Vector Algebra
Parallelogram Diagonal — Projection Condition
nta_pyq_2026_jan
Grade None
Question:
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 $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{AC}$ of the 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:
Step-by-Step Solution
Key Concept: $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=3\hat{i}+6\hat{j}+(\lambda-5)\hat{k}$. Projection $=\frac{|\vec{v}\cdot\overrightarrow{AC}|}{|\overrightarrow{AC}|}=1$.
$2\alpha-\beta=3$.
Correct Answer: 1