JEE Main 2025 Jan 29 Shift 1, Mathematics Q14: Applications of Vectors

JEE Main2025Jan 29, Shift 1Mathematics

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ and $b = 2 \hat{i} + 7 \hat{j} + 3 \hat{k} .$ Let

$L_{1} : r = (- \hat{i} + 2 \hat{j} + \hat{k}) + λ a, λ \in R$ and $L_{2} : r = (\hat{j} + \hat{k}) + μ b, μ \in R$ be two lines.

If the line $L_{3}$ passes through the point of intersection of $L_{1}$ and $L_{2}$ , and is parallel to $a + b$ , then $L_{3}$ passes through the point:

A
$(8, 26, 12)$
B
$(2, 8, 5)$
C
$(- 1, - 1, 1)$
D
$(5, 17, 4)$

Solution

Parameterise the two lines componentwise:

$L_{1} : (λ - 1, 2 λ + 2, λ + 1); L_{2} : (2 μ, 7 μ + 1, 3 μ + 1) .$

Equate to find the intersection: $λ - 1 = 2 μ, 2 λ + 2 = 7 μ + 1, λ + 1 = 3 μ + 1.$ Solving yields $λ = 3, μ = 1$ , and the common point is $(2, 8, 4) .$

Direction of $L_{3}$ : $a + b = 3 \hat{i} + 9 \hat{j} + 4 \hat{k} .$

So $L_{3} : r = (2, 8, 4) + α (3, 9, 4) .$ Setting $α = 2$ gives $(8, 26, 12) .$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q14: Applications of Vectors

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