b}=0 \\ \Rightarrow & -|\mathbf{a}|^2+\mathbf{a} \cdot \mathbf{b}-4 \mathbf{a} \cdot \mathbf{b}+4|\mathbf{b}|^2=0 \\ \Rightarrow & -|\mathbf{a}|^2+4|\mathbf{b}|^2=3 \mathbf{a} \cdot \mathbf{b} \quad \text{... (ii)} \end{array}$

From Eqs. (i) and (ii), we have $ \begin{aligned} & \frac{1}{7}\left(6|\mathbf{a}|^2-5|\mathbf{b}|^2\right)=\frac{1}{3}\left(-|\mathbf{a}|^2+4|\mathbf{b}|^2\right) \\ \Rightarrow & 18|\mathbf{a}|^2-15|\mathbf{b}|^2=-7|\mathbf{a}|^2+28|\mathbf{b}|^2 \\ \Rightarrow & 25|\mathbf{a}|^2=43|\mathbf{b}|^2 \\ \because \quad & 3 \mathbf{a} \cdot \mathbf{b}=-|\mathbf{a}|^2+4|\mathbf{b}|^2 \quad \text{[from Eq. (ii)]}\\ & 3|\mathbf{a}||\mathbf{b}| \cos \theta=-\frac{43}{25}|\mathbf{b}|^2+4|\mathbf{b}|^2 \\ \Rightarrow & 3|\mathbf{a}||\mathbf{b}| \cos \theta=\frac{57}{25}|\mathbf{b}|^2 \\ \Rightarrow & \left.3 \sqrt{\frac{43}{25}}|| \mathbf{b}\right|^2 \cos \theta=\frac{57}{25}|\mathbf{b}|^2 \\ \Rightarrow & \frac{\sqrt{43}}{5} \cos \theta=\frac{19}{25} \Rightarrow \cos \theta=\frac{19}{5 \sqrt{43}} \\ \Rightarrow & \theta=\cos ^{-1}\left(\frac{19}{5 \sqrt{43}}\right) \end{aligned}$

$\begin{aligned} & \mathbf{a}=x \hat{i}+y \hat{j}+z \hat{k}, \\ & \quad x=2 y \text { and }|\mathbf{a}|=5 \sqrt{2} \\ & \therefore \quad x=|a| \cos \alpha \\ & \Rightarrow \quad x=5 \sqrt{2} \cos \alpha, \\ & y=5 \sqrt{2} \cos \beta \end{aligned}$

$\begin{aligned} & \text { and } z=5 \sqrt{2} \cos \gamma=5 \sqrt{2} \cos \left(135^{\circ}\right) \\ & =5 \sqrt{2}\left(-\frac{1}{\sqrt{2}}\right)=-5 \end{aligned}$

Now, $x=2 y$

$\begin{aligned} & \text { Now, } x=2 y \\ & \Rightarrow \quad 5 \sqrt{2} \cos \alpha=2 \times 5 \sqrt{2} \cos \beta \\ & \Rightarrow \quad \cos \alpha=2 \cos \beta \\ & \text { As, } \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad 4 \cos ^2 \beta+\cos ^2 \beta+\frac{1}{2}=1 \\ & \Rightarrow \quad \cos ^2 \beta=\frac{1}{10} \\ & \Rightarrow \quad \cos \beta=\frac{1}{\sqrt{10}} \\ & \therefore \quad x=2 y=2(5 \sqrt{2} \cos \beta)=2\left(5 \sqrt{2} \times \frac{1}{\sqrt{10}}\right)=2 \sqrt{5} \\ & \therefore \quad y=\frac{x}{2}=\sqrt{5} \\ & \therefore \quad a=2 \sqrt{5} \hat{i}+\sqrt{5} \hat{j}-5 \hat{k} \end{aligned}$

Position vectors of the vertices of the $\triangle A B C$ are $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$.

$\therefore$ Centroid of $\triangle A B C$ is $\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$

We know, when through the vertices, lines are drawn parallel to the sides to form a new triangle, then obtained such triangle will have the same centroid as the original triangle, which is also clear from the above graphical figure.

