⇒Vectors

Correct answer option is (B)

The vector perpendicular to both vectors containing $(\hat{i}+\hat{j}+\hat{k})$ and $(\hat{i}+2\hat{j}+3\hat{k})$ is

$\begin{array}{rl}& =(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}+2\hat{j}+3\hat{k})\\ & =\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|\\ & =\hat{i}-2\hat{j}+\hat{k}\end{array}$

Therefore, the magnitude of the projection of vector $(2\hat{i}+\hat{j}+\hat{k})$ on $(\hat{i}-2\hat{j}+\hat{k})$ is

$\begin{array}{rl}& =\left|\frac{(2\hat{i}+\hat{j}+\hat{k})(\hat{i}-2\hat{j}+\hat{k})}{\sqrt{1^2+(-2{)}^2+(1{)}^2}}\right|\\ & =\left|\frac{2-2+1}{\sqrt{6}}\right|\\ & =\frac{1}{\sqrt{6}}\end{array}$

Correct answer option is (D)

Given equation of the line is

$\begin{array}{rl}\bar{\mathrm{r}}& =\bar{\mathrm{a}}+\lambda \bar{\mathrm{b}}\\ \therefore \bar{\mathrm{r}}& =2\hat{\mathrm{i}}-3\hat{\mathrm{j}}-4\hat{\mathrm{k}}+\lambda (6\hat{\mathrm{i}}+3\hat{\mathrm{j}}-4\hat{\mathrm{k}})\end{array}$

To find: Its distance from point $\overline{\alpha }=\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$,

$\begin{array}{rl}& \therefore \text{ Required distance }=\sqrt{|\overline{\alpha }-\bar{\mathrm{a}}{|}^2-{\left[\frac{(\overline{\alpha }-\bar{\mathrm{a}})\cdot \bar{\mathrm{b}}}{|\bar{\mathrm{b}}|}\right]}^2}\\ & \text{ Here, }|\overline{\alpha }-\bar{\mathrm{a}}{|}^2=6\text{ and }{\left[\frac{(\overline{\alpha }-\bar{\mathrm{a}})\cdot \bar{\mathrm{b}}}{|\bar{\mathrm{b}}|}\right]}^2=\frac{25}{61}\\ & \therefore \text{ Required distance }=\sqrt{\frac{341}{61}}\end{array}$

Correct answer option is (A)

$(2\hat{i}+4\hat{j}-5\hat{k})+(\lambda \hat{i}+2\hat{j}+3\hat{k})=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$

Unit vector along the above vector is

$\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+6^2+(-2{)}^2}}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}$

Scalar product of $(\hat{i}+\hat{j}+\hat{k})$ with this unit vector is 1 .

$\begin{array}{ll}\therefore & (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\cdot \frac{(2+\lambda )\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{\sqrt{{\lambda }^2+4\lambda +44}}=1\\ & \frac{(2+\lambda )+6-2}{\sqrt{{\lambda }^2+4\lambda +44}}=1\\ & \sqrt{{\lambda }^2+4\lambda +44}=\lambda +6\\ \therefore & {\lambda }^2+4\lambda +44=(\lambda +6{)}^2\\ & 8\lambda =8\\ & \lambda =1\end{array}$

Correct answer option is (A)

$\begin{array}{rl}\overline{a}+\lambda \overline{b}& =(\hat{i}+2\hat{j}-3\hat{k})+\lambda (3\hat{i}-\hat{j}+2\hat{k})\\ & =(1+3\lambda )\hat{i}+(2-\lambda )\hat{j}+(-3+2\lambda )\hat{k}\end{array}$

Since $\bar{\mathrm{a}}+\lambda \bar{\mathrm{b}}$ is $\perp$ er to $\bar{\mathrm{c}}$, we write

$\begin{array}{rl}& (1)(1+3\lambda )+(3)(2-\lambda )+(1)(-3+2\lambda )=0\\ & \therefore 1+3\lambda +6-3\lambda -3+2\lambda =0\Rightarrow \lambda =-2\end{array}$

Correct answer option is (C)

$\begin{array}{ll}|\bar{\mathrm{a}}\times \bar{\mathrm{b}}|=25& \\ \therefore |\bar{\mathrm{a}}|\cdot |\bar{\mathrm{b}}|\cdot \sin \theta =(5)(13)\sin \theta =25& \\ \therefore \sin \theta =\frac{5}{13}\Rightarrow \cos \theta =\frac{-12}{13}& \dots [\because \frac{\pi }{2}<\theta \le \pi ]\\ \therefore \bar{\mathrm{a}}\cdot \bar{\mathrm{b}}=|\bar{\mathrm{a}}|\cdot |\bar{\mathrm{b}}|\cdot \cos \theta & \\ =(5)(13)\left(\frac{-12}{13}\right)=-60& \end{array}$