Question

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinates axis. The line meet the plane $2x+y+z=9$ at ponit $Q$.
The length of the line segment $PQ$ equals.

• A. $1$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. $2$

Answer 1

Answer: (C)

$\sqrt{3}$

Equation of a line through the point $P(2,-1,2)$, equally inclined to the axes is given by,
$\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=k$ (say)
Any point on the line is $Q(k+2,k-1,k+2)$ which lies on the plane $2x+y+z=9$
$\Rightarrow 2(k+2)+k-1+k+2=9\Rightarrow k=1$,
For this values of $k$, the coordinates of $Q$ are $(3,0,3)$.
So, $PQ=\sqrt{{(3-2)}^2+{(0+1)}^2+{(3-2)}^2}=\sqrt{3}$