Vector Algebra

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 $]

$\overrightarrow{OP}$ . $\overrightarrow{OQ}$ + $\overrightarrow{OR}$ . $\overrightarrow{OS}$ = $\overrightarrow{OR}$ . $\overrightarrow{OP}$ + $\overrightarrow{OQ}$ . $\overrightarrow{OS}$

$ \Rightarrow $ $\overrightarrow{OP}$($\overrightarrow{OQ}$ $-$ $\overrightarrow{OR}$) + $\overrightarrow{OS}$($\overrightarrow{OR}$ $-$ $\overrightarrow{OQ}$) = 0

$ \Rightarrow $ ($\overrightarrow{OP}$ $-$ $\overrightarrow{OS}$)($\overrightarrow{OQ}$ $-$ $\overrightarrow{OR}$) = 0

$ \Rightarrow $ $\overrightarrow{SP}$ . $\overrightarrow{RQ}$ = 0

Similarly $\overrightarrow{SR}$ . $\overrightarrow{PQ}$ = 0 and $\overrightarrow{SQ}$ . $\overrightarrow{PR}$ = 0

$ \therefore \overrightarrow{S}$ is orthocentre.

Now, $\overrightarrow {OX} = {{\overrightarrow {QR} } \over {QR}}$

and $\overrightarrow {OY} = {{\overrightarrow {RP} } \over {RP}}$

Therefore, $(\overrightarrow {OX} \times \overrightarrow {OY} ) = {{\overrightarrow {QR} } \over {QR}} \times {{\overrightarrow {RP} } \over {RP}} = {{\overrightarrow {QR} \times \overrightarrow {RP} } \over {PQ}}$

$ = {{PQ\sin R} \over {PQ}} = \sin R = \sin (\pi - (P + Q) = \sin (P + Q))$

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 $

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

We have $\widehat u = {u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k$

That is, $\left| {\widehat u} \right| = 1 = \sqrt {u_1^2 + u_2^2 + u_3^2} $

$ \Rightarrow u_1^2 + u_2^2 + u_3^2 = 1$

Also, it is given that

$\widehat \omega = {1 \over {\sqrt 6 }}(\widehat i + \widehat j + 2\widehat k)$

That is, $\left| {\widehat \omega } \right| = 1$

Now, $\left| {\widehat u \times \overrightarrow v } \right| = 1$

That is, $\left| {\widehat u} \right|\left| {\overrightarrow v } \right|\sin \theta = 1 \Rightarrow \left| {\overrightarrow v } \right| = {1 \over {\sin \theta }}$

which shows that there are infinitely many possible values exist for $\overrightarrow v $ (here $\theta$ is angle between the vectors $\overrightarrow v $ and $\widehat u$).

Hence, option (B) is correct.

Now, $\widehat \omega \,.\,(\widehat u \times \overrightarrow v ) = 1$

$\left| {\widehat \omega \,.\,(\widehat u \times \overrightarrow v )} \right| = 1$

That is, $\left| {\widehat \omega } \right|\left| {\widehat u \times \overrightarrow v } \right|\cos \alpha = 1$

where $\alpha$ is the angle between $\widehat \omega $ and $\widehat u \times \overrightarrow v $.

Therefore, (1)(1)cos$\alpha$ = 1

$\Rightarrow$ a = 0

which means that $\widehat \omega $ and $\widehat u \times \overrightarrow v $ are parallel vector or $\widehat \omega $ is perpendicular vector to $\widehat u$ and $\overrightarrow v $.

$\widehat u\,.\,\widehat \omega = 0$

$({u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k)\,.\,\left( {{{(\widehat i + \widehat j + 2\widehat k)} \over {\sqrt 6 }}} \right) = 0$

${u_1} + {u_2} + 2{u_3} = 0$

If $\widehat u$ lies in xy-plane then u₃ = 0. Therefore,

${u_1} + {u_2} = 0 \Rightarrow {u_1} = - {u_2} \Rightarrow \left| {{u_1}} \right| = \left| {{u_2}} \right|$

Hence, option (C) is correct.

$\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

Option (A): Let $\vec{a}=\alpha \hat{i}+\beta \hat{j}$ and $\vec{b}=\sqrt{3} \hat{i}+\hat{j}$.

