Two adjacent sides of a parallelogram ABCD are given by AB = 2 i + 10 j + 11 k and AD = - i + 2 j + 2 k . The side AD is rotated by an acute angle in the plane of the parallelogram so that AD becomes AD' . If AD' makes a right angle with the side AB , then the cosine of the angle is...

Step-by-step solution

Let be the angle between the vectors AB and AD . = AB AD | AB | | AD | Given AB = 2 i + 10 j + 11 k and AD = - i + 2 j + 2 k , we have: AB AD = (2)(-1) + (10)(2) + (11)(2) = -2 + 20 + 22 = 40 | AB | = 2^2 + 10^2 + 11^2 = 4 + 100 + 121 = 225 = 15 | AD | = (-1)^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 = 3 = 40 15 3 = 40 45 = 8 9 Since > 0 , is an acute angle. The vector AD is rotated by an acute angle in the plane of the parallelogram to become AD' , which is perpendicular to AB . This means the angle between AB and AD' is 90^