Vector Algebra

Given,

$\left[ {\matrix{ 0 & { - {c_3}} & {{c_2}} \cr {{c_3}} & 0 & { - {c_1}} \cr { - {c_2}} & {{c_1}} & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$

$\therefore$ $ - {b_2}{c_3} + {b_3}{c_2} = 3 - {c_1}$ ..... (1)

${c_3} - {b_3}{c_1} = 1 - {c_2}$ ...... (2)

$ - {c_2} + {b_2}{c_1} = - 1 - {c_3}$ ..... (3)

Adding (1), (2) and (3), we get

$({b_2} - {b_3}){c_1} + ({b_3} - 1){c_2} + (1 - {b_2}){c_3}$

$ = 3 - ({c_1} + {c_2} + {c_3})$

$ \Rightarrow ({c_2}{b_3} - {c_3}{b_2}) - ({c_1}{b_3} - {c_3}) + ({c_1}{b_2} - {c_2})$

$ = 3 - ({c_1} + {c_2} + {c_3})$ ..... (4)

Also given,

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

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

$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$

Now,

$\overrightarrow c \times \overrightarrow b $

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{c_1}} & {{c_2}} & {{c_3}} \cr 1 & {{b_2}} & {{b_3}} \cr } } \right|$

$ = ({c_2}{b_3} - {c_3}{b_2})\widehat i - \widehat j({c_1}{b_3} - {c_3}) + \widehat k({c_1}{b_2} - {c_2})$

And

$\overrightarrow a - \overrightarrow c = (3 - {c_1})\widehat i + (1 - {c_2})\widehat j + ( - 1 - {c_3})\widehat k$

Comparing value of $\overrightarrow c \times \overrightarrow b $ and $\overrightarrow a - \overrightarrow c $ with equation (4), we get

$\overrightarrow c \times \overrightarrow b = \overrightarrow a - \overrightarrow c $ ..... (5)

Multiplying both side with $\overrightarrow b $, we get

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

$ \Rightarrow 0 = 0 - \overrightarrow b \,.\,\overrightarrow c $ [Given $\overrightarrow a \,.\,\overrightarrow b = 0$]

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

$\therefore$ (B) is correct.

Now multiplying both side of equation (5) with $\overrightarrow c $, we get

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

$ \Rightarrow 0 = \overrightarrow a \,.\,\overrightarrow c - |\overrightarrow c {|^2}$

$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$

Now given $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$

$\therefore$ $|\overrightarrow c {|^2} = \sqrt {c_1^2 + c_2^2 + c_3^2} $

$ \Rightarrow |\overrightarrow c {|^2} = c_1^2 + c_2^2 + c_3^2$

$\therefore$ $|\overrightarrow c {|^2} = 0$ when ${c_1} = 0$, ${c_2} = 0$ and ${c_3} = 0$

Now if we put ${c_1} = {c_2} = {c_3} = 0$ in matrix we get,

$\left[ {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right]\left[ {\matrix{ 1 \cr {{b_2}} \cr {{b_3}} \cr } } \right] = \left[ {\matrix{ {3 - {c_1}} \cr {1 - {c_2}} \cr { - 1 - {c_3}} \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right] = \left[ {\matrix{ 3 \cr 1 \cr { - 1} \cr } } \right]$

$\therefore$ Matrix does not satisfy the value of ${c_1} = {c_2} = {c_3} = 0$.

$\therefore$ ${c_1},{c_2},{c_3}$ can't be zero.

$\therefore$ $|\overrightarrow c {|^2} \ne 0$

$ \Rightarrow \overrightarrow a \,.\,\overrightarrow c \ne 0$

$\therefore$ Option (A) is wrong.

As $\overrightarrow a \,.\,\overrightarrow c = |\overrightarrow c {|^2}$

$ \Rightarrow |\overrightarrow a ||\overrightarrow c |\cos \theta = |\overrightarrow c {|^2}$ [$\theta$ = angle between $\overrightarrow a $ and $\overrightarrow c $]

$ \Rightarrow |\overrightarrow a |\cos \theta = |\overrightarrow c |$

$ \Rightarrow (\sqrt {9 + 1 + 1} )\cos \theta = |\overrightarrow c |$

$ \Rightarrow |\overrightarrow c | = \sqrt {11} \cos \theta $

$ \Rightarrow |\overrightarrow c | \le \sqrt {11} $ [as $ - 1 \le \cos \theta \le 1$]

$\therefore$ Option (D) is correct.