Therefore, the magnitude of projection of $\vec{a}$ on $\vec{b}$ is

$ \begin{aligned} &\vec{b}=\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} \\\\ &=\frac{|\sqrt{3} \alpha+\beta|}{\sqrt{3+1}}=\sqrt{3} \\\\ &\Rightarrow \sqrt{3} \alpha+\beta= \pm 2 \sqrt{3} \\\\ &\Rightarrow \sqrt{3}(2+\sqrt{3} \beta)+\beta= \pm 2 \sqrt{2} \\\\ &\Rightarrow \beta=0 \text { or } \beta=-\sqrt{3} \Rightarrow \alpha=2 \text { or } \alpha=-1 \\\\ &\Rightarrow|\alpha|=2 \text { or } 1 \end{aligned} $

Hence, $(\mathrm{A}) \rightarrow(\mathrm{P}),(\mathrm{Q})$.

Option (B): $f(x)=\left\{\begin{array}{r}-3 a x^2-2, x<1 \\ b x+a^2, x>1\end{array}\right.$

Since $f(x)$ is differentiable $\forall x \in \mathbb{R}$, we have $f\left(1^{-}\right)=f\left(1^{+}\right)$

$ \begin{aligned} &\Rightarrow-3 a-2 =b+a^2 \\\\ &\Rightarrow a^2+3 a+2 =-b \\\\ &\Rightarrow(a+2)(a+1) =-b \quad\quad...(1) \end{aligned} $

Also, $f^{\prime}(x)=\left\{\begin{array}{cc}-6 a x ; & x<1 \\ b ; & x>1\end{array}\right.$

$ =f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) $

$ \Rightarrow \quad-6 a=b \quad\quad...(2) $

Therefore, from Eqs. (1) and (2), we get (2) $a^2+3 a+2$ $=6 a$

$ \Rightarrow a=1 \text { or } a=2 $

Hence, (B) $\rightarrow(\mathrm{P})$, (Q).

Also,

$ \begin{aligned} &f^{\prime}(x) =\left\{\begin{array}{cc} -6 a x, & x<1 \\ b, & x \geq 1 \end{array}\right. \\\\ &\Rightarrow f^{\prime}\left(1^{-}\right) =f^{\prime}\left(1^{+}\right) \\\\ &\Rightarrow-6 a =b \\\\ &f^{\prime}\left(1^{-}\right) =f^{\prime}\left(1^{+}\right), a^2+3 a+2=6 a \\\\ &\Rightarrow a =1 \text { or } a=2 \end{aligned} $

Hence, $(\mathrm{B}) \rightarrow(\mathrm{P}),(\mathrm{Q})$

$ \begin{aligned} & \mathbb{\text {Option (C): }}\left(3-3 \omega+2 \omega^2\right)^{4 x+3}+\left(2+3 \omega-3 \omega^2\right)^{4 x+3}+ \\\\ & \left(-3+2 \omega+3 \omega^2\right)^{4 x+3}=0 \\\\ & \Rightarrow\left[1-3 \omega+2\left(1+\omega^2\right)\right]^{4 x+3}+\left[2(1+\omega)+\omega-3 \omega^2\right]+ \\\\ & {\left[-3+\omega^2+2\left(\omega+\omega^2\right)\right]^{4 x+3}=0} \end{aligned} $

$ \begin{aligned} & \Rightarrow {[1-3 \omega-2 \omega]^{4 x+3}+\left[-5 \omega^2+\omega\right]^{4 x+3} } \\\\ & \quad \quad\left(\omega^2\right)^{4 x+3}(1-5 \omega)^{4 x+3}=0 \\\\ & \Rightarrow(1-5 \omega)^{4 x+3}\left(1+\omega^n+\omega^{2 x}\right)=0 \\\\ & \Rightarrow \omega(1-5 \omega)^{4 x+3} \neq 0 \\\\ & \Rightarrow 1+\omega^x+\omega^{2 x}=0 \\\\ & \Rightarrow x=3 k+1 \text { or } x=3 k+2 ; k \in \mathbb{Z} \\\\ & \Rightarrow x \in\{1,2,4,5\} \end{aligned} $

Hence, $(\mathrm{C}) \rightarrow(\mathrm{P}),(\mathrm{Q}),(\mathrm{S}),(\mathrm{T})$.

