Question

Consider the lines

$L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$

$L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$

The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both lines $L_1$ and $L_2$ is

• A. $\frac{2}{\sqrt{75}}$
• B. $\frac{7}{\sqrt{75}}$
• C. $\frac{13}{\sqrt{75}}$
• D. $\frac{23}{\sqrt{75}}$

Answer 1

The correct option is B $\frac{7}{\sqrt{75}}$

Given lines

$L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$

$L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$

SO normal vector obtained by both lines

$n_1=3\hat{i}+\hat{j}+2\hat{k}$

$n_2=\hat{i}+2\hat{j}+3\hat{k}$

planes is perpendicular to both lines SO cross product of normals of both lines gives normal vector of plane

$\Rightarrow n_1\times n_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& 2& 3\end{array}\right|$

$\Rightarrow n_1\times n_2=\hat{i}(3-4)-\hat{j}(9-2)+\hat{k}(6-1)$

$\Rightarrow n_1\times n_2=-\hat{i}-7\hat{j}+5\hat{k}=\overrightarrow{n}$

plane passing through point A(−1,−2,−1)

$\overrightarrow{a}=-\hat{i}-2\hat{j}-\hat{k}$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{n}=(-\hat{i}-2\hat{j}-\hat{k})\cdot (-\hat{i}-7\hat{j}+5\hat{k})$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{n}=1+14-5=10$

eq of plane is

$\overrightarrow{r}\cdot (-\hat{i}-7\hat{j}+5\hat{k})=10$

distance between plane and point(1,1,1)

$\Rightarrow distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

$\Rightarrow distance=\left|\frac{(-1)1+(-7)1+\left(5\right)1+10}{\sqrt{(-1{)}^2+(-7{)}^2+5^2}}\right|$

$\Rightarrow distance=\left|\frac{(-1-7+5+10}{\sqrt{1+49+25}}\right|$

$\Rightarrow distance=\left|\frac{7}{\sqrt{75}}\right|$

$\Rightarrow distance=\frac{7}{\sqrt{75}}$