Skip to content
dronacharya.ai
← JEE Advanced 2014 · Paper 1
MathematicsMultiple correct+3Question 47 of 60

JEE Advanced 2014 · Paper 1, Mathematics question 47

From a point P(λ,λ,λ)P(\lambda,\lambda,\lambda), perpendiculars PQPQ and PRPR are drawn respectively on the lines y=x, z=1y=x,\ z=1 and y=−x, z=−1y=-x,\ z=-1. If PP is such that ∠QPR\angle QPR is a right angle, then the possible value(s) of λ\lambda is(are)

Choose one or more options
Answer and worked solution

Official answer: C

Foot on the first line. Points (t,t,1)(t,t,1), direction (1,1,0)(1,1,0). PQ→⋅(1,1,0)=0\overrightarrow{PQ}\cdot(1,1,0)=0 gives t=λt=\lambda, so Q=(λ,λ,1)Q=(\lambda,\lambda,1) and PQ→=(0,0,1−λ)\overrightarrow{PQ}=(0,0,1-\lambda).

Foot on the second line. Points (s,−s,−1)(s,-s,-1), direction (1,−1,0)(1,-1,0). (s−λ)−(−s−λ)=0(s-\lambda)-(-s-\lambda)=0 gives s=0s=0, so R=(0,0,−1)R=(0,0,-1) and PR→=(−λ,−λ,−1−λ)\overrightarrow{PR}=(-\lambda,-\lambda,-1-\lambda).

Right angle at P.
PQ→⋅PR→=(1−λ)(−1−λ)=0  ⟹  λ=±1.\overrightarrow{PQ}\cdot\overrightarrow{PR}=(1-\lambda)(-1-\lambda)=0\implies\lambda=\pm1.
For λ=1\lambda=1, P=Q=(1,1,1)P=Q=(1,1,1) and there is no angle. So λ=−1\lambda=-1.

(C)

Topics:Angles and distancesEquation of a line in space
← Previous questionNext question →