JEE Main 2023 — Three Dimensional Geometry Question with Solution

Given equations are

$\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ is passing through a point $\left(2,-1,6\right)$ and it's direction ratios are $3,2,2$.

So, 

${\overrightarrow{a}}_1=2\hat{i}-\hat{j}+6\hat{k}$

${\overrightarrow{b}}_1=3\hat{i}+2\hat{j}+2\hat{k}$

And,

$\frac{x-6}{3}=\frac{y-1}{-2}=\frac{z+8}{0}$ is passing through a point $\left(6,1,-8\right)$ and it's direction ratios are $3,-2,0$.

${\overrightarrow{a}}_2=6\hat{i}+\hat{j}-8\hat{k}$

${\overrightarrow{b}}_2=3\hat{i}-2\hat{j}+0\hat{k}$

Now,

${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=4\hat{i}+2\hat{j}-14\hat{k}$

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 2\\ 3& -2& 0\end{array}\right|$

$\Rightarrow {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=4\hat{i}+6\hat{j}-12\hat{k}$

So,

$\Rightarrow \left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{16+36+144}=14$

We know that the shortest distance between two skew lines is $\frac{\left|\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)\right|}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}$

$=\left|\frac{\left(4\hat{i}+2\hat{j}-14\hat{k}\right)·\left(4\hat{i}+6\hat{j}-12\hat{k}\right)}{14}\right|$

$=\frac{\left|16+12+168\right|}{14}=14$ units