Vector Algebra

Angle bisector $\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$

= $\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$

$ \Rightarrow $ $\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$

comparing with $\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$

${{2\lambda } \over {3\sqrt 2 }}$ = 2

$ \Rightarrow $ $\lambda $ = ${3\sqrt 2 }$

$ \therefore $ $\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$

Then $\overrightarrow a .\widehat k - 4 $

= 4 - 4 = 0

Since these vectors are coplanar then,

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

Now, $\alpha (3 - 2\alpha ) - 1\left( {6 + {\alpha ^2}} \right) + 3\left( { - 4 - \alpha } \right) = 0$

$ - 3{\alpha ^2} - 18 = 0$ $ \Rightarrow $ ${\alpha ^2}$ = -6

Required vector is $\overrightarrow r$ = $\lambda \left( {\left( {\overline a + \overline b } \right) \times \left( {\overline a - \overline b } \right)} \right)$

$ \Rightarrow \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 4 & 4 & 0 \cr 2 & 0 & 4 \cr } } \right| = \lambda \left( {16\widehat i - 16\widehat j - 8\widehat k} \right)$

$ \Rightarrow \overrightarrow r = 8\lambda \left( {2\widehat i - 2\widehat j - \widehat k} \right) = \left| {\overrightarrow r } \right|$

$ \Rightarrow \left| {8\lambda } \right|.3 \Rightarrow 8\lambda = \pm 4$

$ \Rightarrow \overrightarrow r = \pm 4\left( {2\widehat i - 2\widehat j - \widehat k} \right)$

$V = \left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{ 1 & \lambda & 1 \cr 0 & 1 & \lambda \cr \lambda & 0 & 1 \cr } } \right|$

$ \Rightarrow 1 - \lambda \left( { - {\lambda ^2}} \right) + 1.\left( {0 - \lambda } \right) = {\lambda ^3} - \lambda + 1$

Whose minimum value occur at $\lambda $ = ${1 \over {\sqrt 3 }}$

$AD = \left| {{{\overrightarrow {AP} .\overrightarrow n } \over {\left| {\overrightarrow n } \right|}}} \right| = \sqrt {61} $

$ \Rightarrow PD = \sqrt {A{P^2} - A{D^2}} $

$ = \sqrt {110 - 61} $

= 7

G is the centroid of $\Delta $ABC



$OG = \sqrt {4 + 16 + 4} $, OA = $\sqrt {9 + 1} $

$AG = \sqrt {1 + 16 + 9} $

$\cos \theta = {{24 + 10 - 26} \over {2\sqrt {24} \sqrt {10} }}$

$ \Rightarrow {8 \over {2\sqrt {8 \times 3 \times 2 \times 5} }}$

$ \Rightarrow {4 \over {4\sqrt {15} }} = {1 \over {\sqrt {15} }}$

A unit vector $\overrightarrow a $ makes angles $\pi $/3 with $\widehat i$

$ \therefore $ $\alpha $ = $\pi $/3

and $\pi $/ 4 with $\widehat j$

$ \therefore $ $\beta $ = $\pi $/ 4

and $\theta $$ \in $(0, $\pi $) with $\widehat k$

$ \therefore $ $\gamma $ = $\theta $

We also know,

${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $ = 1

$ \Rightarrow $ ${\cos ^2}{\pi \over 3} + {\cos ^2}{\pi \over 4} + {\cos ^2}\theta $ = 1

$ \Rightarrow $ ${1 \over 4} + {1 \over 2} + {\cos ^2}\theta $ = 1

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

$ \Rightarrow $ $\theta $ = ${\pi \over 3}$ or ${{2\pi } \over {3}}$

Given $\overrightarrow \alpha = 3\widehat i + \widehat j$

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

${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha $

$ \therefore $ ${\overrightarrow \beta _1}$ = $\lambda $ $\overrightarrow \alpha$

$ \Rightarrow $ ${\overrightarrow \beta _1}$ = $3\lambda \widehat i + \lambda \widehat j$

$\overrightarrow {{\beta _2}} $ is perpendicular to $\overrightarrow \alpha $

$\overrightarrow {{\beta _2}} $ . $\overrightarrow \alpha$ = 0

Let $\overrightarrow {{\beta _2}} $ = $x\widehat i + y\widehat j + z\widehat k$

$ \therefore $ ($x\widehat i + y\widehat j + z\widehat k$).($3\widehat i + \widehat j$) = 0

$ \Rightarrow $ 3x + y = 0

$ \Rightarrow $ y = -3x

$ \therefore $ $\overrightarrow {{\beta _2}} $ = $x\widehat i -3x\widehat j + z\widehat k$

Given $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $

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

= $\left( {3\lambda - x} \right)\widehat i + \left( {\lambda + 3x} \right)\widehat j - z\widehat k$

$ \therefore $ 3$\lambda $ - x = 2 ...(1)

$\lambda $ + 3x = -1.....(2)

z = -3

Solving (1) and (2), we get

$\lambda $ = ${1 \over 2}$ and x = $-{1 \over 2}$

$ \therefore $ ${\overrightarrow \beta _1}$ = ${3 \over 2}\widehat i + {1 \over 2}\widehat j$

and $\overrightarrow {{\beta _2}} $ = $ - {1 \over 2}\widehat i + {3 \over 2}\widehat j - 3\widehat k$

${\overrightarrow \beta _1} \times {\overrightarrow \beta _2} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{3 \over 2}} & {{1 \over 2}} & 0 \cr { - {1 \over 2}} & {{3 \over 2}} & { - 3} \cr } } \right|$

= ${1 \over 2}$($ - 3\widehat i + 9\widehat j + 5\widehat k$)

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

= (2 + x)${\widehat i}$ + (3 - x)${\widehat j}$ - 5${\widehat k}$

$\left| {\overrightarrow a \times \overrightarrow b } \right|$ = r

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

$ \Rightarrow $ r = $\sqrt {4 + 4x + {x^2} + {x^2} + 9 - 6x + 25} $

= $\sqrt {2{x^2} - 2x + 38} $

= $\sqrt {2\left( {{x^2} - x + {1 \over 4}} \right) + 38 - {1 \over 2}} $

= $\sqrt {2{{\left( {x - {1 \over 2}} \right)}^2} + {{75} \over 2}} $

$ \Rightarrow $ r $ \ge $ $\sqrt {{{75} \over 2}} $

$ \Rightarrow $ $ r \ge 5\sqrt {{3 \over 2}} $

$\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = {1 \over 2}\overrightarrow b $

$ \because $  $\overrightarrow b \,\,$ & $\overrightarrow c \,\,$ are linearly independent

$ \therefore $  $\overrightarrow a \,$.$\overrightarrow c \,$ = ${1 \over 2}$ & $\overrightarrow a .\overrightarrow b $ = 0

(All given vectors are unit vectors)

$ \therefore $  $\overrightarrow a $^$\overrightarrow c $ = 60^o & $\overrightarrow a $^$\overrightarrow b $ = 90^o

$ \therefore $  $\left| {\alpha - \beta } \right| = {30^o}$

$\left| {\matrix{ \mu & 1 & 1 \cr 1 & \mu & 1 \cr 1 & 1 & \mu \cr } } \right| = 0$

$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$

${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$

${\mu ^3} - 3\mu + 2 = 0$

${\mu ^3} - 1 - 3\left( {\mu - 1} \right) = 0$

$u = 1,\,\,{\mu ^2} + \mu - 2 = 0$

$\mu = 1,\,\,\mu = - 2$

sum of distinct solutions $=$ $-$ 1

Angle bisector is x $-$ y = 0

$ \Rightarrow $  ${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$

$ \Rightarrow $  $\left| {2\beta - 1} \right| = 3$

$ \Rightarrow $  $\beta $ = 2 or $-$ 1

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

$ \Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$

$ \Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$

$ \Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$

$ \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$

$ \Rightarrow \lambda = 2,3, - 3$

So,   $\lambda $ = 2 (as   $\overrightarrow a $ is parallel to $\overrightarrow c $ for $\lambda $ = $ \pm $3)

