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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Three Dimensional Geometry

The sum of all values of α , for which the shortest distance...

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{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} and xα=y−12=z−12α\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha} is 2\sqrt{2}, is

Options

  1. A6
  2. B8
  3. C-8
  4. D-6

Correct answer

D. -6

Step-by-step solution

The given lines are L1:x+1α=y−2−1=z−4−αL_1: \frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha} and L2:xα=y−12=z−12αL_2: \frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha}.
For L1L_1, point a⃗1=(−1,2,4)\vec{a}_1 = (-1, 2, 4) and direction b⃗1=(α,−1,−α)\vec{b}_1 = (\alpha, -1, -\alpha).
For L2L_2, point a⃗2=(0,1,1)\vec{a}_2 = (0, 1, 1) and direction b⃗2=(α,2,2α)\vec{b}_2 = (\alpha, 2, 2\alpha).
The shortest distance dd is given by d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}.
a⃗2−a⃗1=(1,−1,−3)\vec{a}_2 - \vec{a}_1 = (1, -1, -3).
b⃗1×b⃗2=∣i^j^k^α−1−αα22α∣=i^(−2α+2α)−j^(2α2+α2)+k^(2α+α)=(0,−3α2,3α)\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix} = \hat{i}(-2\alpha + 2\alpha) - \hat{j}(2\alpha^2 + \alpha^2) + \hat{k}(2\alpha + \alpha) = (0, -3\alpha^2, 3\alpha).
∣b⃗1×b⃗2∣=0+9α4+9α2=3∣α∣α2+1|\vec{b}_1 \times \vec{b}_2| = \sqrt{0 + 9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2 + 1}.
(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(1)(0)+(−1)(−3α2)+(−3)(3α)=3α2−9α(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(0) + (-1)(-3\alpha^2) + (-3)(3\alpha) = 3\alpha^2 - 9\alpha.
Given d=2d = \sqrt{2}, so ∣3α2−9α∣3∣α∣α2+1=2⇒∣α−3∣α2+1=2\frac{|3\alpha^2 - 9\alpha|}{3|\alpha|\sqrt{\alpha^2 + 1}} = \sqrt{2} \Rightarrow \frac{|\alpha - 3|}{\sqrt{\alpha^2 + 1}} = \sqrt{2}.
Squaring both sides: (α−3)2=2(α2+1)⇒α2−6α+9=2α2+2(\alpha - 3)^2 = 2(\alpha^2 + 1) \Rightarrow \alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2.
α2+6α−7=0\alpha^2 + 6\alpha - 7 = 0.
The values of α\alpha are the roots of this quadratic equation.
Sum of values of α=−61=−6\alpha = -\frac{6}{1} = -6.
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