Two adjacent sides of a parallelogram `A B C D` are given by ` vec A B=2 hat i+10 hat j+11 hat ka n d vec A D=- hat i+2 hat j+2 hat kdot` The side `A D` is rotated by an acute angle `alpha` in the plane of the parallelogram so that `A D` becomes `A D^(prime)dot` If `A D '` makes a right angle with the side `A B ,` then the cosine of the angel `alpha` is given by a. `8/9` b. `(sqrt(17))/9` c. `1/9` d. `(4sqrt(5))/9`

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`