Question
Let $\overrightarrow{AB} = 2\widehat{i} + 4\widehat{j} - 5k$ and
$\overrightarrow{AD} = \widehat{i} + 2\widehat{j} + \lambda k,\lambda \in \mathbb{R}$. Let the projection of the vector $\overrightarrow{v} = \widehat{i} + \widehat{j} + \widehat{k}$ 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 $\lambda^{2}x^{2} - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to
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Solution
$\overrightarrow{AC} = 3\widehat{i} + 6\widehat{j} + (\lambda - 5)k$
$$\overrightarrow{v} \cdot AC = 1 \Rightarrow 3 + 6 + \lambda - 5 = \sqrt{9 + 36 + (\lambda - 5)^{2}} $$
$${\Rightarrow \lambda^{2} + 8\lambda + 16 = \lambda^{2} - 10\lambda + 70 }{\Rightarrow \lambda = \frac{54}{18} = 3 }$$∴ Quadratic : $9x^{2} - 18x + 5 = 0 \Rightarrow x = \frac{1}{3},\frac{5}{3}$
$$\therefore 2\alpha - \beta = \frac{10 - 1}{3} = 3$$
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