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Mathematics Question on Three Dimensional Geometry

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=y= −x,z=−1-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)

A

2\sqrt{2}

B

11

C

−1-1

D

−2-\sqrt{2}

Answer

−1-1

Explanation

Solution

Line 1: x1=y1=z−10=r,Q(r,r,1)\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r, Q(r, r, 1)
Line 2: x1=y−1=z+10=k,R(k,−k,−1)\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}= k , R(k,-k,-1)
PQ→=(λ−r)i^+(λ−r)j^+(λ−1)k^\overrightarrow{P Q}=(\lambda-r) \hat{i}+(\lambda-r) \hat{j}+(\lambda-1) \hat{k}
and λ−r+λ−r=0\lambda-r+\lambda-r=0 as PQ→\overrightarrow{P Q} is ⊥\perp to L1L_{1}
⇒2λ=2r\Rightarrow 2 \lambda=2 r
⇒λ=r\Rightarrow \lambda=r
PR‾=(λ−k)i^+(λ+k)j^+(λ+1)k^\overline{ PR }=(\lambda- k ) \hat{ i }+(\lambda+ k ) \hat{ j }+(\lambda+1) \hat{ k }
and λ−k−λ−k=0\lambda- k -\lambda- k =0 as PR→\overrightarrow{ PR } is ⊥\perp to L2L _{2}
⇒k=0\Rightarrow k =0
so PQ⊥PRPQ \perp PR
(λ−r)(λ−k)+(λ−r)(λ+k)+(λ−1)(λ+1)=0(\lambda- r )(\lambda- k )+(\lambda- r )(\lambda+ k )+(\lambda-1)(\lambda+1)=0
⇒λ=1,−1\Rightarrow \lambda=1,-1
For λ=1\lambda=1 as points PP and QQ coincide
⇒λ=−1\Rightarrow \lambda=-1