$\therefore$ Centroid of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$.

Suppose $\overrightarrow r = x\overrightarrow a + y\overrightarrow b + 2\overrightarrow c $

and $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k$

$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0 $

$ \Rightarrow {k^2}(\overrightarrow r - \overrightarrow b ) - {k^2}x\overrightarrow a + {k^2}(\overrightarrow r - \overrightarrow c ) - {k^2}y\overrightarrow b + {k^2}(\overrightarrow r - \overrightarrow a ) - {k^2}z\overrightarrow c = \overrightarrow 0 $

$ \Rightarrow 3\overrightarrow r -(\overrightarrow a + \overrightarrow b + \overrightarrow c ) - \overrightarrow r = \overrightarrow 0 $

$ \Rightarrow \overrightarrow r = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$

${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$

$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\left( {2\overrightarrow a + 3\overrightarrow b } \right)$

$9\overrightarrow a .\,\overrightarrow a + 6\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow b = 4\overrightarrow a \,.\,\overrightarrow a + 12\overrightarrow a \,.\,\overrightarrow b + 9\overrightarrow b \,.\,\overrightarrow b $

$5{\left| {\overrightarrow a } \right|^2} - 6\overrightarrow a \,.\,\overrightarrow b = 8{\left| {\overrightarrow b } \right|^2}$

$5{(8)^2} - 6.8\,.\,\left| {\overrightarrow b } \right|\cos 60^\circ = 8{\left| {\overrightarrow b } \right|^2}$ $\because$ $\left( \matrix{ {1 \over 8}\left| {\overrightarrow a } \right| = 1 \hfill \cr \Rightarrow \left| {\overrightarrow a } \right| = 8 \hfill \cr} \right)$

$40 - 3\left| {\overrightarrow b } \right| = {\left| {\overrightarrow b } \right|^2}$

$ \Rightarrow {\left| {\overrightarrow b } \right|^2} + 3\left| {\overrightarrow b } \right| - 40 = 0$

$\left| {\overrightarrow b } \right| = - 8$, $\left| {\overrightarrow b } \right| = 5$

(rejected)

$A(\widehat j)\,.\,B(10\widehat i)$

$H(h\widehat j + 10\widehat k)$

$G(10\widehat i + h\widehat j + 10\widehat k)$

$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$

$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$

$\cos \theta = {{\overrightarrow {AG} \overrightarrow {BH} } \over {\left| {\overrightarrow {AG} } \right|\left| {\overrightarrow {BH} } \right|}}$

${1 \over 5} = {{{h^2}} \over {{h^2} + 200}}$

$4{h^2} = 200 \Rightarrow h = 5\sqrt 2 $

$\left| {\overrightarrow a } \right| = \sqrt 3 $; $\overrightarrow a .\overrightarrow c = 3$; $\overrightarrow a \times \overrightarrow b = - 2\widehat i + \widehat j + \widehat k$, $\overrightarrow a \times \overrightarrow c = \overrightarrow b $

Cross with $\overrightarrow a $,

$\overrightarrow a \times (\overrightarrow a \times \overrightarrow c ) = \overrightarrow a \times \overrightarrow b $

$ \Rightarrow (\overrightarrow a .\overrightarrow c )\overrightarrow a - {a^2}\overrightarrow c = \overrightarrow a \times \overrightarrow b $

$ \Rightarrow 3\overrightarrow a - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$

$ \Rightarrow 3\widehat i + 3\widehat j + 3\widehat k - 3\overrightarrow c = - 2\widehat i + \widehat j + \widehat k$

$ \Rightarrow \overrightarrow c = {{5\widehat i} \over 3} + {{2\widehat j} \over 3} + {{2\widehat k} \over 3}$

$\therefore$ $\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \times \overrightarrow b ).\overrightarrow c = {{ - 10} \over 3} + {2 \over 3} + {2 \over 3} = - 2$

$\overrightarrow a = \left( {\overrightarrow b .\,\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\overrightarrow b } \right)\overrightarrow c $

