JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

Let a line $L$ passing through the point $P (1, 1, 1)$ be perpendicular to the lines $\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1}$ and $\frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0}$ . Let the line $L$ intersect the yz-plane at the point $Q$ . Another line parallel to $L$ and passing through the point $S (1, 0, - 1)$ intersects the yz-plane at the point $R$ . Then the square of the area of the parallelogram $P QR S$ is equal to _____ .

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Solution

Direction vectors of the two given lines: $d_{1} = ⟨ 4, 1, 1 ⟩$ , $d_{2} = ⟨ 1, 1, 0 ⟩$ .

Direction of $L$ :

$d_{L} = d_{1} \times d_{2} = \hat{i} \hat{j} \hat{k} 411110 = ⟨ - 1, 1, 3 ⟩$ .

Find $Q$ : $L$ through $P (1, 1, 1)$ : $(1 - t, 1 + t, 1 + 3 t)$ . yz-plane $\Rightarrow 1 - t = 0 \Rightarrow t = 1 \Rightarrow Q = (0, 2, 4)$ .

Find $R$ : Line through $S (1, 0, - 1)$ with direction $⟨ - 1, 1, 3 ⟩$ : $(1 - μ, μ, - 1 + 3 μ)$ . yz-plane $\Rightarrow μ = 1 \Rightarrow R = (0, 1, 2)$ .

Adjacent edges: $P Q = ⟨ - 1, 1, 3 ⟩$ , $P S = ⟨ 0, - 1, - 2 ⟩$ .

$P Q \times P S = \hat{i} \hat{j} \hat{k} - 113 0 - 1 - 2 = ⟨ 1, - 2, 1 ⟩$ .

$∣ P Q \times P S ∣^{2} = 1 + 4 + 1 = 6$ .

The square of the area of the parallelogram is $6$ .

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