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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Three Dimensional Geometry

If the shortest distance between the lines...

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 _________

Correct answer

0

Step-by-step solution

\begin{aligned} & \overline{\mathrm{a}_1=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overline{\mathrm{a}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & (\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}_1-\overline{\mathrm{a}_2 \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}
4430=∣−6λ−12−105−9∣(−6)2+(−15)2+324430=∣6λ+126∣330132=∣6λ+126∣λ=1,λ=−43∣λ∣=43\begin{aligned} & \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} \\ & \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} \\ & 132=|6 \lambda+126| \\ & \lambda=1, \lambda=-43 \\ & |\lambda|=43 \end{aligned}
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