JEE Main 2025 Apr 7 Shift 2, Mathematics Q16: Vector (or Cross) Product of Two Vectors

JEE Main2025Apr 7, Shift 2Mathematics

Consider the lines $L_{1} : x - 1 = y - 2 = z$ and $L_{2} : x - 2 = y = z - 1$ . Let the feet of the perpendiculars from the point P(5,1,-3) on the lines $L_{1}$ and $L_{2}$ be Q and R respectively. If the area of the triangle PQR is A, then $4 A^{2}$ is equal to :

A
139
B
147
C
151
D
143

Solution

Both lines have direction ratios $(1, 1, 1)$ . Parametrise $L_{1}$ as $(1 + λ, 2 + λ, λ)$ ; the foot of the perpendicular Q from $P (5, 1, - 3)$ satisfies $P Q \cdot (1, 1, 1) = 0$ , giving $(λ - 4) + (λ + 1) + (λ + 3) = 0 \Rightarrow λ = 0$ , so $Q = (1, 2, 0)$ .

Parametrise $L_{2}$ as $(2 + μ, μ, 1 + μ)$ ; similarly $P R \cdot (1, 1, 1) = 0$ gives $(μ - 3) + (μ - 1) + (μ + 4) = 0 \Rightarrow μ = 0$ , so $R = (2, 0, 1)$ .

$P Q = (- 4, 1, 3)$ and $P R = (- 3, - 1, 4)$ .

$P Q \times P R = \hat{i} \hat{j} \hat{k} - 413 - 3 - 14 = 7 \hat{i} + 7 \hat{j} + 7 \hat{k}$ , with magnitude $73$ .

Area $A = \frac{1}{2} \times 73 = \frac{7 3}{2}$ .

$∴ 4 A^{2} = 4 \times \frac{147}{4} = 147$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q16: Vector (or Cross) Product of Two Vectors

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