Two adjacent sides of a parallelogram ABCD are given by `vec(AB)=2hati+10hatj+11hatk` and `vec(AD)=-hati+2hatj+2hatk`. The side AD is rotated by an acute angle `alpha` in the plane of the parallelogram so that AD becomes AD'. If AD' make a right angle withe the side AB then the cosine of the angle `alpha` is given by

Options

A. `8/9`
B. `(sqrt(17))/9`
C. `1/9`
D. `(4sqrt(5))/9`

Answer

B

Explanation

Clearly, AD' is perpendiculat to `vec(AB)` and lies in the plane of `vec(AB)` and `vec(AD)`. Therefore,
`vec(AD)'=vec(AB)xx(vecABxxvecAD)`
`impliesvec(AD')=(vec(AB).vec(AD))vec(AB)-(vec(AB).vec(AB))vec(AD)`
`impliesvec(AD')=5(61hati-10hatj-2hatk)`
`:.cos alpha=(|vec(AD).vec(AD')|)/(|(vec(AD)||vec(AD')|)`
`implies cos alpha=(-hati+2hatj+2hatk).5(61hati-10hatj-2hatk)=(sqrt(17))/9`