$ = 1.2\cos \theta \overrightarrow b - \overrightarrow c $

$ \Rightarrow \overrightarrow a = 2\cos \theta \overrightarrow b - \overrightarrow c $

${\left| {\overrightarrow a } \right|^2} = {(2\cos \theta )^2} + {2^2} - 2.2\cos \theta \overrightarrow b \,.\,\overrightarrow c $

$ \Rightarrow 2 = 4{\cos ^2}\theta + 4 - 4\cos \theta .2\cos \theta $

$ \Rightarrow - 2 = - 4{\cos ^2}\theta $

$ \Rightarrow {\cos ^2}\theta = {1 \over 2}$

$ \Rightarrow {\sec ^2}\theta = 2$

$ \Rightarrow {\tan ^2}\theta = 1$

$ \Rightarrow \theta = {\pi \over 4}$

$ \therefore $ $1 + \tan \theta = 2$

$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$

$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$

$\overrightarrow a + \overrightarrow b = 3\widehat j + 5\widehat k;\overrightarrow a.\overrightarrow b = - 1 + 2 + 6 = 7$

$\left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightarrow b } \right) \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$

$\left( {\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow b } \right)} \right) \times \overrightarrow b } \right)$

$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b - 0} \right)} \right) \times \overrightarrow b $

$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \overrightarrow b $

$\left( {\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow b } \right) \times \overrightarrow b $

$\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a \times \overrightarrow b - \left( {\overrightarrow a .\overrightarrow a } \right)\left( {\overrightarrow b \times \overrightarrow b } \right)$

$\left( {\overrightarrow a .\overrightarrow b } \right)\left( {\overrightarrow a \times \overrightarrow b } \right)$

$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ i & j & k \cr 1 & 1 & 2 \cr { - 1} & 2 & 3 \cr } } \right| = - \widehat i - 5\widehat j + 3\widehat k$

$\therefore$ $7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {7\left( { - \widehat i - 5\widehat j + 3\widehat k} \right)} \right)$

$7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 3 & 5 \cr { - 1} & { - 5} & 3 \cr } } \right|$

$ \Rightarrow 34\widehat i - (5)\widehat j + (3\widehat k)$

$ \Rightarrow 34\widehat i - 5\widehat j + 3\widehat k$

$\therefore$ $7\left( {0\widehat i + 3\widehat j + 5\widehat k} \right) \times \left( { - \widehat i - 5\widehat j + 3\widehat k} \right)$

= $7(34\widehat i - 5\widehat j + 3\widehat k)$

Because vectors are coplanar

Hence, $\left| {\matrix{ a & a & c \cr 1 & 0 & 1 \cr c & c & b \cr } } \right| = 0$

$ \Rightarrow {c^2} = ab \Rightarrow c = \sqrt {ab} $

$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$

$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$

$\sin \theta = \pm \,{4 \over 5}$

$\therefore$ $\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $

$ = 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$

$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$

If the vectors are co-planar,

$\left| {\matrix{ {a + b + 2} & {a + 2b + c} & { - b - c} \cr {b + 1} & {2b} & { - b} \cr {b + 2} & {2b} & {1 - b} \cr } } \right| = 0$

Now, ${R_3} \to {R_3} - {R_2},{R_1} \to {R_1} - {R_2}$

So, $\left| {\matrix{ {a + 1} & {a + c} & { - c} \cr {b + 1} & {2b} & { - b} \cr 1 & 0 & 1 \cr } } \right| = 0$

$ = (a + 1)2b - (a + c)(2b + 1) - c( - 2b)$

$ = 2ab + 2b - 2ab - a - 2bc - c + 2bc$

$ = 2b - a - c = 0$

$\overrightarrow a = \lambda \overrightarrow b + \mu \overrightarrow c = \widehat i(2\lambda + \mu ) + \widehat j(\lambda - \mu ) + \widehat k(\lambda + \mu )$

$\overrightarrow a \,.\,\overrightarrow d = 0 = 3(2\lambda + \mu ) + 2(\lambda - \mu ) + 6(\lambda + \mu )$

