Question

Given lines $L_1:\frac{x}{-2}=\frac{y-1}{0}=\frac{z+2}{1}$

$L_2:\frac{x+1}{0}=\frac{y+1}{2}=\frac{z-2}{-1}$
Find the distance between the given straight lines.

• A. $12$
• B. $\frac{\sqrt{21}}{12}$
• C. $\frac{21}{\sqrt{12}}$
• D. $\frac{11}{\sqrt{8}}$

Answer 1

The correct option is D $\frac{11}{\sqrt{8}}$

$L_1:\frac{x}{-2}=\frac{y-1}{0}=\frac{x+2}{1}$
$L_2:\frac{x+1}{0}=\frac{y+1}{2}=\frac{z-2}{-1}$
$L_1:(0\hat{i}+1\hat{j}-2\hat{k})+\lambda (-2\hat{i}+0\hat{j}+\hat{k})$
$L_2:(-\hat{i}-\hat{j}+2\hat{k})+\mu (0\hat{i}+2\hat{j}-\hat{k})$
S.D. $=[(-1-1)\hat{i}+\left(\right(-1-(+1))\hat{j}(+2-(-2\left)\right)\hat{k}].\left(\right(-2,0,1)\ast (0,2,-1\left)\right)(-2,0,1)\ast (0,2,-1)$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 0& 1\\ 0& 2& -1\end{array}\right|=\frac{-2\hat{i}+2\hat{j}-4\hat{k}}{\sqrt{32}}$
S.D.=$=(-\hat{i}-2\hat{j}+4\hat{k}).\frac{(-2\hat{i}+2\hat{j}-4\hat{k})}{\sqrt{32}}$
$=\frac{(-2-4-16)}{\sqrt{32}}$
$=\left|\frac{-22}{\sqrt{32}}\right|$
$=\left|\frac{-11}{\sqrt{8}}\right|$
$=\frac{11}{\sqrt{8}}$