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Mathematics · 3D Geometry

JEE Main 2024 — 5 April, Shift 2 — Question 24

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} \quad and x+2−1=y+62=z−10\quad \frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}. Then (α−β)2(\alpha-\beta)^{2} is equal to \qquad

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

P(−3λ−2,4λ+2,2λ+5)\mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5)

Q(−μ−2,2μ−6,1)Q(-\mu-2,2 \mu-6,1)

DRS of PQ =(3λ−μ,2μ−4λ−8,−2λ−4)=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4)

Also DRSD R S of PQ=∣i j k −120−342∣PQ=\left| \begin{matrix}\overset{}{\mathop{i}}\, & \overset{}{\mathop{j}}\, & \overset{}{\mathop{k}}\, \\-1 & 2 & 0 \\-3 & 4 & 2 \\\end{matrix} \right|

=(4i^+2j^+2k^)=(4 \hat{i}+2 \hat{j}+2 \hat{k}) OR (2,1,1)(2,1,1)

3λ−μ2=2μ−4λ−81=−2λ−41\frac{3 \lambda-\mu}{2}=\frac{2 \mu-4 \lambda-8}{1}=\frac{-2 \lambda-4}{1}

⇒μ=λ+2&7λ=μ−8\Rightarrow \mu=\lambda+2 \& 7 \lambda=\mu-8

\begin{array}{*{35}{l}}\lambda =-1 & \mu =1 \\\end{array}

LetQ:(−3,−4,1)Let \mathrm{Q}:(-3,-4,1) be a point on one of the line

LPQ=x+32=y+41=z−11\mathrm{L}_{\mathrm{PQ}}=\frac{\mathrm{x}+3}{2}=\frac{\mathrm{y}+4}{1}=\frac{\mathrm{z}-1}{1}

(−1,α,β)(-1, \alpha, \beta); point lie on PQ ⇒1=α+41=β−11\Rightarrow 1=\frac{\alpha+4}{1}=\frac{\beta-1}{1}

⇒α=−3,β=2\Rightarrow \alpha=-3, \beta=2

(α−β)2=25(\alpha-\beta)^{2}=25

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them

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