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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let (L _ 1 : (x-1)/(1)=(y-2)/(-1)=(z-1)/(2)) and (L _ 2 :...

Let (\mathrm{L}}_{1}: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}) and (\mathrm{L}}_{2}: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}) be two lines. Let L3L_3 be a line passing through the point ((\alpha, \beta, \gamma))andbeperpendiculartoboth) and be perpendicular to bothL_1andandL_2.If. IfL_3intersects(Lintersects (\mathrm{L}}_{1}), then (|5 α−\alpha-11 β−\beta-8 γ∣\gamma|) equals :

Options

  1. A20
  2. B18
  3. C25
  4. D16

Correct answer

C. 25

Step-by-step solution

 DR’s of  L3=m→×n→=∣i^j^k^1−12−121∣=−5i^−3j^+k^L3:x−α−5=y−β−3=z−γ1=λA(α−5λ,β−3λ,γ+λ)L1:x−11=y−2−1=z−12=k\begin{aligned} & \text{ DR's\text{ of } } L_3=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ -1 & 2 & 1\end{array}\right| \\ & =-5 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & L_3: \frac{x-\alpha}{-5}=\frac{y-\beta}{-3}=\frac{z-\gamma}{1}=\lambda \\ & \quad A(\alpha-5 \lambda, \beta-3 \lambda, \gamma+\lambda) \\ & L_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}=k\end{aligned}\mathrm{B}(\mathrm{k}+1,-\mathrm{k}+2,2 \mathrm{k}+1)NowNowα−5λ=k+1⇒α=5λ+k+1β−3λ=−k+2⇒β=3λ−k+2γ+λ=2k−1⇒γ=−λ+2k+1∣5α−11β−8γ∣=∣−25∣=25\begin{aligned} & \alpha-5 \lambda=\mathrm{k}+1 \Rightarrow \alpha=5 \lambda+\mathrm{k}+1 \\ & \beta-3 \lambda=-\mathrm{k}+2 \Rightarrow \beta=3 \lambda-\mathrm{k}+2 \\ & \gamma+\lambda=2 \mathrm{k}-1 \Rightarrow \gamma=-\lambda+2 \mathrm{k}+1 \\ & |5 \alpha-11 \beta-8 \gamma|=|-25| \\ & =25 \end{aligned}$
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