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

JEE Main2026Jan 21, Shift 1Mathematics

Let $c$ and $d$ be vectors such that $∣ c + d ∣ = 29$ and $c \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times d$ . If $λ_{1}, λ_{2} (λ_{1} > λ_{2})$ are the possible values of $(c + d) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$ , then the equation

$K^{2} x^{2} + (K^{2} - 5 K + λ_{1}) x y + (3 K + \frac{λ _{2}}{2}) y^{2} - 8 x + 12 y + λ_{2} = 0$

represents a circle, for k equal to :

A
4
B
1
C
$- 1$
D
2

Solution

The vector condition rearranges to $(c + d) \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = 0$ , so $c + d$ is parallel to that vector. With $∣ c + d ∣ = 29$ (and the parallel vector also has magnitude $29$ ), $c + d = \pm (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ .

Dotting with $(- 7 \hat{i} + 2 \hat{j} + 3 \hat{k})$ : $\pm (- 14 + 6 + 12) = \pm 4$ , so $λ_{1} = 4$ , $λ_{2} = - 4$ .

For the conic to be a circle we need (i) coefficient of $x y = 0$ : $K^{2} - 5 K + 4 = 0 \Rightarrow K = 1, 4$ ; (ii) coefficients of $x^{2}$ and $y^{2}$ equal: $K^{2} = 3 K - 2 \Rightarrow K = 1, 2$ .

Common value: $K = 1$ .

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