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Two adjacent sides of a parallelogram $ABCD$ are given by $AB = 2 i + 10 j + 11 k$ and $AD = - i + 2 j + 2 k$ 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,

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\frac{8}{9}

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

\frac{1}{9}

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

answer is B.

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Detailed Solution

Two adjacent sides of a parallelogram

ABCD

are given by,

\vec{AB} = 2 \hat{i} + 10 \hat{j} + 11 \hat{k}

\vec{AD} = - \hat{i} + 2 \hat{j} + 2 \hat{k}

Angle '

θ

' between

\overset{̅}{AB}

and

\overset{̅}{AD}

can be given as,

\cos (θ) = |\frac{\bar{AB} \cdot \bar{AD}}{|\bar{AB}| |\bar{AD}|}|

\cos θ = \frac{2 (- 1) + 10.2 + 11.2}{\sqrt{4 + 100 + 121} \sqrt{1 + 4 + 4}}

\cos θ = |\frac{- 2 + 20 + 22}{(15) (3)}|

\cos θ = \frac{8}{9}

Now,

\sin θ = \frac{\sqrt{9^{2} - 8^{2}}}{9}

\sin θ = \frac{\sqrt{81 - 64}}{9}

\sin θ = \frac{\sqrt{17}}{9}

Here,

θ + α = 90^{\circ}

α = 90^{\circ} - θ

\cos α = \cos (90^{\circ} - θ)

\cos α = \sin θ

(since,

\cos (90 - θ) = \sin θ

)

\cos α = \frac{\sqrt{17}}{9}

Correct option is 2.

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