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

Answer 1

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

Equation of $QR$ is

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}$

Let $P\equiv (2+\lambda ,3+4\lambda ,5+\lambda )$

This point $P$ lies on the plane $5x-4y-z=1$

$\therefore 10+5\lambda -12-16\lambda -5-\lambda =1$
$-7-12\lambda =1$
$\Rightarrow \lambda =\frac{-2}{3}$

then $P\equiv (\frac{4}{3},\frac{1}{3},\frac{13}{3})$

Let $S=(2+\mu ,3+4\mu ,5+\mu )$

Point $T$ is $(2,1,4)$

D.R's of $\overrightarrow{TS}=(\mu ,4\mu +2,\mu +1)$

Direction ratios of $QR:(1,4,1)$

$\overrightarrow{TS}.(\hat{i}+4\hat{j}+\hat{k})=0$
$\mu +16\mu +8+\mu +1=0$
$\mu =\frac{-1}{2}$

$S=(\frac{3}{2},1,\frac{9}{2})$
$PS=\sqrt{{(\frac{4}{3}-\frac{3}{2})}^2+\frac{4}{9}+{(\frac{13}{3}-\frac{9}{2})}^2}$

$=\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}$

$=\sqrt{\frac{9}{18}}=\frac{1}{\sqrt{2}}$