$ \Rightarrow 14\lambda + 7\mu = 0 \Rightarrow \mu = - 2\lambda $

$ \Rightarrow \overrightarrow a = (0)\widehat i - 3\lambda \widehat j + ( - \lambda )\widehat k$

$ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {10} \left| \lambda \right| = \sqrt {10} \Rightarrow \left| \lambda \right| = 1$

$\lambda = 1$ or $ - 1$

$[\overrightarrow a \overrightarrow b \overrightarrow c ]$ = 0

$[\overrightarrow a \overrightarrow b \overrightarrow c ] + [\overrightarrow a \overrightarrow b \overrightarrow d ] + [\overrightarrow a \overrightarrow c \overrightarrow d ] = [\overrightarrow a \overrightarrow b + \overrightarrow c \overrightarrow d ]$

$ = \left| {\matrix{ 0 & { - 3\lambda } & \lambda \cr 3 & 0 & 2 \cr 3 & 2 & 6 \cr } } \right|$

$ = 3\lambda (12) + \lambda (6) = 42\lambda = - 42$

(1) $\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times (\overrightarrow b \times \overrightarrow c )} \right)$

$ = \overrightarrow a ( - \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow b ) = - 2\left( {\overrightarrow a \times (\overrightarrow b \times \overrightarrow c )} \right)$

$ = - 2(\overrightarrow a \times \overrightarrow a ) = \overrightarrow 0 $

(2) Projection of $\overrightarrow a $ on $\overrightarrow b \times \overrightarrow c $

$ = {{\overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )} \over {\left| {\overrightarrow b \times \overrightarrow c } \right|}} = {{\overrightarrow a .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right| = 2$

(3) $\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] + \left[ {\overrightarrow c \overrightarrow a \overrightarrow b } \right] = 2\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = 2\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$

$ = 2\overrightarrow a .\overrightarrow a = 2{\left| {\overrightarrow a } \right|^2} = 8$

(4) $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ and $\overrightarrow b \times \overrightarrow c = \overrightarrow a $

$ \Rightarrow \overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are mutually $ \bot $ vectors.

$\therefore$ $\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| \Rightarrow \left| {\overrightarrow b } \right| = {{\left| {\overrightarrow c } \right|} \over 2}$

Also, $\left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right| \Rightarrow \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| = 2 \Rightarrow \left| {\overrightarrow c } \right| = 2$ & $\left| {\overrightarrow b } \right| = 1$

${\left| {3\overrightarrow a + \overrightarrow b - 2\overrightarrow c } \right|^2} = (3\overrightarrow a + \overrightarrow b - 2\overrightarrow c ).(3\overrightarrow a + \overrightarrow b - 2\overrightarrow c )$

$ = 9{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2}$

$ = (9 \times 4) + 1 + (4 \times 4)$

$ = 36 + 1 + 16 = 53$

Projection of $\overrightarrow {BA} $

on ${\overrightarrow {BC} }$ is equal to

$ = \left| {\overrightarrow {BA} } \right|\cos \angle ABC$

$ = 7\left| {{{{7^2} + {3^2} - {5^2}} \over {2 \times 7 \times 3}}} \right| = {{11} \over 2}$

$\left| {\overrightarrow a } \right| = 3 = a;\overrightarrow a \,.\,\overrightarrow c = c$

Now, $\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $

$ \Rightarrow {c^2} + {a^2} - 2\overrightarrow c \,.\,\overrightarrow a = 8$

$ \Rightarrow {c^2} + 9 - 2(c) = 8$

$ \Rightarrow {c^2} - 2c + 1 = 0 \Rightarrow c = 1 = \left| {\overrightarrow c } \right|$

Also, $\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$

Given, $(\overrightarrow a \times \overrightarrow b ) = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {\pi \over 6}$

$ = (3)(1)(1/2)$

$ = 3/2$

$\overrightarrow a $ is perpendicular to $\overrightarrow b $

$ \therefore $ $\overrightarrow a $ . $\overrightarrow b $ = 0

Given, | $\overrightarrow a $ $\times$ $\overrightarrow b $ | = | $\overrightarrow a $ |

and | $\overrightarrow a $ | = | $\overrightarrow b $ |

$ \therefore $ | $\overrightarrow a $ $\times$ $\overrightarrow b $ | = | $\overrightarrow a $ | = | $\overrightarrow b $ | = k(assume)

