The perpendicular distance, of the line $\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z+3}{2}$ from the point $P(2,-10,1)$, is:
Step-by-Step Solution
Key Concept: Find the foot of perpendicular $A$ from $P$ to the line by setting $\overrightarrow{PA}\cdot\vec{n}=0$ where $\vec{n}=(2,-1,2)$, then compute $|PA|$.
General point $A=(2\lambda+1,-\lambda-2,2\lambda-3)$.
$\overrightarrow{PA}\cdot(2,-1,2)=0$: $(2\lambda-1)\cdot2+(-\lambda+8)\cdot(-1)+(2\lambda-4)\cdot2=0$
$4\lambda-2+\lambda-8+4\lambda-8=0 \Rightarrow 9\lambda=18 \Rightarrow \lambda=2$.
$A=(5,-4,1)$.
$|PA|=\sqrt{(5-2)^2+(-4+10)^2+(1-1)^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}$.
Correct Answer: 4