Vector Algebra

Let us find the area of triangle $OPR$ step by step.

Given:

• Position vector of $P$ is $\vec{a}$

• Position vector of $Q$ is $\vec{b}$

• Position vector of $R$ is $\vec{a}+\vec{b}$

So, for triangle $OPR$, the two side vectors from $O$ are:

$ \overrightarrow{OP}=\vec{a}, \qquad \overrightarrow{OR}=\vec{a}+\vec{b} $

Hence, area of triangle $OPR$ is

$ \frac{1}{2}\left|\vec{a}\times(\vec{a}+\vec{b})\right| $

Now,

$ \vec{a}\times(\vec{a}+\vec{b})=\vec{a}\times\vec{a}+\vec{a}\times\vec{b}=\vec{a}\times\vec{b} $

since $\vec{a}\times\vec{a}=0$.

Therefore,

$ \text{Area of } \triangle OPR=\frac{1}{2}|\vec{a}\times\vec{b}| $

So we only need to find $|\vec{a}\times\vec{b}|$.

We are given:

$ |\vec{a}+\vec{b}|=\sqrt{21} $

Squaring both sides,

$ |\vec{a}+\vec{b}|^2=21 $

So,

$ |\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}=21 \qquad (1) $

Also,

$ |\vec{a}-\vec{b}|=3 $

Squaring,

$ |\vec{a}-\vec{b}|^2=9 $

Thus,

$ |\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}=9 \qquad (2) $

Now subtract $(2)$ from $(1)$:

$ 4\vec{a}\cdot\vec{b}=12 $

So,

$ \vec{a}\cdot\vec{b}=3 $

Now use the condition that $\vec{a}$ is perpendicular to $(\vec{a}-\vec{b})$.

Therefore,

$ \vec{a}\cdot(\vec{a}-\vec{b})=0 $

$ |\vec{a}|^2-\vec{a}\cdot\vec{b}=0 $

Since $\vec{a}\cdot\vec{b}=3$, we get

$ |\vec{a}|^2=3 $

So,

$ |\vec{a}|=\sqrt{3} $

Now use equation $(2)$:

$ |\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}=9 $

Substitute $|\vec{a}|^2=3$ and $\vec{a}\cdot\vec{b}=3$:

$ 3+|\vec{b}|^2-6=9 $

$ |\vec{b}|^2=12 $

Now use the identity

$ |\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2 $

Substitute the values:

$ |\vec{a}\times\vec{b}|^2=(3)(12)-3^2=36-9=27 $

Therefore,

$ |\vec{a}\times\vec{b}|=\sqrt{27}=3\sqrt{3} $

Hence area of triangle $OPR$ is

$ \frac{1}{2}|\vec{a}\times\vec{b}|=\frac{1}{2}(3\sqrt{3}) $

$ \boxed{\frac{3\sqrt{3}}{2}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

Given the matrix

$ M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}, $

and $ MM^{T} = I $, this means that the rows of $ M $ are orthonormal vectors (that is, each has unit length and they are mutually perpendicular).

Let

$ \vec{u} = \alpha \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \gamma \hat{k}, \quad \vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \beta \hat{j} + \delta \hat{k}, \quad \vec{w} = -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \mu \hat{k}. $

Length condition for each row (unit vectors)

For $ \vec{u} $, the condition $ |\vec{u}| = 1 $ gives

$ \alpha^2 + \frac{1}{3} + \gamma^2 = 1 $

Thus,

$ \alpha^2 + \gamma^2 = \frac{2}{3} \tag{1} $

For $ \vec{v} $, the condition $ |\vec{v}| = 1 $ gives

$ \frac{1}{2} + \beta^2 + \delta^2 = 1 $

Therefore,

$ \beta^2 + \delta^2 = \frac{1}{2} \tag{2} $

For $ \vec{w} $, the condition $ |\vec{w}| = 1 $ gives

$ \frac{1}{2} + \frac{1}{3} + \mu^2 = 1 $

Hence,

$ \mu^2 = \frac{1}{6} \tag{3} $

Using the third row’s condition

The third row of $ M $ also has unit length, so

$ \gamma^2 + \delta^2 + \mu^2 = 1 $

Substitute $ \mu^2 = \frac{1}{6} $:

$ \gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6} $

So, for (P), the value of $ \gamma^2 + \delta^2 = \frac{5}{6}. $

Finding $ x $ when $ x\vec{u} + y\vec{v} + z\vec{w} = \hat{j} $

Taking the dot product with $ \vec{u} $ on both sides:

$ x(\vec{u} \cdot \vec{u}) + y(\vec{v} \cdot \vec{u}) + z(\vec{w} \cdot \vec{u}) = \vec{u} \cdot \hat{j} $

Since the vectors are mutually perpendicular, $ \vec{u} \cdot \vec{v} = 0 $ and $ \vec{u} \cdot \vec{w} = 0 $. So,

$ x(1) = \frac{1}{\sqrt{3}} $

Therefore, $ x = \frac{1}{\sqrt{3}}. $

Value of $ |\vec{u} \cdot (\vec{v} \times \vec{w})| $

For an orthogonal matrix, the determinant is $ \pm 1 $, meaning the scalar triple product equals 1 in magnitude. Hence,

$ |\vec{u} \cdot (\vec{v} \times \vec{w})| = 1. $

Value of $ |\vec{u} \times (\vec{v} \times \vec{w})| $

Using vector identity $ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} $, and since the vectors are perpendicular, both $ \vec{u} \cdot \vec{v} = 0 $ and $ \vec{u} \cdot \vec{w} = 0 $. Thus,

$ \vec{u} \times (\vec{v} \times \vec{w}) = 0, $

and its magnitude is 0.