Option (D): HM of $a$ and $b=\frac{2 a b}{a+b}=4$, where $a, b>0$. Now, $a, 5, q, b$ are in AP, where $q>0$.

$ \begin{gathered} \Rightarrow a+b=5+q \\\\ \Rightarrow \frac{a b}{2}=5+q ~~~~~~~~(1) \end{gathered} $

Also

$ \begin{aligned} &a+q= 10 \text { and } q=\frac{5+b}{2} ~~~~~~~~(2)\\\\ & \Rightarrow b=2 q-5(3) \end{aligned} $

Therefore, from Eqs. (1), (2) and (3), $a=\frac{5}{2}$ or $a=6$.

$ \Rightarrow q=\frac{15}{2} \text { or } 4 \Rightarrow|q-a|=5 \text { or } 2 $

Hence, (D) $\rightarrow$ (Q), (T).

For a triangle, we have $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$


$ \begin{aligned} & \Rightarrow \vec{a}=-(\vec{b}+\vec{c}) \\\\ & \Rightarrow|\vec{a}|^2=|\vec{b}|^2+|\vec{c}|^2+2 \vec{b} \cdot \vec{c} \\\\ & \Rightarrow|\vec{a}|^2=48+|\vec{c}|^2+48 \\\\ & \Rightarrow|\vec{c}|^2=|\vec{a}|^2-96=144-96 \\\\ & \Rightarrow|\vec{c}|^2=48 \\\\ & \Rightarrow|\vec{c}|=\sqrt{48}=4 \sqrt{3} \end{aligned} $

Therefore, $\frac{|\vec{c}|^2}{2}-|\vec{a}|=24-12=12$

Option (A) is correct.

$ \frac{|\vec{c}|^2}{2}+|\vec{a}|=24+12=36 $

Also, $\quad \vec{a}+\vec{b}=-\vec{c} \Rightarrow \vec{a} \cdot \vec{b}=-72$

From Eq. (1), we get

$ \begin{aligned} & \vec{a} \times \vec{a}=-(\vec{a} \times \vec{b}+\vec{a} \times \vec{c}) \Rightarrow \vec{a} \times \vec{b}=\vec{c} \times \vec{a} \\\\ & |\vec{a} \times \vec{b}+\vec{a} \times \vec{c}|=|2(\vec{a} \times \vec{b})|=2|\vec{a}||\vec{b}| \sin \theta \\\\ & =96 \sqrt{3} \sqrt{1-\left(\frac{-72}{48 \sqrt{3}}\right)^2} \\\\ & =96 \sqrt{3} \sqrt{1-\left(\frac{\sqrt{3}}{12}\right)^2}=48 \sqrt{3} \end{aligned} $

$L.H.S$ $ = \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 $\,\,\,\,\,\lambda = 1$

$\angle \mathrm{AOB}=\angle \mathrm{BOC}=\angle \mathrm{CCOA}=\frac{\pi}{3}$

According to question, we have

$ \begin{aligned} \vec{a} & =\lambda\{(\vec{x} \cdot \vec{z}) \vec{y}-(\vec{x} \cdot \vec{y}) \vec{z}\} \\\\ & =\lambda\left\{\left(2 \cos \frac{\pi}{3}\right) \vec{y}-\left(2 \cos \frac{\pi}{3}\right) \vec{z}\right\}=\lambda(\vec{y}-\vec{z}) \end{aligned} $

Thus,

$ \begin{aligned} \vec{\alpha} & \times(\vec{\beta} \times \vec{\gamma})=(\vec{\alpha} \cdot \vec{\gamma}) \vec{\beta}-(\vec{\alpha} \cdot \vec{\beta}) \vec{\gamma} \\\\ \vec{b} & =\mu\{(\vec{y} \cdot \vec{x}) \vec{z}-(\vec{y} \cdot \vec{z}) \vec{x}\} \\\\ & =\mu\{\vec{z}-\vec{x}\} \end{aligned} $

Now, $\vec{a} \cdot \vec{y}=\lambda\{2-1\}$; therefore, $\lambda=\vec{a} \cdot \vec{y}$.

Hence, $ \vec{a}=\vec{a} \cdot \vec{y}(\vec{y}-\vec{z})$ ...........(1)

Similarly, $ \vec{b}=\vec{b} \cdot \vec{z}(\vec{z}-\vec{x}) $ ............(2)

Now,

$\begin{aligned} \vec{a} \cdot \vec{b} & =(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})\{\vec{y} \cdot \vec{z}-\vec{y} \cdot \vec{x}-\vec{z} \cdot \vec{z}+\vec{z} \cdot \vec{x}\} \\\\ & =(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})\{1-1-2+1\} \\\\ & =-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z}) .........(3)\end{aligned}$

Hence, from Eqs. (1), (2) and (3), we can conclude that the correct options are $(\mathrm{A}),(\mathrm{B})$ and $(\mathrm{C})$.

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} $

(P) Given, the volume of parallelopiped formed by $\vec{a}, \vec{b}, \vec{c}$ is 2

$\Rightarrow[\vec{a} \vec{b} \vec{c}]=2$

Let V be the volume of parallelopiped formed by $2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$

$\begin{aligned} & \Rightarrow \mathrm{V}=\left[\begin{array}{lll} 2(\vec{a} \times \vec{b}) & 3(\vec{b} \times \vec{c}) & (\vec{c} \times \vec{a}) \end{array}\right] \\ & \Rightarrow \mathrm{V}=2 \times 3[(\vec{a} \times \vec{b})(\vec{b} \times \vec{c})(\vec{c} \times \vec{a})] \\ & \Rightarrow \mathrm{V}=6[\vec{a} \vec{b} \vec{c}]^2 \\ & \Rightarrow \mathrm{V}=6 \times 2^2 \\ & \Rightarrow \mathrm{V}=24 \\ \end{aligned}$

Hence, P match with 3.

(Q) Given, the volume of parallelopiped formed by $\vec{a}, \vec{b}$ and $\vec{c}$ is 5.

$\Rightarrow[\vec{a} \vec{b} \vec{c}]=5 \quad \text{... (i)}$

Let V be the volume of parallelopiped formed by $3(\vec{a}+\vec{b}),(\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$.

$\Rightarrow \mathrm{V}=\left[\begin{array}{lll} 3(\vec{a}+\vec{b}) & (\vec{b}+\vec{c}) & 2(\vec{c}+\vec{a}) \end{array}\right]$

$\begin{aligned} & \Rightarrow \mathrm{V}=3.2[(\vec{a}+\vec{b})(\vec{b}+\vec{c})(\vec{c}+\vec{a})] \\ & \Rightarrow \mathrm{V}=6 \times 2\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right] \\ & \Rightarrow \mathrm{V}=12 \times 5 \\ & \Rightarrow \mathrm{V}=60 \end{aligned}$

Hence, Q match with 4.

(R) Given, the area of a triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is 20 .

$\begin{array}{ll} \Rightarrow & \frac{1}{2}|\vec{a} \times \vec{b}|=20 \\ \Rightarrow & |\vec{a} \times \vec{b}|=40 \quad \text{.... (i)} \end{array}$

Let $\Delta$ be the area of a triangle with adjacent sides $(2 \vec{a}+3 \vec{b})$ and $(\vec{a}-\vec{b})$.

$\begin{aligned} & \Rightarrow \quad \Delta=\frac{1}{2}|(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})| \\ & \Rightarrow \Delta=\frac{1}{2}|2 \vec{a} \times \vec{a}-2 \vec{a} \times \vec{b}+3 \vec{b} \times \vec{a}-3 \vec{b} \times \vec{b}| \\ & \Rightarrow \Delta=\frac{1}{2}|0-2 \vec{a} \times \vec{b}-3 \vec{a} \times \vec{b}-0| \\ & \Rightarrow \Delta=\frac{5}{2}|\vec{a} \times \vec{b}| \\ & \Rightarrow \Delta=\frac{5}{2} \times 40 \\ & \Rightarrow \quad \Delta=100 \end{aligned}$

Hence, R match with 1.

(S) Given, the area of parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is 30.

$\Rightarrow|\vec{a} \times \vec{b}|=30 \quad \text{... (i)}$

Let, $\Delta^{\prime}$ be the area of parallelogram with adjacent sides $(\vec{a}+\vec{b})$ and $\vec{a}$.

