3D Geometry 3 Question 52
####52. In $R^{3}$, let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+z-1=0$. Let $M$ be the locus of the foot of the perpendiculars drawn from the points on $L$ to the plane $P_{1}$. Which of the following point(s) lie(s) on $M$ ?
(2015 Adv.)
(a) $0,-\frac{5}{6},-\frac{2}{3}$
(b) $-\frac{1}{6},-\frac{1}{3}, \frac{1}{6}$
(c) $-\frac{5}{6}, 0, \frac{1}{6}$
(d) $-\frac{1}{3}, 0, \frac{2}{3}$
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Answer:
Correct Answer: 52. (a, b)
Solution:
- Since, $L$ is at constant distance from two planes $P_{1}$ and $P_{2}$. Therefore, $L$ is parallel to the line through intersection of $P_{1}$ and $P_{2}$.
$ \text { DR’s of } \begin{aligned} L & =\left|\begin{array}{ccr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{array}\right| \\ & =\hat{\mathbf{i}}(2-1)-\hat{\mathbf{j}}(1+2)+\hat{\mathbf{k}}(-1-4) \\ & =\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \end{aligned} $
$\therefore \quad$ DR’s of $L$ are $(1,-3,-5)$ passing through $(0,0,0)$.
Now, equation of $L$ is
$ \frac{x-0}{1}=\frac{y-0}{-3}=\frac{z-0}{-5} $
For any point on $L, \frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$
[say]
i.e. $\quad P(\lambda,-3 \lambda,-5 \lambda)$
If $(\alpha, \beta, \gamma)$ is foot of perpendicular from $P$ on $P_{1}$, then
$ \begin{array}{cc} & \frac{\alpha-\lambda}{1}=\frac{\beta+3 \lambda}{2}=\frac{\gamma+5 \lambda}{-1}=k \\ \Rightarrow & \alpha=\lambda+k, \beta=2 k-3 \lambda, \gamma=-k-5 \lambda \\ & \text { which satisfy } P_{1}: x+2 y-z+1=0 \\ \Rightarrow & (\lambda+k)+2(2 k-3 \lambda)-(-k-5 \lambda)+1=0 \\ \therefore & \quad k=-\frac{1}{6} \\ \therefore & x=-\frac{1}{6}+\lambda, y=-\frac{1}{3}-3 \lambda, z=\frac{1}{6}-5 \lambda \end{array} $
which satisfy options (a) and (b).
