JEE Main 2026 Jan 23 Shift 2, Mathematics Q3: Scalar and Vector Triple Product

JEE Main2026Jan 23, Shift 2Mathematics

Let $a = \hat{i} - 2 \hat{j} + 3 \hat{k}$ , $b = 2 \hat{i} + \hat{j} - \hat{k}$ , $c = λ \hat{i} + \hat{j} + \hat{k}$ and $v = a \times b$ . If $v \cdot c = 11$ and the length of the projection of $b$ on $c$ is $p$ , then $9 p^{2}$ is equal to :

A
9
B
6
C
4
D
12

Solution

Compute $v = a \times b = \hat{i} \hat{j} \hat{k} 1 - 23 21 - 1 = - \hat{i} + 7 \hat{j} + 5 \hat{k}$ .

$v \cdot c = - λ + 7 + 5 = 11 \Rightarrow λ = 1$ .

So $c = \hat{i} + \hat{j} + \hat{k}$ , $∣ c ∣ = 3$ .

$p = \frac{b \cdot c}{∣ c ∣} = \frac{2 + 1 - 1}{3} = \frac{2}{3}$ .

$9 p^{2} = 9 \cdot \frac{4}{3} = 12$ .

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JEE Main 2026 Jan 23 Shift 2, Mathematics Q3: Scalar and Vector Triple Product

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