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 $\frac{1}{\sqrt{2}}$

Direction ratio's of $QR$ is $(1,4,1)$.
Co-ordinates of $P\equiv (\frac{4}{3},\frac{1}{3},\frac{13}{3})$
Direction rations of point is $(2,2,-1)$.
Angle between $QR$ and $PT$
$={\cos }^{-1}\left[\frac{1\times 2+4\times 2+1\times -1}{\sqrt{1^2+4^2+1^2}\sqrt{2^2+2^2+(-1{)}^2}}\right]$
$={\cos }^{-1}\left[\frac{9}{3\sqrt{2}\times 3}\right]$
$={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)={45}^o$
$PT=\sqrt{{(2-\frac{4}{3})}^2+{(1-\frac{1}{3})}^2{(4-\frac{13}{3})}^2}$ $=1$
$\Rightarrow PS=TS=\frac{1}{\sqrt{2}}$