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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let the point (-1, α, β) lie on the line of the shortest distance...

Let the point (−1,α,β)(-1, \alpha, \beta) lie on the line of the shortest distance between the lines x+2−3=y−24=z−52\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2} and x+2−1=y+62=z−10\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}. Then (α−β)2(\alpha-\beta)^2 is equal to___________

Correct answer

0

Step-by-step solution

P(−3λ−2,4λ+2,2λ+5)Q(−μ−2,2μ−6,1)DRS of  PQ=(3λ−μ,2μ−4λ−8,−2λ−4)DRS of PQ=∣i^j^k^−120−342∣=(4i^+2j^+2k^)\begin{aligned} & \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) \\ & \mathrm{Q}(-\mu-2,2 \mu-6,1) \\ & \mathrm{DRS} \text{\text{ of } } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4) \\ & \mathrm{DRS} \text{ of } \mathrm{PQ}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 2 & 0 \\ -3 & 4 & 2 \end{array}\right| \\ & =(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \end{aligned}OR (2,1,1)3λ−μ2=2μ−4λ−81=−2λ−41⇒μ=λ+27λ=μ−8λ=−1μ=1Q:(−3,−4,1)LPQ=x+32=y+41=z−11(−1,α,β)⇒1=α+41=β−11⇒α=−3,β=2(α−β)2=25\begin{aligned} & (2,1,1) \\ & \frac{3 \lambda-\mu}{2}=\frac{2 \mu-4 \lambda-8}{1}=\frac{-2 \lambda-4}{1} \\ & \Rightarrow \mu=\lambda+2 & 7 \lambda=\mu-8 \\ & \lambda=-1 \quad \mu=1 \\ & Q:(-3,-4,1) \\ & L_{P Q}=\frac{x+3}{2}=\frac{y+4}{1}=\frac{z-1}{1} \\ & (-1, \alpha, \beta) \Rightarrow 1=\frac{\alpha+4}{1}=\frac{\beta-1}{1} \\ & \Rightarrow \alpha=-3, \beta=2 \\ & (\alpha-\beta)^2=25 \end{aligned}$
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