Verified Solution • 2025–26 Session moderate

QJEEAdv18Maad25: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Maad25 Solution: Let P be a point in the first octant, whose image Q in the…

Let P be a point in the first octant, whose image Q in the plane x + 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 ...............

Step-by-Step Verified Solution

Let P( α\alpha , β\beta , γ\gamma ) and R is image of P in the XY-plane.

∴\therefore R( α\alpha , β\beta , −- γ\gamma )

Also, Q is the image of P in the plane x + y = 3

∴\therefore x−α1=y−β1=z−γ0{{x - \alpha } \over 1} = {{y - \beta } \over 1} = {{z - \gamma } \over 0}

=−2(α+β−3)2= {{ - 2(\alpha + \beta - 3)} \over 2}

x=3−β, y=3−α, z=γx = 3 - \beta ,\,y = 3 - \alpha ,\,z = \gamma

Since, Q is lies on Z-axis

∴\therefore β=3, α=3, z=γ\beta = 3,\,\alpha = 3,\,z = \gamma

∴\therefore P(3,3,γ)P(3,3,\gamma )

Given, distance of P from X-axis be 5

∴\therefore 5=32+γ25 = \sqrt {{3^2} + {\gamma ^2}}

25−9=γ225 - 9 = {\gamma ^2}

⇒γ=±4\Rightarrow \gamma = \pm 4

Then, PR=∣2γ∣=∣2×4∣=8PR = |2\gamma | = |2 \times 4| = 8

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring