Vector Algebra

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

$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \,.\,\overrightarrow c )\overrightarrow b - (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow c = \overrightarrow b + \lambda \overrightarrow c $

If $\overrightarrow b $ & $\overrightarrow c $ are non-parallel

then $\overrightarrow a \,.\,\overrightarrow c = 1$ & $\overrightarrow a \,.\,\overrightarrow b = - \lambda $

but $\overrightarrow a \,.\,\overrightarrow b = 5 \Rightarrow \lambda = - 5$

$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$ and $|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$

${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \widehat b} \right|\left| {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right|}} = \cos \theta $

$ \Rightarrow \cos \theta = {{1 + 3\widehat a\widehat b + 2} \over {|\widehat a + \widehat b||\widehat a + 2\widehat b + 2(\widehat a \times \widehat b)|}}$

$|\widehat a + \widehat b{|^2} = 2 + \sqrt 2 $

$|\widehat a + 2\widehat b + 2(\widehat a \times \widehat b){|^2} = 1 + 4 + 4|\widehat a \times \widehat b{|^2} + 4\widehat a\widehat b$

$ = 5 + 4\,.\,{1 \over 2} + {4 \over {\sqrt 2 }} = 7 + 2\sqrt 2 $

So, ${\cos ^2}\theta = {{{{\left( {3 + {3 \over {\sqrt 2 }}} \right)}^2}} \over {(2 + \sqrt 2 )(7 + 2\sqrt 2 )}} = {{9\sqrt 2 (5\sqrt 2 + 3)} \over {164}}$

$ \Rightarrow 164{\cos ^2}\theta = 90 + 27\sqrt 2 $

$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$

$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$

For acute angle $\overrightarrow u \,.\,\overrightarrow v > 0$

$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$

$\because$ $b > 1$

Let ${\log _e}b = t \Rightarrow t > 0$ as $b > 1$

$a{t^2} + 6at - 12 > 0\,\,\,\,\,\,\,\forall t > 0$

$ \Rightarrow a \in \phi $

Clearly $\overrightarrow a $, $\overrightarrow b $, $\overrightarrow c $ are non-coplanar

$\left| {\matrix{ {1 + t} & {1 - t} & 1 \cr {1 - t} & {1 + t} & 2 \cr t & { - t} & 1 \cr } } \right| \ne 0$

$ \Rightarrow (1 + t)(1 + t + 2t) - (1 - t)(1 - t - 2t) + 1({t^2} - t - t - {t^2}) \ne 0$

$ \Rightarrow (3{t^2} + 4t + 1) - (1 - t)(1 - 3t) - 2t \ne 0$

$ \Rightarrow (3{t^2} + 4t + 1) - (3{t^2} - 4t + 1) - 2t \ne 0$

$ \Rightarrow t \ne 0$

$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$

$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow b = 0$

As given equation is identity

Coefficient of ${x^2} = $ coefficient of ${y^2} = $ coefficient of $xy = 0$

$ \Rightarrow |\overrightarrow a {|^2} = 3|\overrightarrow b {|^2} \Rightarrow |\overrightarrow b | = 3\sqrt 3 $

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

$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a ||\overrightarrow b |\sin \theta $

$ = 9.\,3\sqrt 3 .1 = 27\sqrt 3 $

$\overrightarrow a = \alpha \widehat i + \widehat j + \beta \widehat k$, $\overrightarrow b = 3\widehat i - 5\widehat j + 4\widehat k$

$\overrightarrow a \times \overrightarrow b = - \widehat i + 9\widehat j + 12\widehat k$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 1 & \beta \cr 3 & { - 5} & 4 \cr } } \right| = - \widehat i + 9\widehat j + 12\widehat k$

$4 + 5\beta = - 1 \Rightarrow \beta = - 1$

$ - 5\alpha - 3 = 12 \Rightarrow \alpha = - 3$

$\overrightarrow b - 2\overrightarrow a = 3\widehat i - 5\widehat j + 4\widehat k - 2\left( { - 3\widehat i + \widehat j - \widehat k} \right)$

$\overrightarrow b - 2\overrightarrow a = 9\widehat i - 7\widehat j + 6\widehat k$

