⇒Vectors

Correct answer option is (D)

$\begin{array}{rl}|\bar{\mathrm{u}}\times \bar{\mathrm{v}}|& =|\bar{\mathrm{u}}||\mathrm{v}|\sin \left({150}^{\circ }\right)\\ & =8\times 12\times \frac{1}{2}\\ & =48\end{array}$

Correct answer option is (D)

If angle between $\overline{b}$ and $\overline{c}$ is $\alpha$ and $|\overline{b}\times \overline{c}|=\sqrt{15}$

$\begin{array}{rl}& \Rightarrow |\overline{b}||\overline{c}|\sin \alpha =\sqrt{15}\\ & \Rightarrow \sin \alpha =\frac{\sqrt{15}}{4}\\ & \Rightarrow \cos \alpha =\frac{1}{4}\end{array}$

Now, $\bar{\mathrm{b}}-2\bar{\mathrm{c}}=\lambda \bar{\mathrm{a}}$

$\begin{array}{rl}& \Rightarrow |\overline{b}-2\overline{c}{|}^2={\lambda }^2|\overline{b}{|}^2\\ & \Rightarrow |\overline{b}{|}^2+4|\overline{c}{|}^2-4\overline{b}\cdot \overline{c}={\lambda }^2|\overline{a}{|}^2\\ & \Rightarrow 16+4-4\{|\overline{b}||\overline{c}|\cos \alpha \}={\lambda }^2\\ & \Rightarrow 16+4-4\times 4\times 1\times \frac{1}{4}={\lambda }^2\\ & \Rightarrow {\lambda }^2=16\\ & \Rightarrow \lambda =\pm 4\end{array}$

Correct answer option is (D)

Vector equation of the plane passing through the point $A(\overline{a})$ and parallel to non-zero vectors $\overline{b}$ and $\bar{\mathrm{c}}$ is $\bar{\mathrm{r}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})=\bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}\times \bar{\mathrm{c}})$

Plane $P_1$ is passing through the origin and parallel to vectors $\overline{b_1}=2\hat{j}+3\hat{k}$ and $\overline{c_1}=4\hat{j}-3\hat{k}$

$\therefore \overline{b_1}\times \overline{c_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& 3\\ 0& 4& -3\end{array}\right|=-18\hat{i}$

$\therefore$ Equation of $P_1$ is: $r\cdot (-18\mathrm{i})=0$

Plane $P_2$ is passing through the origin and parallel to vectors $\overline{b_2}=\hat{j}-\hat{k}$ and $\overline{c_2}=3\hat{i}+3\hat{j}$

$\therefore \overline{b_2}\times \overline{c_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& -1\\ 3& 3& 0\end{array}\right|=3\hat{i}-3\hat{j}-3\hat{k}$

$\therefore$ Equation of $P_2$ is : $r_2\cdot (3\hat{i}-3\hat{j}-3\hat{k})=0$

Note that $\bar{\mathrm{A}}$ is parallel to the cross product of $-18\hat{i}$ and $3\hat{i}-3\hat{j}-3\hat{k}$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -18& 0& 0\\ 3& -3& -3\end{array}\right|=-54\hat{j}+54\hat{k}$

Let $\theta$ be the required angle.

$\therefore \theta =$ Angle between $54(-\hat{j}+\hat{k})$ and $2\hat{i}+\hat{j}-2\hat{k}$

$\begin{array}{rl}\therefore \cos \theta & =\frac{54\times (-1-2)}{54\sqrt{0+1+1}\sqrt{4+1+4}}\\ & =\pm \frac{3}{3\sqrt{2}}\\ & =\pm \frac{1}{\sqrt{2}}\\ \therefore \theta =\frac{\pi }{4}& ,\frac{3\pi }{4}\end{array}$

Correct answer option is (B)

Area of the parallelogram with $\overline{a}$ and $\overline{b}$ as two adjacent sides is $|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|$

$\therefore |\bar{\mathrm{a}}\times \bar{\mathrm{b}}|=16$

$\therefore$ Area of the required parallelogram

$\begin{array}{rl}& =|(3\bar{\mathrm{a}}+2\bar{\mathrm{b}})\times (\bar{\mathrm{a}}+3\bar{\mathrm{b}})|\\ & =|3(\bar{\mathrm{a}}\times \bar{\mathrm{a}})+9(\bar{\mathrm{a}}\times \bar{\mathrm{b}})+2(\bar{\mathrm{b}}\times \bar{\mathrm{a}})+(\bar{\mathrm{b}}\times \bar{\mathrm{b}})|\\ & =0+9|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|-2|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|+0\\ & =7|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|\\ & =7\times 16\\ & =112\end{array}$

Correct answer option is (A)

$\begin{array}{rl}& |\overline{c}-\overline{a}|=2\sqrt{2},\\ & \Rightarrow |\overline{c}{|}^2+|\overline{a}{|}^2-2(\overline{a}\cdot \overline{c})=8\\ & \Rightarrow |\overline{c}{|}^2+9-2|\overline{c}|=8\dots [\because \bar{\mathrm{a}}\cdot \bar{\mathrm{c}}=|\bar{\mathrm{c}}|]\\ & \Rightarrow (|\overline{c}|-1{)}^2=0\\ & \Rightarrow |\overline{c}|=1\text{.... (i)}\end{array}$

Now,

$\begin{array}{rl}|(\bar{\mathrm{a}}\times \bar{\mathrm{b}})\times \bar{\mathrm{c}}|& =|(\bar{\mathrm{a}}\times \bar{\mathrm{b}})||\bar{\mathrm{c}}|\sin \frac{\pi }{6}\\ & =|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|(1)\left(\frac{1}{2}\right)\text{... [From (i)]}\\ & =\frac{3}{2}\dots [\because \bar{\mathrm{a}}\times \bar{\mathrm{b}}=2\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}]\end{array}$