3D Geometry 3 Question 55
####55. Let $P$ be a point in the first octant, whose image $Q$ in the plane $x+y=3$ (that is, the line segment $P Q$ is perpendicular to the plane $x+y=3$ and the mid-point of $P Q$ lies in the plane $x+y=3$ ) lies on the $Z$-axis. Let the distance of $P$ from the $X$-axis be 5 . If $R$ is the image of $P$ in the $X Y$-plane, then the length of $P R$ is
(2018 Adv.)
Passage Based Problems
Read the following passage and answer the questions. Consider the lines
$ L_{1}: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_{2}: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3} $
Show Answer
Answer:
Correct Answer: 55. (8)
Solution:
- Let $P(\alpha, \beta, \gamma)$ and $R$ is image of $P$ in the $x y$-plane.
$\therefore R(\alpha, \beta,-\gamma)$
Also, $Q$ is the image of $P$ in the plane $x+y=3$
$ \begin{aligned} \therefore \quad & \frac{x-\alpha}{1}=\frac{y-\beta}{1}=\frac{z-\gamma}{0}=\frac{-2(\alpha+\beta-3)}{2} \\ & x=3-\beta, y=3-\alpha, z=\gamma \end{aligned} $
Since, $Q$ is lies on $Z$-axis
$\therefore \quad \beta=3, \alpha=3, z=\gamma$
$\therefore P(3,3, \gamma)$
Given, distance of $P$ from $X$-axis be 5 .
$ \begin{array}{rlrl} & \therefore & 5 & =\sqrt{3^{2}+\gamma^{2}} \\ 25-9 & =\gamma^{2} \\ \Rightarrow & \gamma & = \pm 4 \\ & \text { Then, } & P R & =|2 \gamma|=|2 \times 4|=8 \end{array} $
