Question

Two adjacent sides of a parallelogram ABCD are given by vector AB = 2i + 10j + 11k  and vector AD = - i + 2j + 2k. The side AD is rotated by an acute angle α 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 α is given by

(A) 8/9

(B) √17/9

(C) 1/9

(D) 4√5/9

Answer 1

Correct option (B) \(\frac{\sqrt{17}}{9}\)

Explanation:

sin θ = \(\frac{\sqrt{81-64}}{9}\) = \(\frac{\sqrt{17}}{9}\)

θ + α = 90°

α = 90° - θ

cos α = cos (90° - θ)

= sin θ

= \(\frac{\sqrt{17}}{9}\)