Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{\mathbf{AB}} = 2\widehat{\dot{\mathbf{i}}} + 10\widehat{\dot{\mathbf{j}}} + 11\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{AD}} = - \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$. The side $AD$ is rotated by an acute angle α in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$, then the cosine of the angle α is given by

Question

Dr. Abubacker · Accepted Answer

(b)
$\overrightarrow{\mathbf{AB}} = 2\widehat{\dot{\mathbf{i}}} + 10\widehat{\dot{\mathbf{j}}} + 11\widehat{\mathbf{k}}$
$\overrightarrow{\mathbf{AD}} = - \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$
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$\overrightarrow{\mathbf{AB}} \bullet \overrightarrow{\mathbf{AD}} = - 2 + 20 + 22 = 40$
$\left| \overrightarrow{\mathbf{AB}} ight| = \sqrt{4 + 100 + 120} = \sqrt{225} = 15$
$\left| \overrightarrow{\mathbf{AD}} ight| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
${\ 	herefore\ cos}{	heta = \frac{40}{45} = \frac{8}{9}}$
$	herefore\ 	heta + \alpha = 90{^\circ}$
$\Longrightarrow \ \alpha = 90{^\circ} - 	heta$
$\Longrightarrow \cos{\alpha = \sin{	heta = \sqrt{1 - \frac{64}{81}}}} = \frac{\sqrt{17}}{9}$

Explanation

Reviewed by Dr. Abubacker

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