Question

If $d$ is the shortest distance between the skew lines $\frac{x}{-1}=\frac{y-2}{2}=\frac{z-3}{-3}$ and $\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{-2},$ then the value of $3d^2+3=$

Answer 1

Given lines :
$L_1:\frac{x}{-1}=\frac{y-2}{2}=\frac{z-3}{-3}$ and $L_2:\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{-2}$
$\Rightarrow D.R^{\prime }s$ of $L_1=(a_1,b_1,c_1)=(-1,2,-3)$ and passes through $A(0,2,3)$
$D.R^{\prime }s$ of $L_2=(a_2,b_2,c_2)=(1,-1,-2)$ and passes through $B(1,0,1)$
$\therefore$ shortest distance $=\left|\frac{\overrightarrow{AB}\cdot \hat{n}}{|\hat{n}|}\right|$
here $\hat{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& -3\\ 1& -1& -2\end{array}\right|=-7\hat{i}-5\hat{j}-\hat{k}\Rightarrow \overrightarrow{AB}=\hat{i}-2\hat{j}-2\hat{k}$
$\therefore$ shortest distance $=\left|\frac{-7+10+2}{\sqrt{49+25+1}}\right|=\frac{1}{\sqrt{3}}$