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 straight line $QR$ is

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

$\Rightarrow \frac{x-2}{-1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda \Rightarrow \frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$ ...$\left(i\right)$

$\therefore P(\lambda +2,4\lambda +3,\lambda +5)$ must lie on $5x-4y-z=1$ $\Rightarrow 5(\lambda +2)-4(4\lambda +3)-(\lambda +5)=1$

$\Rightarrow 5\lambda +10-16\lambda -12-\lambda -5=1$

$\Rightarrow -7-12\lambda =1$

$\therefore \lambda =\frac{-2}{3}$

or $P(\frac{4}{3},\frac{1}{3},\frac{13}{3})$ ...(ii)

Again, we can assume$S$ from Eq. (i) as $S(\mu +2,4\mu +3,\mu +5)$

$\therefore$ dr's of TS $=(\mu +2-2,4\mu +3-1,\mu +5-4)$ $=(\mu ,4\mu +2,\mu +1)$

and dr's of $QR$ $=(1,4,1)$

$\therefore 1\left(\mu \right)+4(4\mu +2)+1(\mu +1)=0$

$\Rightarrow \mu =-\frac{1}{2}$ and $S(\frac{3}{2},1,\frac{9}{2})$ ...$\left(iii\right)$

$\therefore$ Length of $PS$ $=\sqrt{{(\frac{3}{2}-\frac{4}{3})}^2+{(1-\frac{1}{3})}^2+{(\frac{9}{2}-\frac{13}{3})}^2}=\frac{1}{\sqrt{2}}$