JEE Main 2026 Apr 6 Shift 2, Mathematics Q15: Vector (or Cross) Product of Two Vectors

JEE Main2026Apr 6, Shift 2Mathematics

Let $a = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$ and $b = 6 \hat{i} + 3 \hat{j} + 3 \hat{k}$ . Then the square of the area of the triangle with adjacent sides determined by the vectors $(2 a + 3 b)$ and $(a - b)$ is:

A
450
B
900
C
1800
D
2400

Solution

$2 a + 3 b = (4 + 18) \hat{i} + (6 + 9) \hat{j} + (6 + 9) \hat{k} = 22 \hat{i} + 15 \hat{j} + 15 \hat{k}$ .

$a - b = - 4 \hat{i} + 0 \hat{j} + 0 \hat{k}$ .

Cross product: $(22 \hat{i} + 15 \hat{j} + 15 \hat{k}) \times (- 4 \hat{i}) = - 4 (15) (\hat{j} \times \hat{i}) - 4 (15) (\hat{k} \times \hat{i}) = - 4 (15) (- \hat{k}) - 4 (15) (\hat{j}) = - 60 \hat{j} + 60 \hat{k}$ .

Magnitude: $6 0^{2} + 6 0^{2} = 602$ .

Triangle area $= \frac{1}{2} \cdot 602 = 302$ , so area $^{2} = 1800$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q15: Vector (or Cross) Product of Two Vectors

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