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JEE Main 2023 · Mathematics · Three Dimensional Geometry · hard

A line passes through P(0,0,4) and meets the line L1: x=y=z at Q1 and the line L2: vec r=(1,0,2)+μ(0,1,1) at Q2. Then PQ1^2+PQ2^2 equals

Numerical answer

A line passes through P(0,0,4)P(0,0,4) and meets the line L1: x=y=zL_1:\ x=y=z at Q1Q_1 and the line L2: r⃗=(1,0,2)+μ(0,1,1)L_2:\ \vec r=(1,0,2)+\mu(0,1,1) at Q2Q_2. Then PQ12+PQ22PQ_1^2+PQ_2^2 equals
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Answer: 15

Let Q1=(λ,λ,λ)Q_1=(\lambda,\lambda,\lambda) and Q2=(1,μ,2+μ)Q_2=(1,\mu,2+\mu). Collinearity of P,Q1,Q2P,Q_1,Q_2 requires PQ1→=(λ,λ,λ−4)\overrightarrow{PQ_1}=(\lambda,\lambda,\lambda-4) to be parallel to PQ2→=(1,μ,μ−2)\overrightarrow{PQ_2}=(1,\mu,\mu-2). Comparing the first two components gives λ=λ/μ\lambda=\lambda/\mu, so λ=0\lambda=0 or μ=1\mu=1. The case λ=0\lambda=0 makes PQ1→=(0,0,−4)\overrightarrow{PQ_1}=(0,0,-4), impossible because PQ2→\overrightarrow{PQ_2} has xx-component 11. So μ=1\mu=1, PQ2→=(1,1,−1)\overrightarrow{PQ_2}=(1,1,-1) and λ−4=−λ\lambda-4=-\lambda gives λ=2\lambda=2. Then PQ12=12PQ_1^2=12 and PQ22=3PQ_2^2=3, so the sum is 1515.
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