Vector Algebra

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.

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

$\overrightarrow {AB} = 2\widehat i + 10\widehat j + 11\widehat k$

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

Angle '$\theta$' between $\overrightarrow {AB} $ and $\overrightarrow {AD} $ is

$\cos (\theta ) = \left| {{{\overrightarrow {AB} \,.\,\overrightarrow {AD} } \over {\left| {\overrightarrow {AB} } \right|\left| {\overrightarrow {AD} } \right|}}} \right|$

$ = \left| {{{ - 2 + 20 + 22} \over {(15)(3)}}} \right| = {8 \over 9}$

$ \Rightarrow \sin (\theta ) = {{\sqrt {17} } \over 9}$

Since, $\alpha + \theta = 90^\circ $

$\therefore$ $\cos (\alpha ) = \cos (90^\circ - \theta )$

$ = \sin (\theta ) = {{\sqrt {17} } \over 9}$

We have $PS = \sqrt {{1^2} + {3^2}} = \sqrt {10} $

$SR = \sqrt {{6^2} + {1^2}} = \sqrt {37} $, $RQ = \sqrt {{1^2} + {3^2}} = \sqrt {10} $

$QP = \sqrt {{6^2} + {1^2}} = \sqrt {37} $

Then PQRS is a parallelogram. It can be a rectangle, so let's check the product of slopes of PS and SR

$({m_{PS}})\,.\,({m_{SR}}) = {3 \over { - 1}} \times {1 \over 6} = - {1 \over 2} \ne - 1$

Thus PS and SR are not perpendicular. So, it's not a rectangle.

Also, ${m_{PR}}\,.\,{m_{QS}} = {4 \over 5} \times {2 \over { - 7}} = - {8 \over {35}} \ne - 1$

Thus, PR and QS are also not perpendicular. So it's not a rhombus either.

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

$ \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 $\left( {p,q} \right)$

The given equation, $(\overrightarrow a \times \overrightarrow b )\,.\,(\overrightarrow c \times \overrightarrow d ) = 1$, is possible only when $|\overrightarrow a \times \overrightarrow b | = |\overrightarrow c \times \overrightarrow d | = 1$ and $(\overrightarrow a \times \overrightarrow b )||(\overrightarrow c \times \overrightarrow d )$.

Since $\overrightarrow a \,.\,\overrightarrow c = 1/2$ and $\overrightarrow b ||\overrightarrow d $, we get $|\overrightarrow c \times \overrightarrow d | \ne 1$; hence, we conclude that the vectors $\overrightarrow b $ and $\overrightarrow d $ are non-parallel.

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 $\pi .$

Vector in the direction of ${L_1} = \overrightarrow {{n_1}} = 3\widehat i + \widehat j + 2\widehat k$

Vector in the direction of ${L_2} = \overrightarrow {{n_2}} = \widehat i + 2\widehat j + 3\widehat k$

$\therefore$ Vector perpendicular to both L$_1$ and L$_2$

$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

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

$ = - \widehat i - 7\widehat j + 5\widehat k$

$\therefore$ Required unit vector

$ = \widehat n = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {1 + 49 + 25} }} = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$

$|\overrightarrow {OP} | = |\widehat a\cos t + \widehat b\sin t|$

$|\overrightarrow {OP} {|^2} = (\widehat a\cos t + \widehat b\sin t)\,.\,(\widehat a\cos t + \widehat b\sin t)$

$ = |\widehat a{|^2}{\cos ^2}t + 2\widehat a\,.\,\widehat b\cos t\sin t + |\widehat b{|^2}{\sin ^2}t$

$ = {\cos ^2}t + \widehat a\,.\,\widehat b\sin 2t + {\sin ^2}t$

$ = 1 + \widehat a\,.\,\widehat b\sin 2t$

The greatest value of $|\overrightarrow {OP} |$ is reached when $\sin 2t = 1$

i.e., $t = {\pi \over 4}$

$M = |\overrightarrow {OP} |$ greatest $ = \sqrt {1 + \widehat a\,.\,\widehat b} $

The direction of $\overrightarrow {OP} $ is given by

$\widehat u = {{\widehat a + \widehat b} \over {\sqrt 2 \,.\,{{|\widehat a + \widehat b|} \over {\sqrt 2 }}}} = {{\widehat a + \widehat b} \over {|\widehat a + \widehat b|}}$

