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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 |
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Answer» `(8)/(9)` `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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