Mathematics - Vector Question with Solution | TestHub

Given,

$\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$

$\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$

Now we know that,

Shortest distance between two skew lines is given by,

$SD=\left|\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\left(\overrightarrow{n_1}\times \overrightarrow{n_2}\right)}{\left|\overrightarrow{n_1}\times \overrightarrow{n_2}\right|}\right|$

Now finding, $\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 2\\ 1& 2& 0\end{array}\right|$

$\Rightarrow \overrightarrow{n_1}\times \overrightarrow{n_2}=-4\hat{i}+2\hat{j}+4\hat{k}$

And $\overrightarrow{a_2}-\overrightarrow{a_1}=\left(4\hat{i}+\hat{j}-3\hat{k}-\left(-2\hat{i}+0\hat{j}+5\hat{k}\right)\right)=6\hat{i}+\hat{j}-8\hat{k}$

Now putting the value in formula we get,

$SD=\left|\frac{\left(6\hat{i}+\hat{j}-8\hat{k}\right)\left(-4\hat{i}+2\hat{j}+4\hat{k}\right)}{\sqrt{16+16+4}}\right|$

$\Rightarrow SD=\left|\frac{-24+2-32}{6}\right|=\frac{54}{6}=9$