Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Lines are, $\frac{x-1}{2}=\frac{y}{3}=\frac{z}{2}=k_1$(assume) .......(1)

and $\frac{x-3}{2}=\frac{y}{5}=\frac{z-2}{4}=k_2$(assume) ...........(2)

So any point on (1) is $P(2k_1+1,3k_1,2k_1)$

And any point on (2) is $Q(2k_2+3,5k_2,4k_2+2)$

Now direction ratio of $PQ$ are $2k_2-2k_1+2,5k_2-3k_1$ and $4k_2-2k_1+2$

Since $PQ\perp \left(1\right)$
$\Rightarrow 2(2k_2-2k_1+2)+3(5k_2-3k_1)+2(4k_2-2k_1+2)=0$
$\Rightarrow 27k_2-17k_1+8=0$ $⟶\left(3\right)$

Also $PQ\perp \left(2\right)$
$\Rightarrow 2(2k_2-2k_1+2)+5(5k_2-3k_1)+4(4k_2-2k_1+2)=0$
$\Rightarrow 15k_2-9k_1+4=0$ $⟶\left(4\right)$

Solving (3) and (4), we get $k_1=1,k_2=\frac{1}{3}$Therefore, $P=(3,3,2)$ and $Q=(\frac{11}{3},\frac{5}{3},\frac{10}{3})$

ergo, $PQ=\sqrt{{(\frac{11}{3}-3)}^2+{(\frac{5}{3}-3)}^2+{(\frac{10}{3}-2)}^2}=2$