Hence   $\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$

$ = - 10\widehat i + 5\widehat j$

$\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow \alpha + \overrightarrow b $

$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b $

${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$

$3\lambda - 6 = 4\lambda - 2$

$\lambda = - 4$

Given $\overrightarrow b = 2\overrightarrow a $

$ \therefore $ $4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$

$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$

Given $\overrightarrow a $ is perpendicular to $\overrightarrow c $

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

$ \Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$

$ \Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$

Now $\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$

By checking each option you can see,

when ${\lambda _1}$ = $ - {1 \over 2}$

then ${\lambda _2}$ = $3 - 2{\lambda _1}$ = 3 + 1 = 4

and ${\lambda _3}$ = $-1 - 2{\lambda _1}$ = - 1 + 1 = 0

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

$ \therefore $   ${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$

$ \Rightarrow $  ${{{b_1} + {b_2} + 2} \over 2} = 2$

$ \Rightarrow $  ${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$

and $\overrightarrow a $ + $\overrightarrow b $ is perpendicular to $\overrightarrow c $

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

$ \Rightarrow $  $5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$

$ \Rightarrow $  $5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$

solving (1) & (2)

b₁ $=$ $-$ 3 and b₂ $=$ 5

$ \Rightarrow $  $\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$

Given that,

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

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

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

given that

$\overrightarrow a \cdot \overrightarrow c = 4$

and $\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$

$ \Rightarrow $  $4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$

Now  $\overrightarrow a \times \overrightarrow b $

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

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

$ \therefore $  $2\overrightarrow c = 4\left( {\widehat i - \widehat j} \right) + \left( { - \widehat i - \widehat j + \widehat k} \right)$

$ = 4\widehat i - 4\widehat j - \widehat i - \widehat j + \widehat k$

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

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

$ \therefore $  $\left| {\overrightarrow c } \right| = \sqrt {{9 \over 4} + {{25} \over 4} + 1} $

$ = \sqrt {{{38} \over 4}} $

$ = \sqrt {{{19} \over 2}} $

$ \therefore $  ${\left| {\overrightarrow c } \right|^2} = {{19} \over 2}$

$ \because $ $\overrightarrow a $ $=$ $\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $

&   $\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $

Now, $\overrightarrow a $ $ \times $ $\overrightarrow b $ = $\overrightarrow c $     (Given)

$ \Rightarrow $  $\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$

$ \Rightarrow $  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2 $

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

$ \Rightarrow $  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$

Dividing [i] by [iii], we get

tan$\theta $ = ${{\sqrt 2 } \over 3}$

$ \therefore $ sin$\theta $ $=$ ${{\sqrt 2 } \over {\sqrt {11} }}$

Substituting value of sin$\theta $ in [i] we get

$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2 $

$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$

You should know that, when $\overrightarrow u $ is coplanar with $\overrightarrow a $ and $\overrightarrow b $ then we can write $\overrightarrow u = x\overrightarrow a + y\overrightarrow b $

Here, $\overrightarrow u $ is perpendicular with $\overrightarrow a $ then,

$\overrightarrow u .\overrightarrow a = 0$

$ \Rightarrow \,\,\,\,\left( {x\,\overrightarrow a + y\overrightarrow b } \right)\,.\overrightarrow a = 0$

$ \Rightarrow \,\,\,\,x\,.\overrightarrow {\,a} \,.\,\overrightarrow a \, + \,y\,.\,\overrightarrow a \,.\,\overrightarrow b = 0$

$ \Rightarrow \,\,\,\,x\,.\,{\left| {\overrightarrow a } \right|^2} + \,y\overrightarrow a \,.\,\overrightarrow b = 0$

[ As $\left| {\overrightarrow a } \right| = \sqrt {{2^2} + {3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14} $

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

$ = \,\,\,\,\left( {2.0 + 3.1 + \left( { - 1} \right).1} \right)$

$ = \,\,\,\,2$ ]

