Vector Algebra

$\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.

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.$