Mathematics - Vector Question with Solution | TestHub

Given equations of lines are

$L: \frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1}=k$

$L_1: \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1}$

Here,

$\overrightarrow{b_{\mathit{1}}}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 0& 1\\ 1& 1& -1\end{array}\right|$

$\Rightarrow \overrightarrow{b_{\mathit{1}}}\times \overrightarrow{b_2}=-\hat{i}-\hat{j}-2\hat{k}$

And,

$\overrightarrow{a_2}-\overrightarrow{a_1}=6\hat{i}+\left(\lambda -1\right)\hat{j}+\left(-\lambda -4\right)\hat{k}$

$\because$ Shortest distance is $2\sqrt{6}$.

$\Rightarrow 2\sqrt{6}=\left|\frac{-6-\lambda +1+2\lambda +8}{\sqrt{1+1+4}}\right|$

$\Rightarrow |\lambda +3|=12\Rightarrow \lambda =9,-15$

$\Rightarrow \alpha =-2k+5, \gamma =k-\lambda$ where $k\in R$

So,

$\alpha +2\gamma =5-2\lambda =-13, 35$