$ \Rightarrow \,\,\,\,x\,.\,\left( {14} \right) + y\,.\,2 = 0$

$ \Rightarrow \,\,\,\,7x\, + \,y\, = 0........\left( 1 \right)$

Given, $\overrightarrow u \,.\,\overrightarrow b = 24$

$ \Rightarrow \,\,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow b = 24$

$ \Rightarrow \,\,\,\,x\left( {\overrightarrow a .\overrightarrow b } \right) + y{\left| {\overrightarrow b } \right|^2} = 24$

$ \Rightarrow \,\,\,\,x.2 + y.{\left( {\sqrt {{1^2} + {1^2}} } \right)^2} = 24$

$ \Rightarrow \,\,\,\,2x + 2y = 24$

$ \Rightarrow \,\,\,\,x + y = 12.......\left( 2 \right)$

By solvig (1) and (2) we get,

x = - 2 and y = 14

Now, ${\left| {\overrightarrow u } \right|^2} = \overrightarrow u .\overrightarrow u $

$ = \,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow u $

$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $

$ = \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u $

$ = \,\,\,\,0 + 14 \times 24$ [as $\overrightarrow a .\overrightarrow u = 0$ and $\overrightarrow u .\overrightarrow b = 24]$

$=\,\,\,\,$ 336

Suppose angular bisector of A meets BC at D(x, , z)

Using angular bisector theorem,

${{AB} \over {AC}}$ = ${{BD} \over {DC}}$

${{BD} \over {DC}}$ = ${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$

= ${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$ = ${6 \over 3}$ = 2



So, D(x, y, z) $ \equiv $ $\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$, $\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$

D(x, y, z) $ \equiv $ $\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$

Therefore, position vector of point p = ${1 \over 3}$ (6i + 13j + 18k)

Given,

$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $

$ \Rightarrow $ $\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $

Squaring both sides,

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

$ \Rightarrow $ 1 + $4\overrightarrow a .\overrightarrow c $ + 4 = 4   [as $\left| {\overrightarrow a } \right|^2$ = $\left| {\overrightarrow b } \right|^2$ = $\left| {\overrightarrow c } \right|^2$ = 1]

$ \Rightarrow $ $\overrightarrow a .\overrightarrow c = - {1 \over 4}$

$ \Rightarrow $ $\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}$

$ \therefore $ $\cos \theta $ = $ - {1 \over 4}$

$ \therefore $ $\sin ^2 \theta $ = 1 - $\cos ^2 \theta $

= 1 - $1 \over 16$

= $15\over 16$

$ \therefore $ sin$\theta $ = ${{\sqrt {15} } \over 4}$

$ \therefore $ $\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta $

= 1 . 1 . ${{\sqrt {15} } \over 4}$

= ${{\sqrt {15} } \over 4}$

$\overrightarrow {{b_1}} = {{\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)\widehat a} \over 1}$

=   $\left\{ {{{\left( {3\widehat j + 4\widehat k} \right).\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 }}} \right\}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$

=   ${{3\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 \times \sqrt 2 }} = {{3\left( {\widehat i + \widehat j} \right)} \over 2}$

$\overrightarrow {{b_1}} + \overrightarrow {{b_2}} = \overrightarrow b $

$ \Rightarrow $   $\overrightarrow {{b_2}} = \overrightarrow b - \overrightarrow {{b_1}} $

=   $\left( {3\widehat j + 4\widehat k} \right) - {3 \over 2}\left( {\widehat i + \widehat j} \right)$

$ \Rightarrow $   $\overrightarrow {{b_2}} $ = $ - {3 \over 2}\widehat i + {3 \over 2}\widehat j + 4\widehat k$

&  $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{3 \over 2}} & {{3 \over 2}} & 0 \cr { - {3 \over 2}} & {{3 \over 2}} & 4 \cr } } \right|$

$ \Rightarrow $   $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \widehat i\left( 6 \right) - \widehat j\left( 6 \right) + \widehat k\left( { - {9 \over 4} + {9 \over 4}} \right)$

$ \Rightarrow $   $6\widehat i - 6\widehat j + {9 \over 2}\widehat k$

When diagonal ${\overrightarrow {{d_1}} }$ and ${\overrightarrow {{d_2}} }$ are given of a parallelogram then the area of parallelogram = ${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$

Given, ${\overrightarrow {{d_1}} }$ = 8$\widehat i$ $-$ 6$\widehat j$ + 0$\widehat k$

and    ${\overrightarrow {{d_2}} }$ = 3$\widehat i$ + 4$\widehat j$ $-$ 12$\widehat k$

$\therefore\,\,\,$ ${\overrightarrow {{d_1}} }$ $ \times $ ${\overrightarrow {{d_2}} }$   =   $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 6} & 0 \cr 3 & 4 & { - 12} \cr } } \right|$

= 72 $\widehat i$ $-$ ($-$ 96) $\widehat j$ + 50$\widehat k$

= 72 $\widehat i$ + 96 $\widehat j$ + 50 $\widehat k$

$\therefore\,\,\,$ $\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$ = $\sqrt {{{72}^2} + {{96}^2} + {{50}^2}} $

= $\sqrt {16900} $

= 130

$\therefore\,\,\,$ Area of parallelogram = ${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$ = ${1 \over 2}$ $ \times $ 130 = 65

Given:

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

$ \Rightarrow $  $\left| {\overrightarrow a } \right| = 3$

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

$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3$

We have $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ$

$ \Rightarrow $  $\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}$

$ \Rightarrow $  $3 = 3\left| {\overrightarrow c } \right|.{1 \over 2}$

$ \therefore $  $\left| {\overrightarrow c } \right| = 2$

Now  $\left| {\overrightarrow c - \overrightarrow a } \right| = 3$

On squaring, we get

$ \Rightarrow $  ${c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9$

$ \Rightarrow $  $4 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9$

$ \Rightarrow $  $\overrightarrow a .\overrightarrow c = 2$     [$ \because $  $\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c $]

Given,

Position vector of circumcentre, $\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$

We know, position vector of centroid, $\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$

Now, let $\overrightarrow R $ be the orthocentre of the triangle.

We know, $\overrightarrow G $ $ = {{2\overrightarrow C + \overrightarrow R } \over 3}$

$ \Rightarrow $   3$\overrightarrow G $ $ = 2\overrightarrow C + \overrightarrow R $

$ \Rightarrow $   $\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C $

=   $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$

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

Given,

$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$

$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$

$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$

$ \therefore $   $\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$

      $\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$



$\Delta $ABC is a right angle triangle.

Here $\overrightarrow {AB} $ perpendicular to $\overrightarrow {AC} $

$ \therefore $   $\overrightarrow {AB} $  .  $\overrightarrow {AC} $  =  0

$ \Rightarrow $   $-$ 8 + 2(q $-$ 1) $-$ 3(p + 1) = 0

$ \Rightarrow $   3p $-$ 2q + 13 = 0

$ \therefore $   (p, q) lies on the line

3x $-$ 2y + 13 = 0

And slope of the line = ${3 \over 2}$

$ \therefore $   line makes an angle less than 90^o or acute angle with the positive direction of x-axis.

$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right)$

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

On comparing both sides

$\overrightarrow a .\overrightarrow b = - {{\sqrt 3 } \over 2} \Rightarrow \cos \theta = - {{\sqrt 3 } \over 2}$

$\left[ \, \right.$ As $\overrightarrow a $ and $\overrightarrow b $ are unit vectors $\left. \, \right]$

where $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $

$\theta = {{5\pi } \over 6}$

$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

$ \Rightarrow - \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

$ \Rightarrow - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

$ \Rightarrow - \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos \theta \overrightarrow a + \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

$\therefore$ $\,\,\,\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $ are non collinear, the above equation is possible only when

$ - \cos \theta = {1 \over 3}$ and $\overrightarrow c .\overrightarrow a = 0$

$ \Rightarrow \cos \theta = - {1 \over 3}$

$ \Rightarrow \sin \theta = {{2\sqrt 2 } \over 3};\theta \in \,{\rm I}{\rm I}$ quad

