Question

A point $P$ lies on a line through $Q(1,–2,3)$ and is parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$. If $P$ lies on the plane $2x+3y–4z+22=0$, then segment $PQ$ equals to

• A. $\sqrt{42}\text{units}$
• B. $\sqrt{32}\text{units}$
• C. $4\text{units}$
• D. $5\text{units}$

Answer 1

The correct option is A $\sqrt{42}\text{units}$

Equation line on which $P$ lies is $L:\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda$
Any point on $L$
$P(\lambda +1,4\lambda -2,5\lambda +3)$
$\because P$ lies on $2x+3y–4z+22=0$
$\therefore 2(\lambda +1)+3(4\lambda -2)-4(5\lambda +3)+22=0-6\lambda +6=0\lambda =1\Rightarrow P(2,2,8)\therefore PQ=\sqrt{1+16+25}=\sqrt{42}$