Q. The distance of the line $\frac{x-2}2=\frac{y-6}3=\frac{z-3}4$ from the point $(1,4,0)$ along the line $\frac x1=\frac{y-2}2=\frac{z+3}3$ is :
A) $\sqrt{17}$
B) $\sqrt{15}$
C) $\sqrt{14}$
D) $\sqrt{13}$
Explanation
[Solution by Dr. Abubacker] A point of the first line is $(2+2s,6+3s,3+4s)$. The vector from $(1,4,0)$ to it, $(1+2s,2+3s,3+4s)$, must be parallel to $(1,2,3)$: $\frac{1+2s}1=\frac{2+3s}2\Rightarrow s=0$ (and $\frac{3+4s}3=1$ also holds). The point is $(2,6,3)$, so the displacement is $(1,2,3)$ and the distance is $\sqrt{1+4+9}=\sqrt{14}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.