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$Two adjacent sides of a parallelogram A B C D are given by \vec{A B} = 2 \hat{i} + 10 \hat{j} + 11 \hat{k} and \vec{A D} = - \hat{i} + 2 \hat{j} + 2 \hat{k} . The$
$side A D is rotated by an acute angle α in the plane of the$ $parallelogram so that A D becomes A D^{'} . If A D^{'} makes a$
$right angle with the side A B, then the cosine of the angle$ $α is given by$

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$\frac{8}{9}$

$\frac{1}{9}$

$\frac{4 \sqrt{5}}{9}$

$\frac{\sqrt{17}}{9}$

answer is B.

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Detailed Solution

$Angle between vectors \vec{A B} and \vec{A D} is given by$

$\cos θ = \frac{\vec{A B} \cdot \vec{A D}}{| \vec{A B} | \cdot | \vec{A D} |} = \frac{- 2 + 20 + 22}{\sqrt{4 + 100 + 121} \sqrt{1 + 4 + 4}}$

$= \frac{8}{9}$

$\Rightarrow \cos α = \cos (90^{\circ} - θ) = \sin θ = \frac{\sqrt{17}}{9}$

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