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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let the shortest distance between the lines...

Let the shortest distance between the lines x−33=y−α−1=z−31\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1} and x+3−3=y+72=z−β4\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4} be 3303 \sqrt{30}. Then the positive value of 5α+β5 \alpha+\beta is

Options

  1. A42
  2. B46
  3. C48
  4. D40

Correct answer

B. 46

Step-by-step solution

\begin{aligned} & {\mathrm{A}(3, \alpha, 3) & B(-3,-7, \beta)} \\ & \overrightarrow{\mathrm{BA}}=6 \hat{\mathrm{i}}+(\alpha+7) \hat{\mathrm{j}}+(3-\beta) \hat{\mathrm{k}} \\ \end{aligned}
\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|i^j^k^3−11−324\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|
∣BA→⋅(p→×q→)∣∣p→×q→∣=330\frac{|\overrightarrow{\mathrm{BA}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}=3 \sqrt{30}
36+15(α+7)−3(3−β)=(330)236+15(\alpha+7)-3(3-\beta)=(3 \sqrt{30})^2
36+15α+105−9+3β=27015α+3β=1385α+β=46\begin{aligned} & 36+15 \alpha+105-9+3 \beta=270 \\ & 15 \alpha+3 \beta=138 \\ & 5 \alpha+\beta=46\end{aligned}
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