Question

If the shortest distance betwen the lines $\frac{x}{-2}=\frac{y-3}{2}=\frac{z-4}{4}$ and $\frac{x-2}{-1}=\frac{y+2}{1}=\frac{z+4}{2}$ is $d,$ then the value of $(6d^2-20)=$

Answer 1

Given lines :
$L_1:\frac{x}{-2}=\frac{y-3}{2}=\frac{z-4}{4}$ and $L_2:\frac{x-2}{-1}=\frac{y+2}{1}=\frac{z+4}{2}$
$D.r^{\prime }s$ of $L_1=(a_1,b_1,c_1)=(-2,2,4)$
$D.r^{\prime }s$ of $L_2=(a_2,b_2,c_2)=(-1,1,2)$
as $D.r^{\prime }s$ of both lines are proportional; both are parallel.
$\therefore$ shortest distance $=\left|\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\end{array}\right|}{\sqrt{a_1^2+b_1^2+c_1^2}}\right|=\left|\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -5& -8\\ -2& 2& 4\end{array}\right|}{\sqrt{4+4+16}}\right|=\frac{\sqrt{116}}{\sqrt{24}}=\sqrt{\frac{29}{6}}$
$\Rightarrow 6d^2-20=9$