Question

The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,–1,4)$ with the plane $5x–4y–z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $QR$, then the length of the line segment $PS$ is

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

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

The equation of the line $QR$ is
$\frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{-1}=k$
so any arbitrary point is
$(-k+2,-4k+3,-k+5)$
Now this will satisfy the equation of the plane
$5(-k+2)-4(-4k+3)-(-k+5)=1$
$\Rightarrow k=\frac{2}{3}$
So, the point is $(\frac{4}{3},\frac{1}{3},\frac{13}{3})$
Now, let points $S$ be $(-\lambda +2,-4\lambda +3,-\lambda +5)$
So, D.R's of $ST$ is $(-\lambda ,-4\lambda +2,-\lambda +1)$
$\Rightarrow (-\lambda )(-1)-4(-4\lambda +2)-1(-\lambda +1)=0$
$\Rightarrow \lambda =\frac{1}{2}$
The point is $(\frac{3}{2},1,\frac{9}{2})$
So, $PS$ is $\frac{1}{\sqrt{2}}$