$\overrightarrow b + \overrightarrow a = \left( {3\widehat i - 5\widehat j + 4\widehat k} \right) + \left( { - 3\widehat i + \widehat j - \widehat k} \right)$

$\overrightarrow b + \overrightarrow a = - 4\widehat j + 3\widehat k$

Projection of $\overrightarrow b - 2\overrightarrow a $ on $\overrightarrow b + \overrightarrow a $ is $ = {{\left( {\overrightarrow b - 2\overrightarrow a } \right)\,.\,\left( {\overrightarrow b + \overrightarrow a } \right)} \over {\left| {\overrightarrow b + \overrightarrow a } \right|}}$

$ = {{28 + 18} \over 5} = {{46} \over 5}$

Given, $\overrightarrow a = 2\widehat i - \widehat j + 5\widehat k$ and $\overrightarrow b = \alpha \widehat i + \beta \widehat j + 2\widehat k$

Also, $\left( {\left( {\overrightarrow a \times \overrightarrow b } \right) \times i} \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow \left( {\left( {\overrightarrow a \,.\,\widehat i} \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\widehat i} \right)\,.\,\overline a } \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow \left( {2\,.\,\overrightarrow b - \alpha \,.\,\overrightarrow a } \right)\,.\,\widehat k = {{23} \over 2}$

$ \Rightarrow 2\,.\,2 - 5\alpha = {{23} \over 2} \Rightarrow \alpha = {{ - 3} \over 2}$

Now, $\left| {\overrightarrow b \times 2j} \right| = \left| {\left( {\alpha \widehat i + \beta \widehat j + 2\widehat k} \right) \times 2\widehat j} \right|$

$ = \left| {2\alpha \widehat k + 0 - 4\widehat i} \right|$

$ = \sqrt {4{\alpha ^2} + 16} $

$ = \sqrt {4{{\left( {{{ - 3} \over 2}} \right)}^2} + 16} = 5$

Given : $\overrightarrow a = (\alpha ,1, - 1)$ and $\overrightarrow b = (2,1, - \alpha )$

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

$ = ( - \alpha + 1)\widehat i + ({\alpha ^2} - 2)\widehat j + (\alpha - 2)\widehat k$

Projection of $\overrightarrow c $ on $\overrightarrow d = - \widehat i + 2\widehat j - 2\widehat k$

$ = \left| {\overrightarrow c \,.\,{{\overrightarrow d } \over {|d|}}} \right| = 30$ {Given}

$ \Rightarrow \, = \left| {{{\alpha - 1 - 4 + 2{\alpha ^2} - 2\alpha + 4} \over {\sqrt {1 + 4 + 4} }}} \right| = 30$

On solving $\alpha = {{ - 13} \over 2}$ (Rejected as $\alpha > 0$)

and $\alpha = 7$

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

$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$

$\overrightarrow a \,.\,\overrightarrow b = 3$

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

$ \Rightarrow 5 + 9 = 6|\overrightarrow b {|^2}$

$ \Rightarrow |b {|^2} = {7 \over 3}$

$|\overrightarrow a - \overrightarrow b | = \sqrt {|\overrightarrow a {|^2} + |\overrightarrow b {|^2} - 2\overrightarrow a \,.\,\overrightarrow b } = \sqrt {{7 \over 3}} $

projection of $\overrightarrow b $ on $\overrightarrow a - \overrightarrow b = {{\overrightarrow b \,.\,(\overrightarrow a - \overrightarrow b )} \over {|\overrightarrow a - \overrightarrow b |}}$

$ = {{\overrightarrow b \,.\,\overrightarrow a - |\overrightarrow b {|^2}} \over {|\overrightarrow a - \overrightarrow b |}} = {{3 - {7 \over 3}} \over {\sqrt {{7 \over 3}} }}$

$ = {2 \over {\sqrt {21} }}$

$ \because \vec{a}+\vec{b}+\vec{c}=0 $

then $\bar{a}+\vec{c}=-\vec{b}$

then $(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \bar{b}$

$\therefore \quad \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0}\quad\dots(i)$

For $(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$

$ \begin{aligned} &|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1) \\\\ &|\vec{c}|=6-12 \sqrt{2} \text { (not possible) } \end{aligned} $

Hence (S1) is not correct

For (S2) : from (i) $\vec{b}+\vec{c}=-\vec{a}$

$\Rightarrow \vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}=-\vec{a} \cdot \vec{b}$

$\Rightarrow 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-\angle A C B)$

$\therefore \cos (\angle A C B)=\sqrt{\frac{2}{3}}$

$ \begin{aligned} &\therefore \angle A C B=\cos ^{-1} \sqrt{\frac{2}{3}} \\\\ &\therefore S(2) \text { is correct. } \end{aligned} $

Given,

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

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

and let $\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$

Now, $\overrightarrow a + \overrightarrow b = \widehat i + 2\widehat j$

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

$\overrightarrow c $ is coplanar with $\overrightarrow a $ and $\overrightarrow b $

$\therefore$ $\left[ {\overrightarrow c \,\overrightarrow a \,\overrightarrow b } \right] = 0$

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

$ \Rightarrow x( - 2) - y( - 1) + z( - 3) = 0$

$ \Rightarrow - 2x + y - 3z = 0$ ..... (1)

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

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 0 \cr { - 2} & 1 & { - 3} \cr } } \right|$

$ = - 6\widehat i + 3\widehat j + 5\widehat k$

Given, $\overrightarrow c \,.\,\left[ {\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right] = - 42$

$ \Rightarrow \left( {x\widehat i + y\widehat j + z\widehat k} \right)\,.\,\left( { - 6\widehat i + 3\widehat j + 5\widehat k} \right) = - 42$

$ \Rightarrow - 6x + 3y + 5z = - 42$ ...... (2)

Now, $\overrightarrow a - \overrightarrow b = \left( { - \widehat i + \widehat j + \widehat k} \right) - \left( {2\widehat i + \widehat j - \widehat k} \right)$

$ = - 3\widehat i + 0\widehat j + 2\widehat k$

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

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr x & y & z \cr { - 3} & 0 & 2 \cr } } \right|$

$ = 2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k$

Given,

$\left( {\overrightarrow c \times \left( {\overrightarrow a - \overrightarrow b } \right)} \right)\,.\,\widehat k = 3$

$ \Rightarrow \left( {2y\widehat i - \widehat j(2x + 3z) + 3y\widehat k} \right)\,.\,\widehat k = 3$

$ \Rightarrow 3y = 3$

$ \Rightarrow y = 1$

Putting value of $y = 1$ in equation (1) and (2) we get,

$ - 2x + 1 - 3z = 0$ ..... (3)

and $ - 6x + 3 + 5z = - 42$

$ \Rightarrow - 6x + 5z = - 45$ ..... (4)

Solving (3) and (4), we get

$x = 5$ and $z = - 3$

$\therefore$ $\overrightarrow c = 5\widehat i + \widehat j - 3\widehat k$

$ \Rightarrow {\left| {\overrightarrow c } \right|^2} = {5^2} + {1^2} + {( - 3)^2} = 35$

$\overrightarrow{A B} \| \overrightarrow{A C}$ if

$\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2}$

$\Rightarrow \alpha=1$

$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)

Mid point of $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$

$A M=\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$

$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$

$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$

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

Projection of $\overrightarrow a $ on $\overrightarrow c $ is

${{\overrightarrow a \,.\,\overrightarrow c } \over {|\overrightarrow b |}} = {{10} \over 3}$

${{\alpha + 6 + 2} \over {\sqrt {{1^2} + {2^2} + {{( - 2)}^2}} }} = {{\alpha + 8} \over 3} = {{10} \over 3}$

$\therefore$ $\alpha$ = 2

$\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - \beta } & 4 \cr 1 & 2 & { - 2} \cr } } \right| = (2\beta - 8)\widehat i + 10\widehat j + (6 + \beta )\widehat k = - 6\widehat i + 10\widehat j + 7\widehat k$

$2\beta - 8 = - 6$ & $6 + \beta = 7$

$\therefore$ $\beta$ = 1

$\alpha + \beta = 2 + 1 = 3$

$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$

$\therefore$ $\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 2 & { - 1} \cr { - 2} & \alpha & 1 \cr } } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$

