JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

For three unit vectors $a, b, c$ satisfying $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 9$ and $∣2 a + k b + k c ∣ = 3$ , the positive value of $k$ is :

Solution

Expand: $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 2 (∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2}) - 2 (a \cdot b + b \cdot c + c \cdot a) = 6 - 2 (a \cdot b + b \cdot c + c \cdot a)$ .

Setting this equal to 9: $a \cdot b + b \cdot c + c \cdot a = - \frac{3}{2}$ .

Note $∣ a + b + c ∣^{2} = 3 + 2 (- 3/2) = 0$ , so $a + b + c = 0$ . Hence $b + c = - a$ .

$∣2 a + k (b + c) ∣ = ∣2 a - k a ∣ = ∣2 - k ∣ ∣ a ∣ = ∣2 - k ∣$ .

Set $∣2 - k ∣ = 3 \Rightarrow k = 5$ or $k = - 1$ . Positive value: $k = 5$ .

Option (4).

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