$\begin{array}{ll} \Rightarrow & \Delta^{\prime}=|(\vec{a}+\vec{b}) \times \vec{a}| \\ \Rightarrow & \Delta^{\prime}=|\vec{a} \times \vec{a}+\vec{b} \times \vec{a}| \\ \Rightarrow & \Delta^{\prime}=|0-\vec{a} \times \vec{b}| \\ \Rightarrow & \Delta^{\prime}=|\vec{a} \times \vec{b}| \\ \Rightarrow & \Delta^{\prime}=30 \end{array}$

Hence, S match with 2.

Hints :

(i) $[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]=[\vec{a} \vec{b} \vec{c}]^2$

(ii) $\left[\begin{array}{lll}\vec{a}+\vec{b} & \vec{b}+\vec{c} & \vec{c}+\vec{a}\end{array}\right]=2\left[\begin{array}{ll}\vec{a} \vec{b} \vec{c}\end{array}\right]$

(iii) $\vec{A} \cdot(\vec{B} \times \vec{C})=[\vec{A} \vec{B} \vec{C}]$

(iv) $\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}$

(v) $\vec{A} \times \vec{A}=\vec{B} \times \vec{B}=0$

Given that $\overrightarrow{\mathrm{PR}} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\overrightarrow{\mathrm{SQ}} = \hat{i} - 3 \hat{j} - 4 \hat{k}$ are the diagonals of the parallelogram $PQRS$,

we have the following relationships:

$ \begin{aligned} \overrightarrow{\mathrm{PR}} &= \overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{QR}} \\ \overrightarrow{\mathrm{SQ}} &= \overrightarrow{\mathrm{PQ}} - \overrightarrow{\mathrm{PS}} \\ \end{aligned} $

Since $\overrightarrow{\mathrm{QR}} = \overrightarrow{\mathrm{PS}}$,

$ \overrightarrow{\mathrm{PQ}} = \frac{\overrightarrow{\mathrm{PR}} + \overrightarrow{\mathrm{SQ}}}{2} $

and

$ \overrightarrow{\mathrm{PS}} = \frac{\overrightarrow{\mathrm{PR}} - \overrightarrow{\mathrm{SQ}}}{2}. $

Substituting the values of $\overrightarrow{\mathrm{PR}}$ and $\overrightarrow{\mathrm{SQ}}$,

$ \begin{aligned} \overrightarrow{\mathrm{PQ}} &= \frac{(3 \hat{i} + \hat{j} - 2 \hat{k}) + (\hat{i} - 3 \hat{j} - 4 \hat{k})}{2} \\ &= \frac{4 \hat{i} - 2 \hat{j} - 6 \hat{k}}{2} \\ &= 2 \hat{i} - \hat{j} - 3 \hat{k}, \end{aligned} $

$ \begin{aligned} \overrightarrow{\mathrm{PS}} &= \frac{(3 \hat{i} + \hat{j} - 2 \hat{k}) - (\hat{i} - 3 \hat{j} - 4 \hat{k})}{2} \\ &= \frac{2 \hat{i} + 4 \hat{j} + 2 \hat{k}}{2} \\ &= \hat{i} + 2 \hat{j} + \hat{k}. \end{aligned} $

Given $\overrightarrow{\mathrm{PT}} = \hat{i} + 2 \hat{j} + 3 \hat{k}$,

To find the volume $V$ of the parallelepiped formed by $\overrightarrow{\mathrm{PT}}, \overrightarrow{\mathrm{PQ}},$ and $\overrightarrow{\mathrm{PS}}$, we calculate the determinant of the following matrix:

$ V = \left| \begin{array}{ccc} 1 & 2 & 3 \\ 2 & -1 & -3 \\ 1 & 2 & 1 \end{array} \right|. $

Calculating this determinant:

$ \begin{aligned} V &= 1(-1 \times 1 + 6) - 2(2 + 3) + 3(4 + 1) \\\\ &= 1 \times 5 - 2 \times 5 + 3 \times 5 \\\\ &= 5 - 10 + 15\\\\ &= 10. \end{aligned} $

Thus, the volume of the parallelepiped is 10 units.

