Mathematics · 3D Geometry

JEE Advanced 2018 — Paper 2 — Question 39

Let P be a point in the first octant, whose image Q in the plane $\mathrm{x}+\mathrm{y}=3$ (that is, the line segment PQ is perpendicular to the plane $x+y=3$ and the mid-point of PQ 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 xy-plane, then the length of PR is $\_\_\_\_\_$ .

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

Let $\mathrm{Q} \equiv\left(0,0, \mathrm{z}_{1}\right)$ then P is image of Q in $\mathrm{x}+\mathrm{y}=3$

$\Rightarrow \quad \frac{x_{p}-0}{1}=\frac{y_{p}-0}{1}=\frac{-2(-3)}{2}=3$

$\Rightarrow \mathrm{P} \equiv\left(3,3, \mathrm{z}_{1}\right)$

$\Rightarrow(3)^{2}+z_{1}^{2}=25 \Rightarrow z_{1}=4$

Length of $\mathrm{PR}=2 \mathrm{Z}_{1}=8$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Introduction to 3D Geometry