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Mathematics Question on Three Dimensional Geometry

The point PP is the intersection of the straight line joining the points Q(2,3,5)Q(2, 3, 5) and R(1,−1,4)R (1, - 1, 4) with the plane 5x−4y−z=15x - 4y - z = 1. If SS is the foot of the perpendicular drawn from the point T(2,1,4)T (2,1, 4) to QRQR, then the length of the line segment PSPS is

A

12\frac{1}{\sqrt 2}

B

2\sqrt 2

C

22

D

222 \sqrt 2

Answer

12\frac{1}{\sqrt 2}

Explanation

Solution

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 x−x1a=y−y1b=z−z1c=λ\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda intersects the plane Ax+ By+ Cz+ d= 0. 52\frac{5}{2}
Then, (aλ+x1,bλ+y1,cλ+z1)(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(λ+2,4λ+3,λ+5)P(\lambda+2,4\lambda+3,\lambda+5)must lie on 5x−4y−z=15x-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(43,13,133)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=<μ+2−2,4μ+3−1,μ+5−4> \, 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
⇒μ=−12\Rightarrow \mu = - \frac{1}{2} and S(32,1,92)S \bigg(\frac{3}{2},1,\frac{9}{2}\bigg)
∴Length of PS=(32−43)2+(1−13)2+(92−133)2=12\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}