JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

Let $A B = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $A D = \hat{i} + 2 \hat{j} + λ \hat{k}, λ \in R$ . Let the projection of the vector $v = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $A C$ of the parallelogram ABCD be of length one unit. If $α, β$ , where $α > β$ , be the roots of the equation $λ^{2} x^{2} - 6 λ x + 5 = 0$ , then $2 α - β$ is equal to

Solution

In parallelogram ABCD, $A C = A B + A D = 3 \hat{i} + 6 \hat{j} + (λ - 5) \hat{k}$ .

The projection of $v$ on $A C$ has length $\frac{v \cdot A C}{∣ A C ∣} = 1$ , so $v \cdot A C = ∣ A C ∣$ .

$v \cdot A C = 3 + 6 + (λ - 5) = λ + 4$ and $∣ A C ∣^{2} = 9 + 36 + (λ - 5)^{2}$ .

Squaring: $(λ + 4)^{2} = 45 + (λ - 5)^{2}$ , giving $λ^{2} + 8 λ + 16 = λ^{2} - 10 λ + 70$ , so $18 λ = 54$ and $λ = 3$ .

Quadratic becomes $9 x^{2} - 18 x + 5 = 0$ with roots $\frac{1}{3}, \frac{5}{3}$ . Since $α > β$ : $α = \frac{5}{3}, β = \frac{1}{3}$ .

$2 α - β = \frac{10}{3} - \frac{1}{3} = 3$ .

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