Now $\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)} $

$ \Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$

$ \Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$

$\therefore$ $\alpha = \, \pm \,3$

Now, $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$

Let $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$

and $\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$ ..... (i)

and $\overrightarrow a \times (2\widehat i + \widehat k) = 2\widehat i - 13\widehat j - 4\widehat k$

$ \Rightarrow {a_2}\widehat i + (2{a_3} - {a_1})\widehat j - 2{a_2}\widehat k = 2\widehat i - 13\widehat j - 4\widehat k$

$\therefore$ ${a_2} = 2$ ..... (ii)

and ${a_1} - 2{a_3} = 13$ ..... (iii)

From eq. (i) and (iii) : ${a_1} = 3$ and ${a_3} = - 5$

$\therefore$ $\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k$

$\therefore$ projection of $\overrightarrow a $ on $2\widehat i + 2\widehat j + \widehat k = {{6 + 4 - 5} \over 3} = {5 \over 3}$

$\because$ $\overrightarrow a $ and $\overrightarrow b $ be the vectors along the diagonals of a parallelogram having area 2$\sqrt2$.

$\therefore$ ${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2 $

$|\overrightarrow a ||\overrightarrow b |\sin \theta = 4\sqrt 2 $

$ \Rightarrow |\overrightarrow b |\sin \theta = 4\sqrt 2 $ ..... (i)

and $|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$

$|\overrightarrow a ||\overrightarrow b |\cos \theta = |\overrightarrow a ||\overrightarrow b |\sin \theta $

$ \Rightarrow \tan \theta = 1$

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

By (i) $|\overrightarrow b | = 8$

Now $\overrightarrow c = 2\sqrt 2 (\overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b $

$ \Rightarrow \overrightarrow c \,.\,\overrightarrow b = - 2|\overrightarrow b {|^2} = - 128$ ...... (ii)

and $\overrightarrow c \,.\,\overrightarrow c = 8|\overrightarrow a \times \overrightarrow b {|^2} + 4|\overrightarrow b {|^2}$

$ \Rightarrow |\overrightarrow c {|^2} = 8.32 + 4.64$

$ \Rightarrow |\overrightarrow c | = 16\sqrt 2 $ ..... (iii)

From (ii) and (iii)

$|\overrightarrow c ||\overrightarrow b |\cos \alpha = - 128$

$ \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt 2 }}$

$\alpha = {{3\pi } \over 4}$

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

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

Now, $\overrightarrow b \times \overrightarrow c = \overrightarrow a $

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

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

$ \Rightarrow (\widehat i + \widehat j - \widehat k)(2\widehat i - 3\widehat j + 2\widehat k) = 0$

$ = 2 - 3 - 2 = 0$

$ \Rightarrow - 3 = 0$ (Not possible)

$\Rightarrow$ No possible value of $\overrightarrow b $ is possible.

$\because$ $\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = 3\overrightarrow b - \overrightarrow c = \overrightarrow u $

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

$\overrightarrow c \times \left( {\overrightarrow b \times \overrightarrow a } \right) = 3\overrightarrow b - 2\overrightarrow a = \overrightarrow w $

$\therefore$ $\overrightarrow u + \overrightarrow v = \overrightarrow w $

So, vectors $\overrightarrow u $, $\overrightarrow v $ and $\overrightarrow w $ are coplanar, hence their Scalar triple product will be zero.

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$

$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$

${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$

$ \Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$

$\therefore$ $\lambda = 5\sqrt 3 $

$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$

Let $\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$

$\overrightarrow a \,.\,\overrightarrow b = 0$ and $[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$

$ \Rightarrow p + q + r = 0$ ..... (i)

& $\left| {\matrix{ p & q & r \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right| = 0 \Rightarrow \matrix{ {q = r} \cr {p = - 2r} \cr } $

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

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

Now $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$

$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$

$\Rightarrow$ Projection of $\overrightarrow b $ on $3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$

$ = \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$

Projection $ = {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2 $

$\therefore$ B is correct.

$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$

$ \Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4$

$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\widehat a \times \widehat b} \right|^2} + 2\widehat a\,.\,\widehat b = 4$

$\therefore$ $\cos \theta = \cos 2\theta $

$\therefore$ $\theta = {{2\pi } \over 3}$

where $\theta$ is angle between $\widehat a$ and $\widehat b$.

