JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ and $b = λ \hat{j} + 2 \hat{k}$ , $λ \in Z$ be two vectors. Let $c = a \times b$ and $d$ be a vector of magnitude 2 in yz-plane. If $∣ c ∣ = 53$ , then the maximum possible value of $(c \cdot d)^{2}$ is equal to :

A
26
B
104
C
208
D
52

Solution

$c = a \times b = (- 2 - λ) \hat{i} - 4 \hat{j} + 2 λ \hat{k}$ .

$∣ c ∣^{2} = (λ + 2)^{2} + 16 + 4 λ^{2} = 5 λ^{2} + 4 λ + 20 = 53$ , so $5 λ^{2} + 4 λ - 33 = 0$ .

Integer solution: $λ = - 3$ (the other root is $2.2$ ). So $c = \hat{i} - 4 \hat{j} - 6 \hat{k}$ .

Let $d = y \hat{j} + z \hat{k}$ with $y^{2} + z^{2} = 4$ . Then $c \cdot d = - 4 y - 6 z$ .

By Cauchy–Schwarz: $(c \cdot d)^{2} = (- 4 y - 6 z)^{2} \leq (16 + 36) (y^{2} + z^{2}) = 52 \cdot 4 = 208$ .

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