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

$ \Rightarrow (3\widehat i + \widehat j - \widehat k)\,.\,(\widehat i + {b_2}\widehat j + {b_3}\widehat k) = 0$

$ \Rightarrow 3 + {b_2} - {b_3} = 0$

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

Now,

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

$ \Rightarrow |\overrightarrow b | = \sqrt {1 + b_2^2 + b_3^2} $

$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + {(3 + {b_2})^2}$

$ \Rightarrow |\overrightarrow b {|^2} = 1 + b_2^2 + 9 + 6{b_2} + b_2^2$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 6{b_2} + 2b_2^2$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}(3 + {b_2})$

$ \Rightarrow |\overrightarrow b {|^2} = 10 + 2{b_2}{b_3}$

$ \Rightarrow |\overrightarrow b {|^2} > 10$ [as ${b_2}{b_3} > 0$ then $2{b_2}{b_3} > 0$]

$ \Rightarrow |\overrightarrow b | > \sqrt {10} $

$\therefore$ Option (C) is correct.

Given,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

But $\lambda$ > 0

$\therefore$ $\lambda$ = 1

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

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

Option (a)

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

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

Option (b)

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

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

Option (c)

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

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

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

Option (d)

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

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

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

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

Given vectors $PQ = a\widehat i + b\widehat j$ and $PS = a\widehat i - b\widehat j$ are adjacent sides of a parallelogram PQRS, so area of parallelogram PQRS =

|PQ $ \times $ PS| = 2ab = 8 (given)

$ \therefore $ ab = 4 ......(i)

According to the question,

|u| = |projection vector of w = $\widehat i$ + $\widehat j$ along PQ|

$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right| = {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }}$

and, similarly, |v| = |projection vector of $w = \widehat i + \widehat j$ along PS|

$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right|$

$ = {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} \Rightarrow |u| + |v|\, = \,|w|$

$ \Rightarrow {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }} + {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} = \sqrt 2 $

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

If a $ \ge $ b $ \ge $ 0, then $2a = \sqrt 2 \sqrt {{a^2} + {b^2}} $

$ \Rightarrow 4{a^2} = 2{a^2} + 2{b^2} \Rightarrow {a^2} = {b^2}$

$ \Rightarrow a = b$ ....(ii)

From Eqs. (i) and (ii), we get

$a = 2 = b \Rightarrow a + b = 4$

and the length of diagonal PR

$ = |a\widehat i + b\widehat j + a\widehat i - b\widehat j|$

$ = |2a\widehat i| = 2a = 4$

And, the angle bisector of vector PQ and PS is along the vector

$ \pm {\lambda \over {\sqrt {{a^2} + {b^2}} }}((a\widehat i + b\widehat j) \pm (a\widehat i - b\widehat j))$

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.

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} $

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

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 $(\mathrm{C})$.

Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$, $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$ and $\overrightarrow c = \widehat i + \widehat j + \widehat k$.

Any vector in the plane of $\widehat i + \widehat j + 2\widehat k$ and $\widehat i + 2\widehat j + \widehat k$ is given by

$\overrightarrow r = \lambda \overrightarrow a + \mu \overrightarrow b $

$ = \lambda (\widehat i + \widehat j + 2\widehat k) + \mu (\widehat i + 2\widehat j + \widehat k)$

$ = (\lambda + \mu )\widehat i + (\lambda + 2\mu )\widehat j + (2\lambda + \mu )\widehat k$

Also, $\overrightarrow r \,.\,\overrightarrow c = 0$

$ \Rightarrow (\lambda + \mu )\,.\,1 + (\lambda + 2\mu )\,.\,1 + (2\lambda + \mu )\,.\,1 = 0$

$ \Rightarrow 4\lambda + 4\mu = 0$

$ \Rightarrow \lambda + \mu = 0$

$ \Rightarrow \left[ {\matrix{ {\overrightarrow r } & {\overrightarrow a } & {\overrightarrow b } \cr } } \right] = 0$

So, vectors $\widehat j - \widehat k$ and $ - \widehat j + \widehat k$ satisfy this.