JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

A
8
B
$- 6$
C
6
D
$- 8$

Solution

Points: $P_{1} = (- 1, 2, 4)$ , $P_{2} = (0, 1, 1)$ ; directions $d_{1} = (α, - 1, - α)$ , $d_{2} = (α, 2, 2 α)$ .

$P_{1} P_{2} = (1, - 1, - 3)$ .

$d_{1} \times d_{2} = \hat{i} \hat{j} \hat{k} α - 1 - α α 22 α = (- 2 α + 2 α) \hat{i} - (2 α^{2} + α^{2}) \hat{j} + (2 α + α) \hat{k} = (0, - 3 α^{2}, 3 α)$ .

$∣ d_{1} \times d_{2} ∣ = 3∣ α ∣ α^{2} + 1$ .

Shortest distance: $\frac{∣ P _{1} P _{2} \cdot ( d _{1} \times d _{2} ) ∣}{∣ d _{1} \times d _{2} ∣} = \frac{∣0 + 3 α ^{2} - 9 α ∣}{3∣ α ∣ α ^{2} + 1} = \frac{∣ α - 3∣}{α ^{2} + 1} = 2$ .

Square: $(α - 3)^{2} = 2 (α^{2} + 1) \Rightarrow α^{2} + 6 α - 7 = 0 \Rightarrow (α + 7) (α - 1) = 0$ .

$α = - 7$ or $α = 1$ . Sum $= - 6$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →