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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Three Dimensional Geometry

If the shortest distance between the lines...

If the shortest distance between the lines x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and x1=yα=z−51\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} is 56\frac{5}{\sqrt{6}}, then the sum of all possible values of α\alpha is

Options

  1. A32\frac{3}{2}
  2. B−32-\frac{3}{2}
  3. C3
  4. D-3

Correct answer

D. -3

Step-by-step solution

L1:x−12=y−23=z−34L1:x1=yα=z−51x⃗=∣i^j^k^2341α1∣=i^(3−4α)−j^(−2)+k^(2α−3) S.D. =∣BA→⋅n→∣n→∣∣=∣(i^+2j^−2k^)⋅n⃗∣n→∣∣⇒6(13−8α)2=25((4α−3)2+(2α−3)2+16)6(64a2−280α+169)=25(20α2−36α+34)⇒116α2+348α−164=0α1+α2=−348116=−3\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & L_1: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} \\ & \vec{x}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1\end{array}\right|=\hat{\mathrm{i}}(3-4 \alpha)-\hat{\mathrm{j}}(-2)+\hat{\mathrm{k}}(2 \alpha-3) \\ & \text{ S.D. }=\left|\frac{\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \vec{n}}{|\overrightarrow{\mathrm{n}}|}\right| \\ & \Rightarrow 6(13-8 \alpha)^2=25\left((4 \alpha-3)^2+(2 \alpha-3)^2+16\right) \\ & 6\left(64 a^2-280 \alpha+169\right)=25\left(20 \alpha^2-36 \alpha+34\right) \\ & \Rightarrow 116 \alpha^2+348 \alpha-164=0 \\ & \alpha_1+\alpha_2=\frac{-348}{116}=-3\end{aligned}
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