3D Geometry 3 Question 1
####1. The distance of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is
(a) $3 \sqrt{10}$
(b) $10 \sqrt{3}$
(c) $\frac{10}{\sqrt{3}}$
(d) $\frac{20}{3}$
(2016 Main)
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Answer:
(b)
Solution:
- Equation of line passing through the point $(1,-5,9)$ and parallel to $x=y=z$ is
$ \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda $
Thus, any point on this line is of the form $(\lambda+1, \lambda-5, \lambda+9)$.
Now, if $P(\lambda+1, \lambda-5, \lambda+9)$ is the point of intersection of line and plane, then
$ \begin{aligned} & & (\lambda+1)-(\lambda-5)+\lambda+9 & =5 \\ \Rightarrow & & \lambda+15=5 \Rightarrow \lambda & =-10 \end{aligned} $
$\therefore$ Coordinates of point $P$ are $(-9,-15,-1)$.
Hence, required distance
$ \begin{aligned} & =\sqrt{(1+9)^{2}+(-5+15)^{2}+(9+1)^{2}} \\ & =\sqrt{10^{2}+10^{2}+10^{2}} \\ & =10 \sqrt{3} \end{aligned} $