Final Matching:

(P) → (5) $ \dfrac{5}{6} $
(Q) → (4) $ \dfrac{1}{\sqrt{3}} $
(R) → (2) $ 1 $
(S) → (1) $ 0 $

$\begin{aligned} & \vec{u} \times \vec{v}=\vec{w} \text { and } \vec{v} \times \vec{w}=\vec{u} \\ & \Rightarrow \quad \vec{u}, \vec{v} \text { and } \vec{w} \text { are mutually perpendicular. } \\ & (\vec{v} \times \vec{w}) \times \vec{v}=\vec{w} \\ & \Rightarrow \quad \vec{w}(\vec{v} \cdot \vec{v})-\vec{v}(\vec{v} \cdot \vec{w})=\vec{w} \\ & \Rightarrow \quad \vec{w}\left(|\vec{v}|^2-1\right)-\vec{v}(\vec{v} \cdot \vec{w})=\overrightarrow{0}\end{aligned}$

$\begin{aligned} & \Rightarrow \quad|\vec{v}|^2=1 \text { and } \vec{v} \cdot \vec{w}=0 \\ & |\vec{u}||\vec{v}|=|\omega| \Rightarrow|\vec{u}|=\sqrt{6} \\ & \vec{u} \cdot \vec{w}=0 \Rightarrow \alpha+\beta-2 \gamma=0 \\ & \text { and }-t \alpha+\beta+\gamma=0 .......(i) \\ & \alpha-t \beta+\gamma=0 ........(ii) \\ & \alpha+\beta-t \gamma=0 ........(iii) \\ & \text { (ii) }- \text { (i) } \Rightarrow \quad \alpha(1+t)=(t+1) \beta \\ & \text { (iii) }- \text { (ii) } \Rightarrow \quad \beta(1+t)=(1+t) \gamma \\ & \Rightarrow \quad \alpha(1+t)=\beta(1+t)=\gamma(1+t)\end{aligned}$

either $t=-1$ or $\alpha=\beta=\gamma$

$ \begin{aligned} & \Rightarrow \quad \sqrt{\alpha^2+\alpha^2+\alpha^2}=\sqrt{6} \\ & \Rightarrow \quad \alpha=\sqrt{2},-t \alpha=-\alpha-\alpha \\ & \Rightarrow \quad t=2 \end{aligned} $

Since $\alpha=\sqrt{3}$

$ \begin{aligned} \Rightarrow & t=-1 \\ \Rightarrow & \alpha+\beta+\gamma=0 \\ & \alpha+\beta-2 \gamma=0 \\ \Rightarrow & \gamma=0 \end{aligned} $

$(\mathrm{Q}) \rightarrow 1$

$ \begin{aligned} & \alpha+\beta=0 \Rightarrow(\beta+\gamma)=-\alpha \Rightarrow(\beta+\gamma)^2=\alpha^2=3,(\mathrm{R}) \rightarrow 4 \\ & \Rightarrow|\vec{v}|^2=1 \rightarrow(\mathrm{P}) \rightarrow 2 \end{aligned} $

If $\alpha=\sqrt{3} \Rightarrow \gamma^2=0 \rightarrow(\mathrm{Q}) \rightarrow 1$

If $\alpha=\sqrt{3} \Rightarrow(\beta+\gamma)^2=(\sqrt{3})^2=3,(\mathrm{R}) \rightarrow 4$

If $\alpha=\sqrt{2}, t+3=(2)+3=5$, (S) $\rightarrow 5$

$\begin{aligned} & P(\vec{a})=\hat{i}+2 \hat{j}-5 \hat{k} \\\\ & Q(\vec{b})=3 \hat{i}+6 \hat{j}+3 \hat{k} \\\\ & R(\vec{c})=\frac{17}{5} \hat{i}+\frac{16}{5} \hat{j}+7 \hat{k} \\\\ & S(\vec{d})=2 \hat{i}+\hat{j}+\hat{k}\end{aligned}$

From option (A) :


$ \begin{aligned} & {[\overrightarrow{P Q}, \overrightarrow{P R}, \overrightarrow{P S}] \rightarrow \text { S.T.P }} \\\\ & \left|\begin{array}{ccc} 2 & 4 & 6 \\ \frac{12}{5} & \frac{6}{5} & 12 \\ 1 & -1 & 6 \end{array}\right|=0 \end{aligned} $

Hence P, Q, R, S are coplanar.

From option (B) :

$\frac{\vec{b}+2 \vec{d}}{3}=\frac{7 i+8 j+5 k}{3}$


$ \begin{aligned} & \Rightarrow \frac{\frac{17 \lambda}{5}+1}{1+\lambda}=\frac{7}{3} \\\\ & \Rightarrow \frac{(17 \lambda+5)}{1+\lambda}=\frac{35}{3} \\\\ & \Rightarrow 15 \lambda+15=35+35 \lambda \\\\ & \Rightarrow 16 \lambda=20 \\\\ & \Rightarrow \lambda=\frac{5}{4} \end{aligned} $

Hence option (B) is correct.

(D)

$\quad \begin{aligned}|\vec{b} \times \vec{d}|^2 & =|\vec{b}|^2|\vec{d}|^2-(\vec{b} \cdot \vec{d})^2 \\\\ & =54 \times 6-225 \\\\ & =324-225 \\\\ & =99\end{aligned}$

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))$

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

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

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

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

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

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

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

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

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

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

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.

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} $

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

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 $(\mathrm{C})$.

(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.

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

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

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

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

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

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.

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

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

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.

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.

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

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

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

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

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

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.

$ \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} $

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

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