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

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

PLAN
It is based on two concepts one is the intersection of straight line and plane and other is the foot of the perpendicular from a point to the straight line.
Description of situation
(i) if the straight line
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda$

Intersects the plane $Ax+By+Cz+d=0.$
Then$(a\lambda +x_1,b\lambda +y_1,c\lambda +z_1)$would satisfy
$Ax+By+Cz+d=0$
(ii) If A is the foot perpendicular from P to l, then (DR's of PA is perpendicular to DR's of l.


$\Rightarrow PA.I=0$
Equation of straight line QR, is
$\frac{x-2}{1-2}=\frac{y-3}{-1-3}=\frac{z-5}{4-5}\Rightarrow \frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{-1}\Rightarrow \frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda \therefore P(\lambda +2,4\lambda +3,\lambda +5)$ must lie on $5x-4y-z=1.\Rightarrow 5(\lambda +2)-4(4\lambda +3)-(\lambda +5)=1\Rightarrow 5\lambda +10-16\lambda -12-\lambda -5=1\Rightarrow -7-12\lambda =1\therefore \lambda =\frac{-2}{3}orP(\frac{4}{3},\frac{1}{3},\frac{13}{3})$
Again, we can assume S from Eq. (i).
as S lie on line QR $(\mu +2,4\mu +3,\mu +5)\therefore D{R}^{\prime }sofTS=<\mu +2-2,4\mu +3-1,\mu +5-4>=<\mu ,4\mu +2,\mu +1>$
And DR's of QR = < 1, 4,1>
Since QR is perpendicular to TS .
$\therefore 1\left(\mu \right)+4(4\mu +2)+1(\mu +1)=0\Rightarrow \mu =-\frac{1}{2}$ and S $(\frac{3}{2},1,\frac{9}{2})$
Length of PS = $\sqrt{{(\frac{3}{2}-\frac{4}{3})}^2+{(1-\frac{1}{3})}^2+{(\frac{9}{2}-\frac{13}{3})}^2}=\frac{1}{\sqrt{2}}$