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 $

$\begin{aligned} & \text { Given, } \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} \\ & \begin{aligned} & \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}=0 \\ & \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})+\vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \\ & \therefore(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}) \\ & \Rightarrow \quad (\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \end{aligned} \end{aligned}$

Hence, $\vec{a}+\vec{b}$ is collinear to $2 \hat{i}+3 \hat{j}+4 \hat{k}$

Let $\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\quad \text{... (i)}$

$\begin{array}{l} \Rightarrow & |\vec{a}+\vec{b}|=|\lambda| \sqrt{2^2+3^2+4^2} \\ \Rightarrow & \sqrt{29}=|\lambda| \sqrt{29} \quad \text { (Given, }|\vec{a}+\vec{b}|=\sqrt{29}) \\ \Rightarrow & |\lambda|=1 \\ \Rightarrow & |\lambda|= \pm 1 \\ \therefore & \vec{a}+\vec{b}= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \end{array}$

$\begin{aligned} \text { Now, } &(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\ &= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\ &= \pm(-14+6+12) \\ &= \pm 4 \end{aligned}$

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

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 : $\overrightarrow 0$.

We have,

$\overrightarrow v = \lambda \overline a + \mu \overline b $

$ = \lambda (\widehat i + \widehat j + \widehat k) + \mu (\widehat i - \widehat j + \widehat k)$

Projection of $\overrightarrow v $ on $\overline c $

${{\overline v \,.\,\overline c } \over {\left| {\overline c } \right|}} = {1 \over {\sqrt 3 }}$

$ \Rightarrow {{\left[ {(\lambda + \mu )\widehat i + (\lambda - \mu )\widehat j + (\lambda + \mu )\widehat k} \right]\,.\,\left( {\widehat i - \widehat j - \widehat k} \right)} \over {\sqrt 3 }} = {1 \over {\sqrt 3 }}$

$ \Rightarrow \lambda + \mu - \lambda + \mu - \lambda - \mu = 1 \Rightarrow \mu - \lambda = 1 \Rightarrow \lambda = \mu - 1$

$\overline v = (\mu - 1)(\widehat i + \widehat j + \widehat k) + \mu (\widehat i - \widehat j + \widehat k) = \mu (2\widehat i + 2\widehat k) - \widehat i - \widehat j - \widehat k$

$\overline v = (2\mu - 1)\widehat i - \widehat j + (2\mu - 1)\widehat k$

At $\mu = 2$, $\overline v = 3\widehat i - \widehat j + 3\widehat k$.

(A) We have, $\overrightarrow a - \overrightarrow b = - 1 + 3 = 2$, where $|\overrightarrow a | = 2$ and $|\overrightarrow b | = 2$.

$\cos \theta = {2 \over {2 \times 2}} = {1 \over 2}$

$\theta = {\pi \over 3},\,{{2\pi } \over 3}$ ; however, it is ${{2\pi } \over 3}$ as its opposite to side of maximum length.

(B) $\int\limits_a^b {(f(x) - 3x)dx = {a^2} - {b^2}} $

$\int\limits_a^b {f(x)dx = {3 \over 2}({b^2} - {a^2}) + {a^2} - {b^2} = {{ - {a^2} + {b^2}} \over 2} \Rightarrow f(x) = x} $

Therefore, $f(\pi /6) = ({\pi ^2}/6)$

(C) ${{{\pi ^2}} \over {\ln 3}}\left( {{{\ln |(\sec \pi x + \tan \pi x)|_{7/6}^{5/6}} \over \pi }} \right)$

$ = {\pi \over {\ln 3}}\left( {\ln \left| {\sec {{5\pi } \over 6} + \tan {{5\pi } \over 6}} \right| - \ln \left| {\sec {{7\pi } \over 6} + \tan {{7\pi } \over 6}} \right|} \right) = \pi $

(D) Let us consider,

$u = {1 \over {1 - z}} \Rightarrow z = 1 - {1 \over u}$

$|z| = 1 \Rightarrow \left| {1 - {1 \over u}} \right| = 1 \Rightarrow |u - 1| = |u|$

Hence, the locus of u is perpendicular bisector of line segment joining 0 and 1.

Therefore, the maximum arg(u) approaches $\pi$/2, but it will not attain.