$\therefore$ $2\left| {\widehat a \times \widehat b} \right| = \sqrt 3 = \left| {\widehat a - \widehat b} \right|$

(S1) is correct.

And projection of $\widehat a$ on $(\widehat a + \widehat b) = \left| {{{\widehat a\,.\,(\widehat a + \widehat b)} \over {\left| {\widehat a + \widehat b} \right|}}} \right| = {1 \over 2}$

(S2) is correct.

$\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$

$ \begin{aligned} &\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\ &\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a}) \end{aligned} $

$ \begin{aligned} &\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\ &\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\ &\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\ &\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3}) \end{aligned} $

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 $

We know, Volume(V) = $\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]$

$ \Rightarrow $ 158 = $\left| {\matrix{ 1 & 1 & n \cr 2 & 4 & { - n} \cr 1 & n & 3 \cr } } \right|$

$ \Rightarrow $ (12 + n²) – (6 + n) + n(2n–4)=158

$ \Rightarrow $ 3n² –5n + 6 –158 = 0

$ \Rightarrow $ 3n² – 5n – 152 = 0

$ \Rightarrow $ 3n² – 24n + 19n – 152 = 0

$ \Rightarrow $ (3n + 19) (n–8) = 0

$ \Rightarrow $ n = 8, $ - {{19} \over 3}$ (rejected)

$ \therefore $ $\overrightarrow a = \widehat i + \widehat j + 8\widehat k$,

$\overrightarrow b = 2\widehat i + 4\widehat j - 8\widehat k$ and

$\overrightarrow c = \widehat i + 8\widehat j + 3\widehat k$

Now ${\overrightarrow a .\overrightarrow c }$ = 1 + 8 + 24 = 33

${\overrightarrow b .\overrightarrow c }$ = 2 + 32 - 24 = 10

$f(x) = \overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )$

$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$

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

$ = x({x^2} + 2) + 2( - 2a + 7) + 3(4 - 7x)$

$ \Rightarrow f(x) = {x^3} - 27x + 26$

$f'(x) = 3{x^2} - 27$

For maxima or minima $f'(x) = 0$

$ \therefore $ $3{x^2} - 27 = 0$

$ \Rightarrow x = \pm \,3$

Now, $f''(x) = 6x$

f''(x) at x = $-$3 is 6($-$3) = $-$18 < 0

$ \therefore $ At x = $-$3 f(x) is maximum.

So, x₀ = $-$3

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

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

$\overrightarrow c = 7\widehat i - 2\widehat j - 3\widehat k$

Now, $\overrightarrow a \,.\,\overrightarrow b \, + \,\overrightarrow b \,.\,\overrightarrow c \, + \,\overrightarrow c \,.\,\overrightarrow a $

$ = (6 + 6 - 3) + ( - 14 + 6 + 3) + ( - 21 + 4 - 9)$

$ = 9 - 5 - 26$

$ = - 22$

Let, $\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$

and $\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$

We know, Angle between two vectors

$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,\overrightarrow {{a_2}} |}}$

$ = {{ab + bc + ac} \over {\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{a^2} + {b^2} + {c^2}} }}$

$ = {{ab + bc + ac} \over {({a^2} + {b^2} + {c^2})}}$

Given, ${a^2} + {b^2} + {c^2} = 1$

$ \therefore $ $\cos \alpha = ab + bc + ac$

$ = ab\left( {{1 \over a} + {1 \over b} + {1 \over c}} \right)$ .....(1)

Given, $a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right) = \lambda $ (Assume)

$ \therefore $ ${1 \over a} = {{\cos \theta } \over \lambda }$

${1 \over b} = {{\cos \left( {\theta + {{2\pi } \over 3}} \right)} \over \lambda }$

${1 \over c} = {{\cos \left( {\theta + {{4\pi } \over 3}} \right)} \over \lambda }$

$ \therefore $ ${1 \over a} + {1 \over b} + {1 \over c} = {1 \over \lambda }\left[ {\cos \theta + \cos \left( {\theta + {{2\pi } \over 3}} \right) + \cos \left( {\theta + {{4\pi } \over 3}} \right)} \right]$

$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos \left( {{{2\theta + 2\pi } \over 2}} \right)\cos \left( {{{{{2\pi } \over 3}} \over 2}} \right)} \right]$