Now, angle between $\overrightarrow a $ and ($\overrightarrow a $ + $\overrightarrow b $ + ($\overrightarrow a $ $\times$ $\overrightarrow b $))

$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

[Note $\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$]

$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$

Now, $|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$

$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $

$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$

$ = {k^2} + {k^2} + {k^2}$

$ \therefore $ $|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$

$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$

$ \therefore $ $\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$

$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$

$|\overrightarrow a | = 8,|\overrightarrow b | = 7,|\overrightarrow c | = 10$

Projection of $\overrightarrow {AB} $ on $\overrightarrow {AC} $

= $|\overrightarrow {AB} |cos\theta $

$ = 10\left( {{{|\overrightarrow {AB} {|^2} + |\overrightarrow {CA} {|^2} - |\overrightarrow {BC} {|^2}} \over {2.|\overrightarrow {CA} ||\overrightarrow {AB} {|}}}} \right)$

$ = 10\left( {{{{{10}^2} + {7^2} - {8^2}} \over {2(10)(7)}}} \right)$

$ = 10\left( {{{85} \over {140}}} \right)$

$ = {{85} \over {14}}$

${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$

(3p)² + 1 = (p + 1)² + 10

$ \Rightarrow $ 9p² $-$ p² $-$ 2p $-$ 10 = 0

$ \Rightarrow $ 8p² $-$ 2p $-$ 10 = 0

$ \Rightarrow $ 4p² $-$ p $-$ 5 = 0

$ \Rightarrow $ 4p² $-$ 5p + 4p $-$ 5 = 0

$ \Rightarrow $ (4p $-$ 5) (p + 1) = 0

$ \Rightarrow $ p = ${5 \over 4}$, $-$ 1

$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k\,$

$\overrightarrow {OP} \bot \overrightarrow {OQ} $

$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$

$\overrightarrow {PQ} = \left( { - 1 - x} \right)\widehat i + \left( {2 - y} \right)\widehat j + \left( {3x + 1} \right)\widehat k$

$\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $

$\sqrt {20} = \sqrt {{{\left( { - 1 - x} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} $

20 = 1 + x² + 2x + 4 + y² $-$ 4y + 9x² + 1 + 6x

20 = 10x² + y² + 8x + 6 $-$ 4y

20 = 10x² + 4x² + 8x + 6 $-$ 8x

14 = 14x² $ \Rightarrow $ x² = 1

Also, $\overrightarrow {OP} .\,\overrightarrow {OQ} = 0$

$ - x + 2y - 3x = 0$

$4x = 2y$

y = 2x

$ \therefore $ y² = 4x² $ \Rightarrow $ y² = 4

x = 1 as x > 0 and y = 2

$ \therefore $ $\left| {\matrix{ x & y & { - 1} \cr { - 1} & 2 & {3x} \cr 3 & z & { - 7} \cr } } \right| = 0$

$ \Rightarrow $ $\left| {\matrix{ 1 & 2 & { - 1} \cr { - 1} & 2 & 3 \cr 3 & z & { - 7} \cr } } \right|$ = 0

$ \Rightarrow $ 1($-$14 $-$3z) $-$ 2(7 $-$ 9) $-$ 1($-$z $-$6) = 0

$ \Rightarrow $ $-$14 $-$3z + 4 + z + 6 = 0

$ \Rightarrow $ 2z = $-$4 $ \Rightarrow $ z = $-$2

$ \therefore $ x² + y² + z² = 9

$\overrightarrow a = (2, - 3,4)$, $\overrightarrow b = (7,1, - 6)$

$\overrightarrow r \times \overrightarrow a - \overrightarrow r \times \overrightarrow b = 0$

