Question

If the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}$ intersects the plane $2x+3y-z+13=0$ at a point $P$ and the plane $3x+y+4z=16$ at a point $Q,$ then $PQ$ is equal to :

• A. $\sqrt{14}$
• B. $2\sqrt{14}$
• C. $14$
• D. $2\sqrt{7}$

Answer 1

The correct option is B $2\sqrt{14}$

Given line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda \left(\text{let}\right)$
Therefore general point of line:
$(3\lambda +2,2\lambda -1,1-\lambda )$ which intersect the plane $2x+3y-z+13=0$ at a point $P$
then $2(3\lambda +2)+3(2\lambda -1)-(1-\lambda )+13=0\Rightarrow \lambda =-1$
So, point $P(-1,-3,2)$
Also line intersect the plane $3x+y+4z=16$ at a point $Q$
therefore $3(3\lambda +2)+(2\lambda -1)+4(1-\lambda )=16\Rightarrow \lambda =1$
So, point $Q(5,1,0)$
therefore $PQ=\sqrt{(5+1{)}^2+(1+3{)}^2+(0-2{)}^2}$
$\therefore PQ=2\sqrt{14}$