The shortest distance between L$_1$ and L$_2$ is

${{(\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} )} \over {|\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} |}} = (\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )\,.\,\widehat n$

Where, ${a_1} = - \widehat i - 2\widehat j - \widehat k$

${a_2} = 2\widehat i - 2\widehat j + 3\widehat k$

$\therefore$ $\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = 3\widehat i + 4\widehat k$

$\therefore$ $({\widehat a_2} - {\widehat a_1})\,.\,\widehat n$

$ = (3\widehat i + 4\widehat k)\,.\,\left( {{{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}} \right)$

$ = {{ - 3 + 20} \over {5\sqrt 3 }} = {{17} \over {5\sqrt 3 }}$

The important thing to remember in this is the formula

${\left[ {\overrightarrow x \,.\,\overrightarrow y \,.\,\overrightarrow z } \right]^2} = \left| {\matrix{ {\overrightarrow x \,.\,\overrightarrow x } & {\overrightarrow x \,.\,\overrightarrow y } & {\overrightarrow x \,.\,\overrightarrow z } \cr {\overrightarrow y \,.\,\overrightarrow x } & {\overrightarrow y \,.\,\overrightarrow y } & {\overrightarrow y \,.\,\overrightarrow z } \cr {\overrightarrow z \,.\,\overrightarrow x } & {\overrightarrow z \,.\,\overrightarrow y } & {\overrightarrow z \,.\,\overrightarrow z } \cr } } \right|$

Volume of the parallelopiped $v = \left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right]$

$\therefore$ ${v^2} = {\left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right]^2} = \left| {\matrix{ {\widehat a\,.\,\widehat a} & {\widehat a\,.\,\widehat b} & {\widehat a\,.\,\widehat c} \cr {\widehat b\,.\,\widehat a} & {\widehat b\,.\,\widehat b} & {\widehat b\,.\,\widehat c} \cr {\widehat c\,.\,\widehat a} & {\widehat c\,.\,\widehat b} & {\widehat c\,.\,\widehat c} \cr } } \right|$

$ = \left| {\matrix{ 1 & {{1 \over 2}} & {{1 \over 2}} \cr {{1 \over 2}} & 1 & {{1 \over 2}} \cr {{1 \over 2}} & {{1 \over 2}} & 1 \cr } } \right| = {1 \over 2}$ (on evaluation)

$ \Rightarrow v = \left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right] = {1 \over {\sqrt 2 }}$

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$

Given, $\quad \vec{a}+\vec{b}+\vec{c}=0$

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

$\Rightarrow \vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\overrightarrow{0}$

$-\vec{c} \times \vec{a}=-\vec{b} \times \vec{c}$

$\vec{c} \times \vec{a}=\vec{b} \times \vec{c}$

Similarly, we shall obtain

$\vec{b} \times \vec{c}=\vec{a} \times \vec{b}$

Note that no two of $\vec{a}, \vec{b}, \vec{c}$ are parallel.

If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ then

$\begin{aligned}& \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \\ & \vec{a}+\vec{b}=-\vec{c} \\ & (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{c}|^{2} \\ & \vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+2 \vec{a} \cdot \vec{b}=|\vec{c}|^{2} \\ & 1+1+2 \times 1 \times 1 \cos \theta \approx 1 \\ & \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \\ & \text { so } \vec{a} \neq \vec{b} \text { similarly, } \vec{b} \neq \vec{c}, \vec{a} \neq \vec{c} \\ & \therefore \vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \neq 0 \end{aligned}$

We know that for coplanar of vector

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

${R_1} \to {R_1} + {R_2} + {R_3}$

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

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

$(2 - {\lambda ^2})\left| {\matrix{ 1 & 1 & 1 \cr 0 & { - (1 + {\lambda ^2})} & 0 \cr 0 & 0 & { - (1 + {\lambda ^2})} \cr } } \right| = 0$

$({R_1} - {R_1},{R_3} - {R_1})$

$ \Rightarrow (2 - {\lambda ^2}){(1 + {\lambda ^2})^2} = 0$

$\lambda = \, \pm \sqrt 2 $

$\therefore$ Two real solutions.

Statement 1 : $\overrightarrow {PQ} \times (\overrightarrow {RS} + \overrightarrow {ST} ) \ne 0$ since $\overrightarrow {PQ} $ is not parallel to $\overrightarrow {RT} $,

We get,

$\overrightarrow {PQ} \times \overrightarrow {RT} $

That is statement 1 is true.

