3D Geometry 3 Question 39
####39. Perpendicular are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of perpendiculars lie on the line
(a) $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$
(c) $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$
(b) $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$
(d) $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$
(2013 Adv.)
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Answer:
Correct Answer: 39. (d)
Solution:
- Key Idea To find the foot of perpendiculars and find its locus. Formula used
Footof perpendicular from $\left(x_{1}, y_{1}, z_{1}\right)$ to
$a x+b y+c z+d=0$ be $\left(x_{2}, y_{2}, z_{2}\right)$, then
$ \frac{x_{2}-x_{1}}{a}=\frac{y_{2}-y_{1}}{b}=\frac{z_{2}-z_{1}}{c}=\frac{-\left(a x_{1}+b y_{1}+c z_{1}+d\right)}{a^{2}+b^{2}+c^{2}} $
Any point on $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$
$ \Rightarrow \quad x=2 \lambda-2, y=-\lambda-1, z=3 \lambda $
Let foot of perpendicular from $(2 \lambda-2,-\lambda-1,3 \lambda)$
to $x+y+z=3$ be $\left(x_{2}, y_{2}, z_{2}\right)$
$ \begin{aligned} \therefore \quad \frac{x_{2}-(2 \lambda-2)}{1} & =\frac{y_{2}-(-\lambda-1)}{1}=\frac{z_{2}-(3 \lambda)}{1} \\ & =-\frac{(2 \lambda-2-\lambda-1+3 \lambda-3)}{1+1+1} \end{aligned} $
$ \begin{array}{rlrl} \Rightarrow & x_{2}-2 \lambda+2 =y_{2}+\lambda+1=z_{2}-3 \lambda=2-\frac{4 \lambda}{3} \\ \therefore & x_{2} =\frac{2 \lambda}{3}, y_{2}=1-\frac{7 \lambda}{3}, z_{2}=2+\frac{5 \lambda}{3} \\ \Rightarrow & \lambda =\frac{x_{2}-0}{2 / 3}=\frac{y_{2}-1}{-7 / 3}=\frac{z_{2}-2}{5 / 3} \end{array} $
Hence, foot of perpendicular lie on
$ \frac{x}{2 / 3}=\frac{y-1}{-7 / 3}=\frac{z-2}{5 / 3} \Rightarrow \frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5} $
