Mathematics · 3D Geometry

JEE Main 2024 — 6 April, Shift 2 — Question 27

If the shortest distance between the lines x−λ3=y−2−1=z−11\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} and x+2−3=y+52=z−44\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} is 4430\frac{44}{\sqrt{30}}, then the largest possible value of ∣λ∣|\lambda| is equal to \qquad

Answer: 43

Numerical answer — enter this value.

Step-by-step solution

a‾1=λi^+2j^+k^\quad \overline{\mathrm{a}}_{1}=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

a‾2=−2i^−5j^+4k^\overline{\mathrm{a}}_{2}=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}

p→−=3i^−j^+k^\overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}

q⃗−=−3i^+2j^+4k^\vec{q}-=-3 \hat{i}+2 \hat{j}+4 \hat{k}

(λ+2)i^+7j^−3k^=a‾1−a‾2(\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_{1}-\overline{\mathrm{a}}_{2}

p→×q→−=−6i^−15j^+3k^\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

4430=∣−6λ−12−105−9∣(−6)2+(−15)2+32\frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^{2}+(-15)^{2}+3^{2}}}

4430=∣6λ+126∣330\frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}}

132=∣6λ+126∣132=|6 \lambda+126|

λ=1,λ=−43\lambda=1, \lambda=-43

∣λ∣=43|\lambda|=43

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Skew lines & shortest distance between them