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
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
Updated On: Jun 14, 2022
- $\frac{1}{\sqrt 2}$
- $\sqrt 2$
- $2$
- $2 \sqrt 2$
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The Correct Option is A
Solution and Explanation
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$$ \, DR's\, of\, TS = < \mu + 2 -2, 4 \mu+3-1,\mu+5-4>$
$ \hspace30mm = < \mu , 4 \mu+ 2,\mu+1>$
and DR's of QR = < 1,4,1 >
Since, perpendicular
$ \therefore$ $\hspace 10mm 1(\mu) +4( 4 \mu+ 2) + 1 (\mu+1)=0$
$\Rightarrow \mu = - \frac{1}{2}$ and $S \bigg(\frac{3}{2},1,\frac{9}{2}\bigg)$
$\therefore Length\, of\, PS = \sqrt{\bigg(\frac{3}{2}-\frac{4}{3}\bigg)^2+\bigg(1-\frac{1}{3}\bigg)^2 +\bigg(\frac{9}{2}-\frac{13}{3}\bigg)^2}= \frac{1}{\sqrt 2}$
Description of situation (i) If the straight line $\hspace30mm$ $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\lambda$ intersects the plane Ax+ By+ Cz+ d= 0. $\frac{5}{2}$
Then, $(a \lambda + x_1,b \lambda+ y_1, c\lambda+z_1)$ would satisfy $\hspace30mm Ax+By+Cz+d=0$
(ii) If A is the foot of perpendicular from P to I.
Then, (DR's of PA) is perpendicular to DR's of I.
Equation of straight line QR, is
$\hspace30mm \frac{x-2}{1-2}=\frac{y-3}{-1-3}=\frac{z-5}{4-5}$
$\Rightarrow\hspace30mm \frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{-1}$
$\Rightarrow\hspace30mm \frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}= \lambda \hspace10mm ...(i) $
$\therefore$ $P(\lambda+2,4\lambda+3,\lambda+5)$must lie on $5x-4y-z=1$
$\Rightarrow \hspace30mm 5(\lambda+2)-4(4\lambda+3)-(\lambda+5)=1$
$\Rightarrow \hspace35mm 5\lambda+10-16\lambda-12 -\lambda-5=1$
$\Rightarrow \hspace 60mm -7-12 \lambda =1$
$\therefore$ $\hspace 80mm \lambda = \frac{-2}{3}$
or $P \bigg(\frac{4}{3},\frac{1}{3},\frac{13}{3}\bigg) $
Again, we can assume S from E(i),
$ \hspace 40mm as\, S (\mu + 2, 4 \mu+3,\mu+5)$
$ \therefore$$ \, 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 $S \bigg(\frac{3}{2},1,\frac{9}{2}\bigg)$
$\therefore Length\, of\, PS = \sqrt{\bigg(\frac{3}{2}-\frac{4}{3}\bigg)^2+\bigg(1-\frac{1}{3}\bigg)^2 +\bigg(\frac{9}{2}-\frac{13}{3}\bigg)^2}= \frac{1}{\sqrt 2}$
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Three Dimensional Geometry
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Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
