JEE Main 2025 Jan 29 Shift 1, Mathematics Q17: Lines in Space

JEE Main2025Jan 29, Shift 1Mathematics

Let $L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}$ and $L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}$ be two lines.

Let $L_{3}$ be a line passing through the point $(α, β, γ)$ and be perpendicular to both $L_{1}$ and $L_{2}$ . If $L_{3}$ intersects $L_{1}$ , then $∣5 α - 11 β - 8 γ ∣$ equals :

A
$18$
B
$16$
C
$25$
D
$20$

Solution

Direction of $L_{3}$ : $m \times n$ where $m = (1, - 1, 2), n = (- 1, 2, 1) .$

$m \times n = ((- 1) (1) - (2) (2), (2) (- 1) - (1) (1), (1) (2) - (- 1) (- 1)) = (- 5, - 3, 1) .$

So $L_{3} : \frac{x - α}{- 5} = \frac{y - β}{- 3} = \frac{z - γ}{1} = t .$ A general point is $(α - 5 t, β - 3 t, γ + t) .$

A general point of $L_{1}$ is $(k + 1, - k + 2, 2 k + 1) .$ For intersection equate coordinates:

$α - 5 t = k + 1, β - 3 t = - k + 2, γ + t = 2 k + 1.$

Eliminating $t$ and $k$ , we evaluate the target combination:

$5 α - 11 β - 8 γ = 5 (k + 1 + 5 t) - 11 (- k + 2 + 3 t) - 8 (2 k + 1 - t) = (5 k + 5 + 25 t) + (11 k - 22 - 33 t) - (16 k + 8 - 8 t) = 0 \cdot k + 0 \cdot t - 25 = - 25.$

Therefore $∣5 α - 11 β - 8 γ ∣ = 25.$

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q17: Lines in Space

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