Question

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

Answer 1

Answer: 2

$a_1=(-1,0,3)$
$a_2=(0,-1,2)$
$b_1=(1,-a,0)$ dr's of line (1)
$b_2=(1,-1,1)$ dr's of line (2)
${\overline{a}}_2-{\overline{a}}_1=(1,-1,-1)$
${\overline{b}}_1\times {\overline{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ 1& -a& 0\\ 1& -1& 1\end{array}\right|$
${\overline{b}}_1\times {\overline{b}}_2=\hat{i}(-a)-\hat{j}+k(a-1)$
$\left|{\overline{b}}_1\times {\overline{b}}_2\right|=\sqrt{a^2+1+(a-1{)}^2}$
$a_2-a_1\cdot {\overline{b}}_1\times {\overline{b}}_2=2-2a$
$\frac{2(1-a)}{\sqrt{a^2+1+(a-1{)}^2}}=\sqrt{\frac{2}{3}}$
Squaring an both the side
After solving $a=2,\frac{1}{2}$