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

PLAN It is based on two concepts one is intersection of straight line and plane and other is the foot of perpendicular from a point to the straight line.
Description of situation (i) If the straight line $\hspace30mm$ $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda$ intersects the plane Ax+ By+ Cz+ d= 0. $\frac{5}{2}$
Then, $(a \lambda + x_1,b \lambda+ y_1, c\lambda+z_1)$ would satisfy $\hspace30mm Ax+By+Cz+d=0$
(ii) If A is the foot of perpendicular from P to I.
Then, (DR's of PA) is perpendicular to DR's of I.
Equation of straight line QR, is
$\hspace30mm \frac{x-2}{1-2}=\frac{y-3}{-1-3}=\frac{z-5}{4-5}$
$\Rightarrow\hspace30mm \frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{-1}$
$\Rightarrow\hspace30mm \frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}= \lambda \hspace10mm ...(i) $
$\therefore$ $P(\lambda+2,4\lambda+3,\lambda+5)$must lie on $5x-4y-z=1$
$\Rightarrow \hspace30mm 5(\lambda+2)-4(4\lambda+3)-(\lambda+5)=1$
$\Rightarrow \hspace35mm 5\lambda+10-16\lambda-12 -\lambda-5=1$
$\Rightarrow \hspace 60mm -7-12 \lambda =1$
$\therefore$ $\hspace 80mm \lambda = \frac{-2}{3}$
or $P \bigg(\frac{4}{3},\frac{1}{3},\frac{13}{3}\bigg) $
Again, we can assume S from E(i),
$ \hspace 40mm as\, S (\mu + 2, 4 \mu+3,\mu+5)$
$ \therefore$$ \, DR's\, of\, TS = < \mu + 2 -2, 4 \mu+3-1,\mu+5-4>$
$ \hspace30mm = < \mu , 4 \mu+ 2,\mu+1>$
and DR's of QR = < 1,4,1 >
Since, perpendicular
$ \therefore$ $\hspace 10mm 1(\mu) +4( 4 \mu+ 2) + 1 (\mu+1)=0$
$\Rightarrow \mu = - \frac{1}{2}$ and $S \bigg(\frac{3}{2},1,\frac{9}{2}\bigg)$
$\therefore Length\, of\, PS = \sqrt{\bigg(\frac{3}{2}-\frac{4}{3}\bigg)^2+\bigg(1-\frac{1}{3}\bigg)^2 +\bigg(\frac{9}{2}-\frac{13}{3}\bigg)^2}= \frac{1}{\sqrt 2}$