$\overrightarrow {PQ} \times \overrightarrow {RT} \ne 0$

Statement 2 : $\overrightarrow {PQ} \times \overrightarrow {RS} \ne \overrightarrow 0 $ and $\overrightarrow {PQ} \times \overrightarrow {ST} \ne \overrightarrow 0 $ since $\overrightarrow {PQ} $ is not parallel to $\overrightarrow {RS} $.

Statement 2 is false.

$\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$

$ \begin{aligned} \vec{a} & =\hat{i}+2 \hat{j}+\hat{k} \\ \vec{b} & =\hat{i}-\hat{j}+\hat{k} \\ \vec{c} & =\hat{i}-\hat{j}-\hat{k} \end{aligned} $

Step 1: Write the vector in the plane of $\vec{a}$ and $\vec{b}$

Any vector in the plane made by $\vec{a}$ and $\vec{b}$ can be written as:

$ \vec{r} = \vec{a} + \lambda \vec{b} $

Substituting the values:

$ \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) $

$ \vec{r} = (1+\lambda)\hat{i} + (2-\lambda)\hat{j} + (1+\lambda)\hat{k} $

Step 2: Use the projection condition

The question asks for a vector whose projection onto $\vec{c}$ is $\frac{1}{\sqrt{3}}$. The formula for the projection of $\vec{r}$ on $\vec{c}$ is:

$ \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}} $

First, find $|\vec{c}|$:

$ \vec{c} = \hat{i} - \hat{j} - \hat{k} $ So, $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$

Step 3: Calculate the dot product

Find $\vec{r} \cdot \vec{c}$:

$ \left[ (1+\lambda)\hat{i} + (2-\lambda)\hat{j} + (1+\lambda)\hat{k} \right] \cdot (\hat{i} - \hat{j} - \hat{k}) $

Multiply out each part:
$\hat{i}$ part: $1+\lambda$
$-\hat{j}$ part: $- (2-\lambda)$
$-\hat{k}$ part: $- (1+\lambda)$

So, $ (1+\lambda) - (2-\lambda) - (1+\lambda) $

Simplify:

$(1+\lambda) - 2 + \lambda - 1 - \lambda = 1+\lambda - 2 + \lambda - 1 - \lambda$

Combine like terms:
$1 - 2 - 1 + \lambda + \lambda - \lambda = -2 + \lambda$

Step 4: Set the projection equal to the given value and solve for $\lambda$

$ \frac{-2 + \lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}} $

Multiply both sides by $\sqrt{3}$: $ -2 + \lambda = 1 $

So, $ \lambda = 3 $

Step 5: Find the required vector

Plug $\lambda = 3$ back into the vector expression:

$ \vec{r} = (1+3)\hat{i} + (2-3)\hat{j} + (1+3)\hat{k} = 4\hat{i} - \hat{j} + 4\hat{k} $

$ \begin{aligned} &\begin{array}{ll} (i)& \quad x+y=(a) \text { and } a x-y=1 \\ & \quad x+a x-1=|a| \\ \Rightarrow & \quad x=\frac{|a|+1}{a+1}>0 \quad y=\frac{|a|-1}{a+1}>0 \\ & \quad a+1>0 \text { also }|a|-1>0 \quad|a|>1 \\ \therefore & \quad a_0=1 \end{array}\\ &\text { (ii) since, } \vec{a} \cdot \hat{k}=\gamma \end{aligned} $

$ \begin{aligned} & \text { Now, } \quad \hat{k} \times(\hat{k} \times \vec{a})=(\hat{k} \cdot \vec{a}) \hat{k}-(\hat{k} \cdot \hat{k}) \cdot \vec{a}=0 \\ & \Rightarrow \quad y \hat{k}-(\alpha \hat{i}+\beta \hat{j}+y \hat{k})=0 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & -(\alpha \hat{i}+\beta \hat{j}) & =0 \\ \Rightarrow & \alpha=\beta & =0 \\ \text { Also } & \alpha+\beta+\gamma & =2 \\ \Rightarrow & \gamma & =2 \end{array} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} (iii) & \left|\int_0^1\left(1-y^2\right) d y\right|+\left|\int_0^1\left(y^2-1\right) d y\right| \\ & =2\left|\int_0^1\left(1-y^2\right) d y\right|=\frac{4}{3} \\ & \left|\int_0^1 \sqrt{1-x} d x\right|+\left|\int_{-1}^0 \sqrt{1+x} d x\right| \\ & =2 \int_0^1 \sqrt{1-x} d x \\ & =2 \int_0^1 \sqrt{x} d x \end{aligned} \end{aligned} $

