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

Now ,AB `= 2 hati + 10 hatj +11hatk`
`AD=- hati + 2 hatj + 2hatk`

`implies |AB|=sqrt(4+100+121)=sqrt(225)=15`
` and |AD|=sqrt(1+4+4)=sqrt(9)=3`
Now `AB.AD =( 2hati + 10 hatj +11 hatk ).( -hati +2hatj +2hatk)`
` =-2+20 +22=40`
`therefore cos theta =(AB.AD)/(|AB||AD|)=(40)/(45)=(8)/(9)`
` :' theta + alpha =90^(@)implies alpha =90^(@)-theta `
`implies cos alpha =sin theta = sqrt(1-(64)/(81))=(sqrt(17))/(9)`