Question

From the point $P(3,-1,11)$, a perpendicular is drawn on the line $L$ given by the equation $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$. Let $Q$ be the foot of the perpendicular.

What is the length of the line segment $PQ$?

• A. $\sqrt{47}$ units
• B. $7$ units
• C. $\sqrt{53}$ units
• D. $8$ units

Answer 1

The correct option is C $\sqrt{53}$ units

The general point on the line $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is $(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$

Let $Q(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$

The directional cosines of $PQ=2\lambda -3,3\lambda +3,4\lambda -8$

Since the line $L$ and $PQ$ are perpendicular

$(2\lambda -3)\left(2\right)+(3\lambda +3)\left(3\right)+(4\lambda -8)4=0⟹\lambda =1⟹Q(2,5,7)$

The length of the line segment $PQ=\sqrt{{(3-2)}^2+{(-1-5)}^2+{(11-7)}^2}=\sqrt{53}$