1.

Two adjacent sides of a parallelogram ABCD are given by AB=2hati+10hatj+11hatk and 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' makes a right angle with the sides AB, then the cosine of the angle alpha is given by

Answer»

`(8)/(9)`
`(sqrt(17))/(9)`
`(1)/(9)`
`(4sqrt(5))/(9)`

Solution :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`
`thereforecos THETA=(AB.AD)/(|AB||AD|)=(40)/(45)=(8)/(9)`
` :'theta+ alpha=90^(@)implies alpha =90^(@)-theta`
`impliescos alpha=sintheta= sqrt(1-(64)/(81))=(sqrt(17))/(9)`


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