Question

The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$, is

• A. $\frac{1}{\sqrt{6}}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{6}}$

Lines are, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k_1$(assume) $⟶\left(1\right)$
and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}=k_2$(assume) $⟶\left(2\right)$
So any point on (1) is $P(2k_1+1,3k_1+2,4k_1+3)$
and any point on (2) is $Q(3k_2+2,4k_2+4,5k_2+5)$
Now direction ratio of $PQ$ are $3k_2-2k_1+1,4k_2-3k_1+2$ and $5k_2-4k_1+2$
Since $PQ\perp \left(1\right)$
$\Rightarrow 2(3k_2-2k_1+1)+3(4k_2-3k_1+2)+4(5k_2-4k_1+2)=0$
$\Rightarrow 38k_2-29k_1+16=0$ $⟶\left(3\right)$
Also $PQ\perp \left(2\right)$
$\Rightarrow 3(3k_2-2k_1+1)+4(4k_2-3k_1+2)+5(5k_2-4k_1+2)=0$
$\Rightarrow 50k_2-38k_1+21=0$ $⟶\left(4\right)$
Solving (3) and (4), we get $k_2=-\frac{1}{6},k_1=\frac{1}{3}$
Therefore, $P=(\frac{5}{3},3,\frac{13}{3})$ and $Q=(\frac{3}{2},\frac{10}{3},\frac{25}{6})$
, $PQ=\sqrt{{(\frac{3}{2}-\frac{5}{3})}^2+{(\frac{10}{3}-3)}^2+{(\frac{25}{6}-\frac{13}{3})}^2}=\frac{1}{\sqrt{6}}$
Hence, option 'A' is correct.