Aspire Faculty ID #17935 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

Let $\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda \hat{k}$, $\lambda \in \mathbb{R}$. Let the projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the diagonal $\overrightarrow{AC}$ of parallelogram $ABCD$ be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to:

Solution

$\overrightarrow{AC} = 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}$ $|\vec{v} \cdot \overrightarrow{AC}| = 1 \Rightarrow 3 + 6 + \lambda - 5 = \sqrt{9 + 36 + (\lambda - 5)^2}$ $\Rightarrow \lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 70$ $\Rightarrow \lambda = 3$ Quadratic: $9x^2 - 18x + 5 = 0 \Rightarrow x = \frac{1}{3},; \frac{5}{3}$ $\Rightarrow 2\alpha - \beta = \frac{10 - 1}{3} = 3$
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