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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2018 · Paper 2 · Mathematics · Three Dimensional Geometry

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

Let PP be a point in the first octant, whose image QQ in the plane x+y=3x + y = 3 (that is, the line segment PQPQ is perpendicular to the plane x+y=3x + y = 3 and the mid-point of PQPQ lies in the plane x+y=3x + y = 3) lies on the z-axis. Let the distance of PP from the x-axis be 55. If RR is the image of PP in the xy-plane, then the length of PRPR is ______.

Correct answer

0

Step-by-step solution

Let $P(α,β,γ)P\left(\alpha, \beta, \gamma\right), Q(0,0,γ)Q\left(0,0, \gamma\right) and R(α,β,−γ)R\left(\alpha, \beta, - \gamma\right)
Now, \overset{\to}{PQ} ∥ \left(\overset{^}{i} + \overset{^}{j}\right) ⇒ - \left(\alpha\overset{^}{i} + \beta\overset{^}{j}\right) ∥ \left(\overset{^}{i} + \overset{^}{j}\right)
⇒α=β⇒ \alpha = \beta
Also, mid point of PQ lies on the plane
⇒α2+β2=3⇒α+β=6⇒α=3⇒ \dfrac{\alpha}{2} + \dfrac{\beta}{2} = 3 ⇒ \alpha + \beta = 6 ⇒ \alpha = 3
Now, distance of point PP from x-axis is β2+γ2=5\sqrt{{\beta}^{2} + {\gamma}^{2}} = 5
⇒β2+γ2=25⇒γ2=16⇒ {\beta}^{2} + {\gamma}^{2} = 25 ⇒ {\gamma}^{2} = 16
As β=α=3\beta = \alpha = 3
As γ=4\gamma = 4
Hence, $PR = 2\gamma = 8
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