$\overrightarrow r \times (\overrightarrow a - \overrightarrow b ) = 0$

$\overrightarrow r = \lambda (\overrightarrow a - \overrightarrow b )$

$\overrightarrow r = \lambda ( - 5\widehat i - 4\widehat j + 10\widehat k)$

$\overrightarrow r \,.\,(2, - 3,1) = ?$

Given $\overrightarrow r \,.\,(1,2,1) = - 3$

$\lambda ( - 5 - 8 + 10) = - 3 \Rightarrow \lambda = 1$

$ \therefore $ $( - 5, - 4,10)\,.\,(2, - 3,1)$

= - 10 + 12 + 10 = 12

Given $\overrightarrow r $ $\times$ $\overrightarrow a $ = $\overrightarrow b $ $\times$ $\overrightarrow r $

$ \Rightarrow $ $\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $

$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$

$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$

$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$

$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$

$ \because $ $\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$

$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$

$ \Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$

$\lambda (1 - 2\alpha ) = - 1$ .... (1)

$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$

$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$

$ \Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$ .... (2)

From (1) & (2)

$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$

$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$

$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$

$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$

($\alpha$, $\beta$) $ \equiv $ (2 cos 75$^\circ$, 2 sin 75$^\circ$)

Area = ${1 \over 2}$ (2 cos 75$^\circ$) (2 sin 75$^\circ$)

= sin(150$^\circ$) = ${1 \over 2}$ square unit

$\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}} $

$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$

$1 = \lambda x,y = - \lambda ,z = \lambda $

$x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k$

Unit vector $ = {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}$

$ = {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}$

Let ${\lambda ^2} = 1$, possible unit vector $ = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$

$\overrightarrow a \,.\,\overrightarrow b = 0$

$\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) = (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow a \,.\,\overrightarrow a )\overrightarrow b = - |\overrightarrow a {|^2}\overrightarrow b $

Now, $\overrightarrow a \times (\overrightarrow a \times ( - |\overrightarrow a {|^2}\overrightarrow b ))$

$ = - |\overrightarrow a {|^2}(\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ))$

$ = - |\overrightarrow a {|^2}( - |\overrightarrow a {|^2}\overrightarrow b ) = |\overrightarrow a {|^4}\overrightarrow b $

Given,

$\overrightarrow {OA} = 2\widehat i + 2\widehat j + 2\widehat k$

$\overrightarrow {OB} = \widehat i - \widehat j + 2\widehat k$

and $\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$

Also, $\left| {OB \times OC} \right| = 9/2$ ..... (i)

Now, $\overrightarrow {OB} \times \overrightarrow {OC} = \overrightarrow {OB} \times {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$

$ = {{ - \lambda } \over 2}\overrightarrow {OB} \times \overrightarrow {OA} = {\lambda \over 2}(\overrightarrow {OA} \times \overrightarrow {OB} )$

We need to find $\overrightarrow {OA} \times \overrightarrow {OB} $ for this,

($\because$ $\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a $ and $\overrightarrow a \times \overrightarrow a $ = 0)

$\overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 1 \cr 1 & { - 2} & 2 \cr } } \right|$

$ = 6\widehat i - 3\widehat j - 6\widehat k = 3(2\widehat i - \widehat j - 2\widehat k)$

So, $\overrightarrow {OB} \times \overrightarrow {OC} = {{3\lambda } \over 2}(2\widehat i - \widehat j - 2\widehat k)$

From Eq. (i), $\left| {{{3\lambda } \over 2}\sqrt {4 + 1 + 4} } \right| = 9/2$

$ \Rightarrow \left| {{{9\lambda } \over 2}} \right| = 9/2 \Rightarrow {9 \over 2}\left| \lambda \right| = {9 \over 2}$

$ \Rightarrow \left| \lambda \right| = 1 \Rightarrow \lambda = \pm 1$

But $\lambda$ > 0

$\therefore$ $\lambda$ = 1

So, $\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \overrightarrow {OA} )$ (By putting $\lambda$ = 1)

$ \Rightarrow \overrightarrow {OC} = {1 \over 2}( - \widehat i - 4\widehat j + \widehat k) = - {1 \over 2}\widehat i - 2\widehat j + {1 \over 2}\widehat k$

Option (a)

Projection of ${\overrightarrow {OC} }$ and ${\overrightarrow {OA} }$ $ = {{\overrightarrow {OC} \,.\,\overrightarrow {OA} } \over {\left| {\overrightarrow {OA} } \right|}}$

$ = {{{1 \over 2}( - 2 - 8 + 1)} \over 3} = {{ - 3} \over 2}$

Option (b)

Area of $\Delta OAB = {1 \over 2}\left| {\overrightarrow {OA} \times \overrightarrow {OB} } \right| = {1 \over 2}\left| {\overrightarrow {OA} } \right|\left. {\overrightarrow {OB} } \right|\sin 90^\circ $

$ = {1 \over 2}\left| {3 \times 3} \right| = {9 \over 2}$

Option (c)

Area of the $\Delta ABC = {1 \over 2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|$

$ = {1 \over 2}\left\| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 4} & 1 \cr { - 5/2} & { - 4} & {{{ - 1} \over 2}} \cr } } \right\|$

$ = {1 \over 2}\left| {6\widehat i - 3\widehat j - 6\widehat k} \right| = {1 \over 2} \times 3 \times 3 = {9 \over 2}$

Option (d)

The acute angle between the diagonals of the parallelogram with adjacent sides ${\overrightarrow {OA} }$ and ${\overrightarrow {OC} }$ = $\theta$

$ \Rightarrow {{(\overrightarrow {OA} + \overrightarrow {OC} )\,.\,(\overrightarrow {OA} - \overrightarrow {OC} )} \over {\left| {\overrightarrow {OA} + \overrightarrow {OC} } \right|\left| {\overrightarrow {OA} - \overrightarrow {OC} } \right|}} = \cos \theta $

$ \Rightarrow \cos \theta = {{\left( {{3 \over 2}\widehat i + {3 \over 2}\widehat k} \right)\,.\,\left( {{5 \over 2}\widehat i + 4\widehat j + {1 \over 2}\widehat k} \right)} \over {{3 \over 2}\sqrt 2 \times \sqrt {{{90} \over 4}} }}$

$ = {{18} \over {3\sqrt 2 \times \sqrt {90} }} = {1 \over {\sqrt 5 }} \ne {1 \over 2}$ [$\therefore$ $\theta$ $\ne$ $\pi$/3]

$\begin{aligned} \mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \mathbf{c} & =x \hat{\mathbf{i}}+(x-2) \hat{\mathbf{j}}-\hat{\mathbf{k}}\end{aligned}$

Given, $\mathbf{c}$ lies in the plane of $\mathbf{a}$ and $\mathbf{b}$.

Then, $[\mathbf{a}, \mathbf{b}, \mathbf{c}]=0$

$\begin{aligned} & \Rightarrow \quad[\mathbf{a}, \mathbf{b}, \mathbf{c}]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{array}\right|=0 \\ & 1(1-2(x-2)-\mathrm{l}(-1-2 x)+\mathrm{l}(x-2+x)=0 \\ & \Rightarrow \quad 1-2 x+4+1+2 x+x-2+x=0 \\ & \Rightarrow \quad 2 x+4=0 \\ & \Rightarrow \quad x=-2 \\ \end{aligned}$

The equation of line passing through $u$ and $v$ is $=\vec{u}+\lambda(\vec{v}-\vec{u})$

$\begin{aligned} \mathbf{r} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}) \\ & =(2+\lambda) \hat{\mathbf{i}}+(1-6 \lambda) \hat{\mathbf{j}} \end{aligned}$

For the point $P$,

$\begin{aligned} & \frac{5}{2} \hat{\mathbf{i}}-2 \hat{\mathbf{j}}=(2+\lambda) \hat{\mathbf{i}}+(1-6 \lambda) \hat{\mathbf{j}} \\ & \Rightarrow \quad 2+\lambda=\frac{5}{2} \Rightarrow \lambda=\frac{1}{2} \\ & \text { and } \quad 1-6 \lambda=1-6 \times \frac{1}{2}=-2 \\ \end{aligned}$

Thus, point $P$ lie in the line through $u$ and $v$.

