Question

The distance of the point $(2,3,4)$ 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

Answer: (D)

Given, point $P(2,3,4)$ and line
$\frac{x-1}{-1}=\frac{y-0}{2}=\frac{z+1}{3}=r$ (say) ..(i)
Then, coordinates of any point N on the line (i) are $(-r+1,2r,3r-1)$ ..(ii)
Let N be the foot of the perpendicular to line (i).
$\therefore$ Direction ratios of PN are
$(-r+1-2,2r-3,3r-1-4)$
$=(r-1,2r-3,3r-5)$ ..(iii)
$\because$ PN is perpendicular to line (i).
$\therefore$ Using the condition,
$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\Rightarrow$ $-1(-r-1)+2(2r-3)+3(3r-5)=0$
$\Rightarrow$ $r+1+4r-6+9r-15=0$
$\Rightarrow$ $14r=20\Rightarrow r=\frac{20}{14}=\frac{10}{7}$
Then, from Eq. (ii), we get $N=\left(-\frac{10}{7}+1,\frac{20}{7},\frac{30}{7}-1\right)=\left(\frac{3}{7},\frac{20}{7},\frac{23}{7}\right)$
Now, perpendicular distance $PN=\sqrt{{\left(-\frac{3}{7}-2\right)}^2+{\left(\frac{20}{7}-3\right)}^2+{\left(\frac{23}{7}-4\right)}^2}$
$=\frac{1}{7}\sqrt{289+1+25}=\frac{\sqrt{315}}{7}=\frac{3}{7}\sqrt{35}$