$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos (\theta + \pi )\cos \left( {{\pi \over 3}} \right)} \right]$

$ = {1 \over \lambda }\left[ {\cos \theta + 2( - \cos \theta ) \times {1 \over 2}} \right]$

$ = {1 \over \lambda } \times 0$

$ = 0$

Putting value of $\lambda $ in equation (1),

cos$\alpha $ = ab(0) = 0

$ \Rightarrow $ $\alpha = {\pi \over 2}$

L₁ = $\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$

= $\widehat i\left( {1 + 2l} \right) + \widehat j\left( { - 1} \right) + \widehat k\left( l \right)$

L₂ = $\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$

= $\widehat i\left( {2 + m} \right) + \widehat j\left( {m - 1} \right) + \widehat k\left( { - m} \right)$

Equating coefficient of $\widehat i$, $\widehat j$ and $\widehat k$ of L₁ and L₂

2l + 1 = m + 2 ... (1)

–1 = –1 + m ...(2)

l = –m ...(3)

from (ii) m = 0

from (iii) $l$ = 0

These values of m and $l$ do not satisfy equation (1).

Hence the two lines do not intersect for any values of $l$ and m.

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

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

$\left| {\overrightarrow a } \right|$ = $\sqrt 6 $, $\left| {\overrightarrow b } \right|$ = $\sqrt 3 $

and ${\overrightarrow a .\overrightarrow b }$ = 4

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

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

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

$ \therefore $ ${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$

$ \Rightarrow $ ${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$

$ \Rightarrow $ ${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$

$ \Rightarrow $ ${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$

$ \Rightarrow $ 0 = ${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$

$ \Rightarrow $ $\lambda $ = ${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$ = ${{ - 6} \over 4} = - {3 \over 2}$

$ \therefore $ $\overrightarrow c $ = ${\overrightarrow a - {3 \over 2}\overrightarrow b }$

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

$ \Rightarrow $ $\overrightarrow c $ = $ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$

$ \therefore $ $\overrightarrow c .\overrightarrow b $ = $ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$($\widehat i - \widehat j + \widehat k$)

$ \therefore $ $\overrightarrow c .\overrightarrow b $ = $ - {1 \over 2}$

Volume of parallelopiped = 1

$\left| {\left[ {\matrix{ {\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr } } \right]} \right|$ = 1

$ \Rightarrow $ $\left| {\matrix{ 1 & 1 & \lambda \cr 1 & 1 & 3 \cr 2 & 1 & 1 \cr } } \right|$ = $ \pm $1

$ \Rightarrow $ $\lambda $ = 2, 4

$\overrightarrow u = \widehat i + \widehat j + 2 \widehat k$ or
$\overrightarrow u = \widehat i + \widehat j + 4 \widehat k$

$ \therefore $ cos $\theta $ = ${{{\overrightarrow u .\overrightarrow w } \over {\left| {\overrightarrow u } \right|\left| {\overrightarrow w } \right|}}}$

= ${{2 + 1 + 4} \over {\sqrt {18} \sqrt 6 }}$ or ${{2 + 1 + 2} \over {\sqrt 6 \sqrt 6 }}$

= ${7 \over {6\sqrt 3 }}$ or ${5 \over 6}$

$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $

$ \Rightarrow $ ${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$ = 0

$ \Rightarrow $ ${{{\left| {\overrightarrow a } \right|}^2}}$ + ${{{\left| {\overrightarrow b } \right|}^2}}$ + ${{{\left| {\overrightarrow c } \right|}^2}}$ +
${2\left( {\overrightarrow a .\overrightarrow b } \right)}$ + ${2\left( {\overrightarrow b .\overrightarrow c } \right)}$ + ${2\left( {\overrightarrow c .\overrightarrow {a} } \right)}$ = 0

$ \Rightarrow $ 3 + ${2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)}$ = 0

$ \Rightarrow $ 3 + 2$\lambda $ = 0

$ \Rightarrow $ $\lambda $ = $ - {3 \over 2}$

$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $

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

= ${\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + }$ ${\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }$

= ${\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }$

= ${3\left( {\overrightarrow a \times \overrightarrow b } \right)}$