Question

The distance of the point $P(2,3,4)$ from the line $1-x=\frac{y}{2}=\frac{1}{3}(1+z)$ is

• A. $\frac{1}{7}\sqrt{35}$
• B. $\frac{4}{7}\sqrt{35}$
• C. $\frac{2}{7}\sqrt{35}$
• D. $\frac{3}{7}\sqrt{35}$

Answer 1

The correct option is D $\frac{3}{7}\sqrt{35}$

Given line is $\frac{x-1}{-1}=\frac{y-0}{2}=\frac{z+1}{3}=r\left(say\right)...\left(i\right)$
Then, coordinates of any point $N$ on the line are
$(-r+1,2r,3r-1)....\left(ii\right)$
Let $N$ be the foot of the perpendicular to line (i)
$\therefore$ Direction ratios of $PN$ are
$(-r+1-2,2r-3,3r-1-4)=(-r-1,2r-3,3r-5)....\left(iii\right)$
$\because PN$ is perpendicular to line (i).
$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\Rightarrow -(-r-1)+2(2r-3)+3(3r-5)=0$
$\Rightarrow r+1-4r-6+9r-15=0$
$\Rightarrow r=\frac{10}{7}$
$\therefore$ From Eq. (iii)
$N=(-\frac{10}{7}+1,\frac{20}{7},\frac{30}{7}-1)$
$=(\frac{-3}{7},\frac{20}{7},\frac{23}{7})$
$\therefore$ Perpendicular distance
$PN=\sqrt{{(-\frac{3}{7}-2)}^2+{(\frac{20}{7}-3)}^2+{(\frac{23}{7}-4)}^2}$
$=\frac{1}{7}\sqrt{289+1+25}=\frac{3}{7}\sqrt{35}$.