Question

The shortest distance between the line $\frac{x-3}{2}=\frac{y}{0}=\frac{z}{4}$ and the $y$-axis is

• A. $\frac{1}{5}$
• B. $1$
• C. $0$
• D. $\frac{\sqrt{180}}{5}$

Answer 1

The correct option is D $\frac{\sqrt{180}}{5}$

Given line is, $\frac{x-3}{2}=\frac{y}{0}=\frac{z}{4}=a$(say) $⟶\left(1\right)$
and equation of $y$-axis is given by, $\frac{x}{0}=\frac{y}{1}=\frac{z}{0}=b$(say) $⟶\left(2\right)$
So any point on (1) is $P(2a+3,0,4a)$
and any point on (2) is $Q(0,b,0)$
Now direction ratio of $PQ$ are $2a+3,-b$ and $4a$
Since $PQ\perp \left(1\right)$
$\Rightarrow 2(2a+3)+0(-b)+4\left(4a\right)=0$
$\Rightarrow a=\frac{-3}{10}$
Also $PQ\perp \left(2\right)$
$\Rightarrow 0(2a+3)+1(-b)+0\left(4a\right)=0$
$\Rightarrow b=0$
Therefore, $P=(\frac{12}{5},0,\frac{-6}{5})$ and $Q=(0,0,0)$
$PQ=\sqrt{{(\frac{12}{5}-0)}^2+(0-0{)}^2+{(\frac{-6}{5}-0)}^2}=\frac{\sqrt{180}}{5}$