$ =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^1=\frac{4}{3} $

(concept)

$ \because \int_{-1}^0 \sqrt{1+x} $

Put $x=-t$

$ \begin{aligned} \Rightarrow \quad d x= & -d t \\ & \int_1^0 \sqrt{1-t}(-d t) \\ = & \int_0^1 \sqrt{1-t} d t \\ = & \int_1^1 \sqrt{1-t} d x \end{aligned} $

(iv) $\sin A \sin B \sin C+\cos A \cos B$

$ \begin{aligned} & \leq \sin \mathrm{A} \sin \mathrm{~B}+\cos \mathrm{A} \cos \mathrm{~B} \\ & \leq \cos (\mathrm{A}-\mathrm{B}) \\ & \cos (\mathrm{A}-\mathrm{B}) \geq 1 \quad \Rightarrow \quad \cos (\mathrm{~A}-\mathrm{B})=1 \\ & \sin \mathrm{C}=1 \end{aligned} $

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$

Since $\hat{v}$ unit vector along the incident ray and $\widehat{w}$ unit vector along the reflected ray

Hence, $\hat{a}$ is a unit vector along the external bisector of $\hat{v}$ and $\widehat{w}$

$\therefore \quad \hat{w}-\hat{v}=\lambda \hat{a}$

Squaring on both sides

${(\widehat w - \widehat v)^2} = {(\lambda \widehat a)^2}$

$\Rightarrow 1+1-2\cos2\theta=\lambda^2$

$\Rightarrow \quad \lambda^2=2(1-cos2\theta)$

$\Rightarrow \quad \lambda^{2}=2 \cdot 2 \sin ^{2} \theta$

$\Rightarrow \quad \lambda=2 \sin \theta$

Hence, $\quad \widehat{w}-\hat{v}=2 \sin \theta \cdot \hat{a}$

$\begin{aligned} & =2 \cos \left(90^{\circ}-\theta\right) \cdot \hat{a} \\ & =-(2 \hat{a} \cdot \hat{v}) \cdot \hat{a} \\ \widehat{w} & =\hat{v}-2(\hat{a} \cdot \hat{v}) \cdot \hat{a} \end{aligned}$

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 $ $\left. {} \right]$

If $\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ is collinear with $\overrightarrow{\mathbf{c}}$, then

$ \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}=t \overrightarrow{\mathbf{c}} $

Also, if $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$, then

$ \begin{aligned} & \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \\\\ & \Rightarrow \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}} \end{aligned} $

On putting this value in Eq. (i), we get

$ \overrightarrow{\mathbf{a}}+2(\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}} $

$ \Rightarrow \overrightarrow{\mathbf{a}}+2 \lambda \overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}}=t \overrightarrow{\mathbf{c}} $

$ \Rightarrow (\overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}-2 \lambda \overrightarrow{\mathbf{a}} $

On comparing, we get

and $-6=t $

$\Rightarrow t=-6$

From Eq. (i),

$\vec{a}+2 \vec{b}=-6 \vec{c} $

$\Rightarrow \vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$

The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:

$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 4\widehat i + 2\widehat j - 2\widehat k$

The total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:

$\mathbf{work\ done\ by\ force\ 1} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 4 \cdot 4 + 1 \cdot 2 - 3 \cdot -2 = 16 + 2 + 6 = 24$

$\mathbf{work\ done\ by\ force\ 2} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 3 \cdot 4 + 1 \cdot 2 - 1 \cdot -2 = 12 + 2 + 2 = 16$

The total work done by the forces is the sum of the work done by each force:

$\mathbf{total\ work\ done} = 24 + 16 = 40$

Therefore, the total work done by the forces is 40 J (joules) or 40 units.

