Question

Assertion :$L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$, $L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$
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 the lines $L_1$ and $L_2$ is $\frac{13}{5\sqrt{3}}$. Reason: The unit vector perpendicular to both the lines $L_1$ and $L_2$ is $\frac{-\overrightarrow{i}-7\overrightarrow{j}+5\overrightarrow{k}}{5\sqrt{3}}$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Normal vector of the plane is given by,
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& 2& 3\end{array}\right|=-\hat{i}-7\hat{j}+5\hat{k}$
Hence, equation of plane is,
$(\overrightarrow{r}-(-\hat{i}-2\hat{j}-1\hat{k}\left)\right)\cdot \overrightarrow{n}=0$
$\Rightarrow -(x+1)-7(y+2)+5(z+1)=0$
$\Rightarrow -x-7y+5z-10=0....\left(p\right)$
Hence, perpendicular distance from $(1,1,1)$ is
$=\mid \frac{-1-7+5-10}{\sqrt{1^2+7^2+5^2}}\mid =\frac{13}{5\sqrt{3}}$
Also normal vector perpendicular to both $L_1$ and $L_2$ is
$=\frac{-\hat{i}-7\hat{j}+5\hat{k}}{5\sqrt{3}}$
Since shortest distance between two line is along unit vector perpendicular to both line. Hence, Reason is correct explanation of Assertion.