Two adjacent sides of a parallelogram ABCD are given by \overset{\rightarrow}{AB} \textrm{ }\textrm{ } = \textrm{ }\textrm{ } 2 \hat{i} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } 10 \hat{j} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } 11 \textrm{ }\textrm{ } \hat{k} \textrm{ }\textrm{ } and \textrm{ }\textrm{ } \overset{\rightarrow}{AD} \textrm{ }\textrm{ } = \textrm{ }\textrm{ } - \textrm{ } \hat{i} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } 2 \hat{j} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } 2 \textrm{ } \hat{k} The side AD is rotated by an acute angle \alpha 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 a is given by

Engineering

Mathematics

Scalar Product of Two Vectors

Question

Two adjacent sides of a parallelogram ABCD are given by

$\vec{AB} = 2 \hat{i} + 10 \hat{j} + 11 \hat{k} and \vec{AD} = - \hat{i} + 2 \hat{j} + 2 \hat{k}$

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 a is given by

$\frac{1}{9}$

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

$\frac{8}{9}$

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

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$\cos (90 - α) = \frac{- 2 + 20 + 22}{15 \times 3} \Rightarrow sinα = \frac{8}{9}$

$\Rightarrow cosα = \frac{\sqrt{17}}{9}$

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