Projection of $\overrightarrow v $ along $\overrightarrow u = {{\overrightarrow v .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow v .\overrightarrow u } \over 2}$

projection of $\overrightarrow w $ along $\overrightarrow u = {{\overrightarrow w .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow w .\overrightarrow u } \over 2}$

Given ${{\overrightarrow v .\overrightarrow u } \over 2} = {{\overrightarrow w .\overrightarrow u } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Also, $\overrightarrow v .\overrightarrow w = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Now ${\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|^2}$

$ = {\left| {\overrightarrow u } \right|^2} + {\left| {\overrightarrow v } \right|^2} + {\left| {\overrightarrow w } \right|^2} - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\overrightarrow u .\overrightarrow v - 2\overrightarrow v .\overrightarrow w + 2\overrightarrow u .\overrightarrow w $

$ = 1 + 4 + 9 + 0$ [ From $(1)$ and $(2)$ ] $=14$

$\therefore$ $\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right| = \sqrt {14} $

Vectors $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\lambda \overrightarrow b + 4\overrightarrow c ,\,\,\,$

and $\left( {2\lambda - 1} \right)\overrightarrow c $ are

coplanar if $\left| {\matrix{ 1 & 2 & 3 \cr 0 & \lambda & 4 \cr 0 & 0 & {2\lambda - 1} \cr } } \right| = 0$

$ \Rightarrow \lambda \left( {2\lambda - 1} \right) = 0 \Rightarrow \lambda = 0$ or ${1 \over 2}$

$\therefore$ Forces are noncoplanar for all $\lambda ,$

except $\lambda = 0,{1 \over 2}$

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

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

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

$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = {{ - 1 - 4 - 9} \over 2}$

$ = - 7$

Vector perpendicular to the face $OAB$

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

Vector perpendicular to the face $ABC$

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

Angle between the faces $=$ angle between their normals

$\cos \theta = \left| {{{5 + 5 + 9} \over {\sqrt {35} \sqrt {35} }}} \right| = {{19} \over {35}}$

or $\theta = {\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$

$\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w - \overrightarrow v \times \overrightarrow v + \overrightarrow v \times \overrightarrow w } \right)$

$ = \left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w + \overrightarrow v \times \overrightarrow w } \right) = \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow v } \right)$

$ - \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow w } \right) + \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) + \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow v } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right)$

$ + \overrightarrow v .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right) + \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right)$

$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right)$

$ = \left[ {\left. {\overrightarrow u \overrightarrow v \overrightarrow w } \right)} \right] + \left[ {\left. {\overrightarrow v \overrightarrow w \overrightarrow u } \right)} \right] - \left[ {\overrightarrow w \overrightarrow u \overrightarrow v } \right]$

$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$

$\left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ a & {{a^2}} & 1 \cr b & {{b^2}} & 1 \cr c & {{c^2}} & 1 \cr } } \right| + \left| {\matrix{ a & {{a^2}} & {{a^3}} \cr b & {{b^2}} & {{b^3}} \cr c & {{c^2}} & {{c^3}} \cr } } \right| = 0$

$ \Rightarrow \left( {1 + abc} \right)\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| = 0$

As $\,\,\,\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0$ (given condition)

$\therefore$ $abc=-1$

$A = \left( {7, - 4,7} \right),B = \left( {1, - 6,10} \right),$

$C = \left( { - 1, - 3,4} \right)$ and $D = \left( {5, - 1,5} \right)$

$AB = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( { - 4 + 6} \right)}^2} + {{\left( {7 - 10} \right)}^2}} $

$ = \sqrt {36 + 4 + 9} = 7$

Similarly $BC = 7,\,\,CD = \sqrt {41} ,\,\,DA = \sqrt {17} $

$\therefore$ None of the options is satisfied

Since, $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{u}}$ and $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{v}}$

$ \begin{aligned} \Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \\\\ \therefore|\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}| & =\left|\frac{\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}\right| \\\\ & =\frac{|\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \end{aligned} $

$ \begin{aligned} & \text { Now, } \overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})=\left|\begin{array}{rrr} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{array}\right|=-6 \\\\ & \begin{array}{ll} \Rightarrow |\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|=6 \\\\ \text { and } \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-2 \hat{\mathbf{k}} \\\\ \Rightarrow |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2 \\\\ \therefore |\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}|=\frac{6}{2}=3 \end{array} \end{aligned} $

$PV\,\,$ of $\,\,\overrightarrow {AD} $ $ = {{\left( {3 + 5} \right)i + \left( {0 - 2} \right)j + \left( {4 + 4} \right)k} \over 2}$

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

or $\,\,\,\left| {\overrightarrow {AD} } \right| = \sqrt {16 + 16 + 1} = \sqrt {33} $