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Mathematics · Straight lines

JEE Main 2026 — 24 January, Evening Shift — Question 11

The sum of all values of α\alpha, for which the shortest distance between the lines x+1α=y−2−1=z−4−α\frac{\mathrm{x}+1}{\alpha}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-4}{-\alpha} and xα=y−12=z−12α\frac{\mathrm{x}}{\alpha}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-1}{2 \alpha} is 2\sqrt{2}, is

  1. Option A:

    88

  2. Option B:

    −6-6

    Correct
  3. Option C:

    66

  4. Option D:

    −8-8

Answer: B

Step-by-step solution

2=∣−113α−1−αα22α∣∣i^j^k^α−1−αα22α∣\sqrt{2} = \frac{ \left| \begin{array}{ccc} -1 & 1 & 3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{array} \right| }{ \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{array} \right| } 2=−1(−2α+2α)−1(2α2+α2)+3(2α+α)∣i^(−2α+2α)−j^(2α2+α2)+k^(2α+α)∣\sqrt{2} = \frac{ -1(-2\alpha+2\alpha) -1(2\alpha^{2}+\alpha^{2}) +3(2\alpha+\alpha) }{ \left| \hat{i}(-2\alpha+2\alpha) -\hat{j}(2\alpha^{2}+\alpha^{2}) +\hat{k}(2\alpha+\alpha) \right| } 2=−3α2+9α9α4+9α2=−α+3α2+1\sqrt{2} = \frac{-3\alpha^{2}+9\alpha}{\sqrt{9\alpha^{4}+9\alpha^{2}}} = \frac{-\alpha+3}{\sqrt{\alpha^{2}+1}} ⇒2α2+2=α2+9−6α\Rightarrow 2\alpha^{2}+2=\alpha^{2}+9-6\alpha α2+6α−7=0\alpha^{2}+6\alpha-7=0 (α+7)(α−1)=0(\alpha+7)(\alpha-1)=0 α=−7, 1\alpha=-7,\,1 Sum=−7+1=−6\text{Sum}=-7+1=-6 Option (2)\text{Option }(2)

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Evening Shift
Subject
Mathematics
Chapter
Straight lines
Topic
Cartesian Coordinates and Basic Coordinate Geometry

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