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Question:

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 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

89

4√59

√179

19

Solution:

→AB ⋅ →AD > 0
So, ′θ′ is acute
cosθ = →AB ⋅ →AD / |→AB||→AD| = 89
cos(α + θ) = 0 ⇒ cosα ⋅ cosθ - sinα ⋅ sinθ = 0
8cosα = √17sinα
64cos²α = 17(1 - cos²α)
cos²α = 17/81
cosα = √179