⇒Vectors

Correct answer option is (A)

Given that, $\bar{\mathrm{a}}+2\bar{\mathrm{b}}$ and $5\bar{\mathrm{a}}-4\bar{\mathrm{b}}$ are perpendicular to each other.

$\begin{array}{rl}& \therefore (\overline{a}+2\overline{b})\cdot (5\overline{a}-4\overline{b})=0\\ & \Rightarrow 5|\overline{a}{|}^2-8|\overline{b}{|}^2-4\overline{a}\cdot \overline{b}+10\overline{b}\cdot \overline{a}=0\\ & \Rightarrow -3+6\overline{a}\cdot \overline{b}=0...[|\overline{a}|=|\overline{b}|=1]\\ & \Rightarrow 6|\overline{a}||\overline{b}|\cos \theta =3\\ & \Rightarrow \cos \theta =\frac{1}{2}\\ & \Rightarrow \theta =\frac{\pi }{3}\end{array}$

Correct answer option is (B)

Let $\overline{a}$ be the vector joining $\mathrm{A}(-2,0,3)$ and $\mathrm{B}(1,4,2)$.

$\begin{array}{rl}& \therefore \bar{\mathrm{a}}=(1-(-2))\hat{\mathrm{i}}+(4-0)\hat{\mathrm{j}}+(2-3)\hat{\mathrm{k}}\\ & =3\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}}\\ & \text{ and }\overline{b}=6\hat{i}-2\hat{j}+3\hat{k}\\ & \therefore \text{ Projection }=\frac{\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}}{|\bar{\mathrm{b}}|}=\frac{3\times 6+4\times (-2)-1\times 3}{\sqrt{6^2+(-2{)}^2+3^2}}\\ & =\frac{18-8-3}{\sqrt{49}}\\ & =\frac{7}{7}\\ & =1\end{array}$

Correct answer option is (C)

Let $\bar{\mathrm{d}}=\bar{\mathrm{a}}+\lambda \bar{\mathrm{b}}$

$\begin{array}{rl}\bar{\mathrm{d}}& =(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}})+\lambda (2\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\\ & =2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}+2\lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}\\ & =(2\lambda +2)\hat{\mathrm{i}}+(3+\lambda )\hat{\mathrm{j}}+(2-\lambda )\hat{\mathrm{k}}\end{array}$

Now, $\bar{\mathrm{d}}$ is perpendicular to $\bar{\mathrm{c}}$.

$\begin{array}{rl}\therefore & \overline{c}\cdot \overline{d}=0\\ & \Rightarrow (\hat{i}+3\hat{j})\cdot [(2\lambda +2)\hat{i}+(3+\lambda )\hat{j}+(2-\lambda )\hat{k}]=0\\ & \Rightarrow 1(2\lambda +2)+3(3+\lambda )=0\\ & \Rightarrow 2\lambda +2+9+3\lambda =0\\ & \Rightarrow 5\lambda +11=0\\ & \Rightarrow \lambda =\frac{-11}{5}\end{array}$

Correct answer option is (C)

$\begin{array}{rl}& \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}-2\hat{\mathrm{k}}\\ & \overline{\mathrm{CD}}=3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}}\end{array}$

Vector projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{CD}}$

$\begin{array}{rl}& =(\overline{\mathrm{AB}}\cdot \overline{\mathrm{CD}})\frac{\overline{\mathrm{CD}}}{|\overline{\mathrm{CD}}{|}^2}\\ & =(-3+6-4)\frac{(3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}})}{{\left(\sqrt{3^2+(-6{)}^2+2^2}\right)}^2}\\ & =\frac{-1}{49}(3\hat{\mathrm{i}}-6\hat{\mathrm{j}}+2\hat{\mathrm{k}})\end{array}$

Correct answer option is (C)

Let $\bar{\mathrm{r}}$ be the vector coplanar to $\bar{\mathrm{a}}$ and $\bar{\mathrm{b}}$. Then,

$\begin{array}{rl}\bar{\mathrm{r}}& =\bar{\mathrm{a}}+\mathrm{m}\bar{\mathrm{b}}\\ & =(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ & =\hat{\mathrm{i}}(1+\mathrm{m})+\hat{\mathrm{j}}(2-\mathrm{m})+\hat{\mathrm{k}}(1+\mathrm{m})\text{.... (i)}\end{array}$

Since the projection of $\bar{\mathrm{r}}$ along $\bar{\mathrm{c}}$ is $\frac{1}{\sqrt{3}},\frac{\bar{\mathrm{r}}\cdot \bar{\mathrm{c}}}{|\bar{\mathrm{c}}|}=\pm \frac{1}{\sqrt{3}}$

$\begin{array}{rl}& \Rightarrow \frac{(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})}{\sqrt{3}}=\pm \frac{1}{\sqrt{3}}\\ & \Rightarrow (1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})=\pm 1\\ \therefore & \mathrm{m}=3\text{ or }\mathrm{m}=1\end{array}$

Substituting $\mathrm{m}=3$ in equation (i), we get

$\begin{array}{rl}& \overline{r}=\hat{i}(1+3)+\hat{j}(2-3)+\hat{k}(1+3)\\ & \Rightarrow \overline{r}=4\hat{i}-\hat{j}+4\hat{k}\end{array}$