Question

Choose the correct statement(s) in respect of the lines
$L_1:\overrightarrow{r}=\left(\begin{array}{c}\hat{i}+\hat{j}+\hat{k}\end{array}\right)+\lambda \left(\begin{array}{c}-2\hat{i}+\hat{j}-\hat{k}\end{array}\right)and{L}_2:\overrightarrow{r}=\hat{j}+\mu \left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$

• A. The acute angle between the lines is ${30}^{\circ }$
• B. The acute angle between the lines is ${60}^{\circ }$
• C. The equation of plane containing the lines $L_1\&L_2$ is $\overrightarrow{r}.\left(\begin{array}{c}\hat{i}+\hat{j}-\hat{k}\end{array}\right)=1$
• D. The image of point $P(1,2,3)$ w.r.t to the plane containing $L_1\&L_2$ is $(-1,0,-3)$.

Answer 1

Answer: (A), (C)

The acute angle between the lines is ${30}^{\circ }$
The equation of plane containing the lines $L_1\&L_2$ is $\overrightarrow{r}.\left(\begin{array}{c}\hat{i}+\hat{j}-\hat{k}\end{array}\right)=1$

Given eq of lines

$L_1:\overrightarrow{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda (-2\hat{i}+\hat{j}-\hat{k})$

$L_2:\overrightarrow{r}=\hat{j}+\mu (\hat{i}-\hat{j})$

angle between two lines

$\cos \theta =\widehat{n_1}\cdot \widehat{n_2}$

$\cos \theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}$

$\cos \theta =\frac{(-2\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-\hat{j})}{\sqrt{(-2{)}^2+1^2+(-1{)}^2}\sqrt{(1^2+(-1{)}^2}}$

$\cos \theta =\frac{-2-1}{\sqrt{4+1+1}\sqrt{1+1}}$

$\cos \theta =\frac{-3}{\sqrt{6}\sqrt{2}}$

$\cos \theta =\frac{-3}{2\sqrt{3}}$

$\cos \theta =\frac{\sqrt{3}}{2}$ $\cos (-\theta )=\cos \theta$

$\theta ={30}^{.}$

Eq of plane containing two lines

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ -2& 1& -1\\ 1& -1& 0\end{array}\right|=0$

$x-x_1(-1)-(y-y_1)\left(1\right)+(z-z_1)\left(1\right)=0$

$-(x-1)-(y-1)+(z-1)=0$

$-x+1-y+1+z-1=0$

$x+y-z-1=0$

in vector form

$\overrightarrow{r}\cdot (\hat{i}+\hat{j}-\hat{k})=1$

formula to find image of point(x_{1},y_{1},z_{1}) here point is P(1,2,3)

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}\right)$

$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{-1}=-2\left(\frac{1+2-3-1}{1^2+1^2+(-1{)}^2}\right)$

$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{-1}=-2\left(\frac{-1}{3}\right)$

$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{-1}=\frac{2}{3}$

$\frac{x-1}{1}=\frac{2}{3}$

$x=\frac{5}{3}$

$\frac{y-2}{1}=\frac{2}{3}$

$y=\frac{8}{3}$

$\frac{y-3}{-1}=\frac{2}{3}$

$z=\frac{7}{3}$

$P^{\prime }(\frac{5}{3},\frac{8}{3},\frac{7}{3})$