Question

If the shortest distance between the line $r=(-i+3 k)+\lambda(i-a j) $ and $r =(- j +2 k )+\mu( i - j + k )$ is $\sqrt{\frac{2}{3}}$, then the integral value of a is equal to

Answer 1

$a _{1}=(-1,0,3)$
$a _{2}=(0,-1,2)$
$b _{1}=(1,- a , 0)$ dr's of line (1)
$b _{2}=(1,-1,1)$ dr's of line (2)
$\overline{ a }_{2}-\overline{ a }_{1}=(1,-1,-1)$
$\overline{ b }_{1} \times \overline{ b }_{2}=\begin{vmatrix} \hat{i }& \hat{j} & k \\ 1 & - a & 0 \\ 1 & -1 & 1\end{vmatrix}$
$\overline{ b }_{1} \times \overline{ b }_{2}= \hat{i} (- a )- \hat{j }+ k ( a -1) $
$\left|\overline{ b }_{1} \times \overline{ b }_{2}\right|=\sqrt{ a ^{2}+1+( a -1)^{2}} $
$ a _{2}- a _{1} \cdot \overline{ b }_{1} \times \overline{ b }_{2}=2-2 a $
$\frac{2(1- a )}{\sqrt{ a ^{2}+1+( a -1)^{2}}}=\sqrt{\frac{2}{3}}$
Squaring an both the side
After solving $a =2, \frac{1}{2}$