JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let $a = - \hat{i} + 2 \hat{j} + 2 \hat{k}, b = 8 \hat{i} + 7 \hat{j} - 3 \hat{k}$ and $c$ be a vector such that $a \times c = b$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $∣ a + c ∣^{2}$ is equal to :

A
33
B
30
C
35
D
27

Solution

Write $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ . Expanding $a \times c$ and equating to $b$ gives three equations:

$2 c_{3} - 2 c_{2} = 8, c_{3} + 2 c_{1} = 7, c_{2} + 2 c_{1} = 3$

The constraint $c_{1} + c_{2} + c_{3} = 4$ closes the system. Solving: $c_{1} = 2, c_{2} = - 1, c_{3} = 3$ .

$a + c = \hat{i} + \hat{j} + 5 \hat{k}$ , so $∣ a + c ∣^{2} = 1 + 1 + 25 = 27$ .

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