$L.H.S$ $ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right]$

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

$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left[ {\overrightarrow b \,\overrightarrow c \,\overrightarrow a } \right]\overrightarrow c } \right]$ $\,\,\,\,\,\,\left[ \, \right.$As $\overrightarrow b \times \overrightarrow c .\overrightarrow c = 0$ $\left. \, \right]$

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

$\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\overrightarrow c \times \overrightarrow a } \right] = {\left[ {\overrightarrow a \,\,\,\,\,\overrightarrow b \,\,\,\,\,\overrightarrow c } \right]^2}$

So $\,\,\,\,\,\lambda = 1$

As $M$ is mid point of $BC$

$\therefore$ $\overrightarrow {AM} = {1 \over 2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)$

$ = 4\overrightarrow i + \overrightarrow j + 4\overrightarrow k $

Length of median $AM$

$ = \sqrt {16 + 1 + 16} = \sqrt {33} $

Let $\overrightarrow c = \widehat a + 2\widehat b$ and $\overrightarrow d = 5\widehat a - 4\widehat b$

Since $\overrightarrow c $ and $\overrightarrow d $ are perpendicular to each other

$\therefore$ $\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \right).\left( {5\widehat a - 4\widehat b} \right) = 0$

$ \Rightarrow 5 + 6\widehat a.\widehat b - 8 = 0$ $\,\,\,\,\,\,$ (as $\widehat a.\widehat a = 1$)

$ \Rightarrow \widehat a.\widehat b = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$

Let $ABCD$ be a parallelogram such that

$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $ and $\angle BAD$ be an acute angle.

We have



$\overrightarrow {AX} = \left( {{{\overrightarrow p .\overrightarrow q } \over {\left| {\overrightarrow p } \right|}}} \right)\left( {{{\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}} \right) = {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $

Let $\overrightarrow r = \overrightarrow {BX} = \overrightarrow {BA} + \overrightarrow {AX} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \overrightarrow q + {{\overrightarrow p .\overrightarrow q } \over {{{\left| {\overrightarrow p } \right|}^2}}}\overrightarrow p $

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

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

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

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

$ \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow d = - \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b + \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c $

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

We have $\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow a .\overrightarrow a = 1,\,\,\overrightarrow b .\overrightarrow b = 1$

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

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

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

$ = \left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\overrightarrow b - 2\overrightarrow a } \right]$

$ = 4\overrightarrow a .\overrightarrow b - \overrightarrow b .\overrightarrow b - 4\overrightarrow a .\overrightarrow a $

$ = - 5$

We are given that $\overrightarrow a + 3 \overrightarrow b$ is collinear with $\overrightarrow c$, and $\overrightarrow b + 2 \overrightarrow c$ is collinear with $\overrightarrow a$. This means we can write:

1. $\overrightarrow a + 3 \overrightarrow b = \lambda \overrightarrow c \quad ...(i)$
2. $\overrightarrow b + 2 \overrightarrow c = \mu \overrightarrow a \quad ...(ii)$

for some scalars $\lambda$ and $\mu$.

We are trying to find $\overrightarrow a + \overrightarrow b + 6\overrightarrow c$ in terms of $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$. We can also express this as :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (\lambda + 6) \overrightarrow c \quad ...(iii)$

by adding $6\overrightarrow c$ to both sides of equation (i).

Now, from equation (ii), multiplying by 3 gives us :

$3\overrightarrow b + 6 \overrightarrow c = 3\mu \overrightarrow a \quad ...(iv)$

Adding $\overrightarrow a$ to both sides of equation (iv) gives :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(v)$

Now, we have two expressions for $\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c$, one in terms of $\overrightarrow c$ (from equation iii) and one in terms of $\overrightarrow a$ (from equation v). Setting these equal to each other gives :

$(\lambda + 6) \overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(vi)$

Since $\overrightarrow a$ and $\overrightarrow c$ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :

$\lambda + 6 = 0$ and $1 + 3\mu = 0$

This gives $\lambda = -6$ and $\mu = -\frac{1}{3}$.