For the point $Q$

$\begin{aligned} & 2+\lambda=7 / 3 \Rightarrow \lambda=1 / 3 \\ & \text{and} \quad 1-6 \lambda=-1 \end{aligned}$

Thus, point $Q$ also lie in the line passes through $u$ and $v$.

Points $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}),(2 \hat{\mathbf{i}}-\hat{\mathbf{j}})$ i.e. $(1,2)$ and $(2,-1)$.

Line joining points $(1,2)$ and $(2,-1)$ is

$y-2=\frac{-1-2}{2-1}(x-1)$

or $\quad y-2=-3 x+3$

$y+3 x=5$ ..... (i)

Points $(-\hat{\mathbf{i}}),(2 \hat{\mathbf{i}}) \text { i.e. }(-1,0),(2,0)$

Line joining points $(-1,0)(2,0)$ is

$y-0 =\frac{0-0}{2+1}(x+1)$

$\Rightarrow \quad y =0$ ..... (ii)

Solve Eqs. (i) and (ii) for $x$ and $y$, we get

$y=0, x=5 / 3$

$\therefore$ Point of intersection $\left(\frac{5}{3}, 0\right)$,

i.e. $\frac{5}{3} \hat{\mathbf{i}}+0 \hat{\mathbf{j}}=\frac{5}{3} \hat{\mathbf{i}}$

$\begin{aligned} & \frac{(\mathbf{a} \times \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{b})^2}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{(|\mathbf{a}| \cdot|\mathbf{b}| \sin \theta \hat{\mathbf{n}})^2+(|\mathbf{a}||\mathbf{b}| \cos \theta)^2}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta \hat{\mathbf{n}} \cdot \hat{\mathbf{n}}+|\mathbf{a}|^2|\mathbf{b}|^2 \cos ^2 \theta}{2|\mathbf{a}|^2|\mathbf{b}|^2} \\ & =\frac{\sin ^2 \theta+\cos ^2 \theta}{2}=\frac{1}{2}\end{aligned}$

$\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$

Let $\mathbf{d}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$

Given, $\mathbf{a} \cdot \mathbf{d}=0$

$ \begin{aligned} \Rightarrow & (\hat{\mathbf{i}}-\hat{\mathbf{j}}) \cdot(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}) =0 \\ \Rightarrow & p-q =0 \Rightarrow p=q \end{aligned}$

Given, $[\mathbf{b} \mathbf{c} \mathbf{d}]=0$

$\begin{array}{ll} \Rightarrow \left|\begin{array}{ccc} 0 & \mathrm{l} & -\mathrm{l} \\ -\mathrm{l} & 0 & \mathrm{l} \\ p & p & r \end{array}\right|=0 \\ \Rightarrow(-1)(-r-p)-1(-p)=0 \\ \Rightarrow r+p+p=0 \\ \Rightarrow r=-2 p \\ \therefore \mathbf{d}=p \hat{\mathbf{i}}+p \hat{\mathbf{j}}-2 p \hat{\mathbf{k}} \end{array}$

Also, $|\mathbf{d}|=1$

$\begin{aligned} & \Rightarrow \quad \sqrt{p^2+p^2+(-2 p)^2}=1 \\ & \Rightarrow \quad 6 p^2=1 \\ & \Rightarrow \quad p= \pm \frac{1}{\sqrt{6}} \\ & \therefore \quad d=\frac{ \pm \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{6}} \\ \end{aligned}$

$\begin{aligned}|\mathbf{u} \times \mathbf{v}| & =\| \mathbf{u}|\cdot| \mathbf{v}|\sin \theta \mathbf{n}| \\ & =\| \mathbf{u}|\cdot||\mathbf{v}| \sin 45^{\circ} \hat{\mathbf{n}} \mid \\ & =|| \mathbf{u}|\cdot| \mathbf{v}\left|\cdot \frac{1}{\sqrt{2}} \hat{\mathbf{n}}\right| \\ & =|\mathbf{u}| \cdot|\mathbf{v}| \cos 45^{\circ} \cdot 1=\mathbf{u} . \mathbf{v}\end{aligned}$