JEE Main 2026 Jan 23 Shift 1, Mathematics Q3: Scalar and Vector Triple Product

JEE Main2026Jan 23, Shift 1Mathematics

Let $a = - \hat{i} + \hat{j} + 2 \hat{k}$ , $b = \hat{i} - \hat{j} - 3 \hat{k}$ , $c = a \times b$ and $d = c \times a$ . Then $(a - b) \cdot d$ is equal to :

A
$4$
B
$- 4$
C
$- 2$
D
$2$

Solution

Use the vector triple product. $d = (a \times b) \times a = ∣ a ∣^{2} b - (a \cdot b) a$ .

$∣ a ∣^{2} = 1 + 1 + 4 = 6$ ; $a \cdot b = - 1 - 1 - 6 = - 8$ .

$∴ d = 6 b + 8 a$ .

$(a - b) \cdot d = (a - b) \cdot (6 b + 8 a) = 8∣ a ∣^{2} - 6∣ b ∣^{2} - 2 (a \cdot b)$

$∣ b ∣^{2} = 1 + 1 + 9 = 11$ .

$= 8 (6) - 6 (11) - 2 (- 8) = 48 - 66 + 16 = - 2$ .

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JEE Main 2026 Jan 23 Shift 1, Mathematics Q3: Scalar and Vector Triple Product

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