Finally, substituting $\lambda = -6$ into equation (iii) gives :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = 0$

So, $\overrightarrow a + \overrightarrow b + 6\overrightarrow c = \overrightarrow 0$.

Therefore, the correct answer is Option D : $\overrightarrow 0$.

Since, $\overrightarrow a ,\overrightarrow b $ and $\overrightarrow c $ are mutually orthogonal

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

$ \Rightarrow 2\lambda + 4 + \mu = 0\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$ \Rightarrow \lambda - 1 + 2\mu = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

On solving $(i)$ and $(ii)$, we get $\lambda = - 3,\mu = 2$

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

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

$ \Rightarrow \left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right).\left( {\widehat i - \widehat j - \widehat k} \right) = 0,$

where $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$

${b_1} - {b_2} - {b_3} = 0\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

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

$ \Rightarrow \left( {\widehat j - \widehat k} \right).\left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right) = 3$

$ \Rightarrow {b_2} - {b_3} = 3$

From equation $(i)$

${b_1} = {b_2} + {b_3} = \left( {3 + {b_3}} \right) + {b_3} = 3 + 2{b_3}$

$\overrightarrow b = \left( {3 + 2{b_3}} \right)\widehat i + \left( {3 + {b_3}} \right)\widehat j + {b_3}\widehat k$

From the option given, it is clear that ${b_3}$ equal to either $2$ or $-2.$

${b_3} = 2$

then $\overrightarrow b = 7\widehat i + 5\widehat j + 2\widehat k$ which is not possible

If ${b_3} = - 2,$ then $\overrightarrow b = - \widehat i + \widehat j - 2\widehat k$

$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow \omega } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow \omega \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow \omega \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$

$ \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] = 0$

$ \Rightarrow 3{p^2} - pq + 2{q^2} = 0\,\,$ $\,\,\,\,\left( \, \right.$ As $\,\,\,\,\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] \ne 0$ $\left. {} \right)$

$ \Rightarrow 2{p^2} + {p^2} - pq + {{{q^2}} \over 4} + {{7{q^2}} \over 4} = 0$

$ \Rightarrow 2{p^2} + {\left( {p - {q \over 2}} \right)^2} + {7 \over 4}{q^2} = 0$

$ \Rightarrow p = 0,q = 0,p = {q \over 2}$ $ \Rightarrow p = 0,q = 0$

$\therefore$ Exactly one value of $\left( {p,q} \right)$

As $\overrightarrow a $ lies in the plane of $\overrightarrow b $ and $\overrightarrow c $

$\therefore$ $\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c $

$ \Rightarrow \alpha \widehat i + 2\widehat j + \beta \widehat k = \widehat i + \widehat j + \lambda \left( {\widehat j + \widehat k} \right)$

$ \Rightarrow \alpha = 1,2 = 1 + \lambda ,\,\beta = \lambda \Rightarrow \alpha = 1,\beta = 1$

Clearly $\overrightarrow a = - {8 \over 7}\overrightarrow c $

$ \Rightarrow \overrightarrow a ||\overrightarrow c $ and are opposite in direction

$\therefore$ Angle between $\overrightarrow a $ and $\overrightarrow c $ is $\pi .$

Given $\left| {2\widehat u \times 3\widehat v} \right| = 1$

and $\theta $ is acute angle between $\widehat u$

and $\widehat v,\,\,\left| {\widehat u} \right| = 1,\,\,\left| {\widehat v} \right| = 1\,\,\,$

$ \Rightarrow \,\,\,6\left| {\widehat u} \right|\left| {\widehat v} \right|\left| {\sin \theta } \right| = 1$

$ \Rightarrow 6\left| {\sin \theta } \right| = 1 \Rightarrow \sin \theta = {1 \over 6}$

Hence, there is exactly one value of $\theta $

for which $2\widehat u \times 3\widehat v$ is a unit vector.

