Question

The distance of the point $(2,1,3)$ from the line $\frac{x-5}{4}=\frac{y+1}{3}=\frac{z-2}{6}$ is

• A. $\sqrt{14}$
• B. $\sqrt{26}$
• C. $7$
• D. $\sqrt{21}$

Answer 1

The correct option is A $\sqrt{14}$


$\text{Let}R$ be the foot of the perpendicular.
$\text{Let}\frac{x-5}{4}=\frac{y+1}{3}=\frac{z-2}{6}=k$
Then, any point on the line $AB$ is of the form $(4k+5,3k-1,6k+2)$
Direction ratios of $PR$ are $(4k+3,3k-2,6k-1).$
Also, direction ratios of the line $AB$ are $(4,3,6).$
$\because PR\perp AB$
$4(4k+3)+3(3k-2)+6(6k-1)=0$
$\Rightarrow k=0$
Thus, coordinates of $R$ are $(5,-1,2).$
$\therefore PR=\sqrt{(5-2{)}^2+(-1-1{)}^2+(2-3{)}^2}$
$=\sqrt{14}$