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

and $\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k$

If $\overrightarrow c $ lies in the plane of $\overrightarrow a $ and $\overrightarrow b ,$

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

i.e.$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - 1} & 2 \cr x & {\left( {x - 2} \right)} & { - 1} \cr } } \right| = 0$

$ \Rightarrow 1\left[ {1 - 2\left( {x - 2} \right)} \right] - 1\left[ { - 1 - 2x} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1\left[ {x - 2 + x} \right] = 0$

$ \Rightarrow 1 - 2x + 4 + 1 + 2x + 2x - 2 = 0$

$ \Rightarrow 2x = - 4\,\,\,\,\,\,\,\,\, \Rightarrow x = - 2$

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

$\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\overrightarrow a .\overrightarrow b \ne 0,\,\,\overrightarrow b .\overrightarrow c \ne 0$

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

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

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

$ \Rightarrow \overrightarrow a ||\overrightarrow c .$

$\overrightarrow {CA} = \left( {2 - a} \right)\widehat i + 2\widehat j;$

$\overrightarrow {CB} = \left( {1 - a} \right)\widehat i - 6\widehat k$

$\overrightarrow {CA} .\overrightarrow {CB} = 0$

$\,\,\,\,\,\,\,\, \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0$

$ \Rightarrow a = 2,1$

Vector $a\overrightarrow i + a\overrightarrow j + c\overrightarrow k ,\,\,\overrightarrow i + \overrightarrow k $

and $c\overrightarrow i + c\overrightarrow j + b\overrightarrow k $ are coplanar

$\left| {\matrix{ a & a & c \cr 1 & 0 & 1 \cr c & c & b \cr } } \right| = 0 \Rightarrow {c^2} = ab$

$ \Rightarrow c = \sqrt {ab} $

$\therefore$ $c$ is $G.M.$ of $a$ and $b.$

$\overrightarrow a = \widehat j - \widehat k,\overrightarrow b = x\widehat i + \overrightarrow j + \left( {1 - x} \right)\widehat k$

and $\overrightarrow c = y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k$

$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = \overrightarrow a .\overrightarrow b \times \overrightarrow c = \left| {\matrix{ 1 & 0 & { - 1} \cr x & 1 & {1 - x} \cr y & x & {1 + x - y} \cr } } \right|$

$ = 1\left[ {1 + x - y - x + {x^2}} \right] - \left[ { - {x^2} - y} \right]$

$ = 1 - y + {x^2} - {x^2} + y = 1$

Hence $\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$ is independent of $x$ and $y$ both.

$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right){\lambda ^2}\overrightarrow b \,\,\,\lambda \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a + \overrightarrow b \,\,\overrightarrow b \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left\{ {\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] + \left[ {\overrightarrow b \,\overrightarrow b \,\overrightarrow c } \right]} \right\} = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow b } \right] + \left[ {\overrightarrow a \,\overrightarrow c \,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = - \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$

$ \Rightarrow {\lambda ^4} = - 1$

$ \Rightarrow \lambda \,\,$ has no real values.

$\overrightarrow {PA} + \overrightarrow {AP = 0} $ and $\overrightarrow {PC} + \overrightarrow {CP} = 0$

$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$

and

$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0$

Adding, we get

$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0.$

Since

$\overrightarrow {AC} = - \overrightarrow {BC} $ $\,\,\,\,\,\,$ & $\,\,\,\,\,\,$ $\overrightarrow {CP} = - \overrightarrow {PC} $

$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0.$

Let $\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $

$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $

$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$

Similarly, ${\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {x^2} + {z^2}\,\,$

and ${\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {x^2} + {y^2}$

$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow j } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow k } \right)^2}$

$ = 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2\overrightarrow a 2$

Given $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

Clearly $\overrightarrow a $ and $\overrightarrow b $ are noncollinear

$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

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

and $ - \overrightarrow b .\overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| \Rightarrow \cos \theta = {{ - 1} \over 3}$

$\therefore$ $\sin \theta = \sqrt {1 - {1 \over 9}} = {{2\sqrt 2 } \over 3}$

$\,\,\,\,\,\,\,\left[ {} \right.$ $\theta $ is acute angle between $\overrightarrow b $ and $\overrightarrow c $ $\left. {} \right]$