Question

Two adjacent sides of a parallelogram ABCD are given by AB→=2i^+10j^+11k^ and AD→=-i^+2j^+2k^.The side AD is rotated by an acute angel 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. (A) 89
• B. (B) 179
• C. (C) 19
• D. (D) 459

Answer 1

Answer: (B)

(B) 179

Answer 2

Explanation:
Given: Two adjacent sides of a parallelogram ABCD areAB→=2i^+10j^+11k^ and AD→=-i^+2j^+2k^ .....(i)We have to find cosine of the acute angel α through which side AD of parallelogram is rotated to AD' such that AD' becomes perpendicular to AB. Let θ be the angle between AB→,AD→ and AD' is perpendicular to AB.Then, α + θ = ππ2[Clear from the above diagram] ⇒ α = π2-θUsing the definition of dot product , we havecos⁡θ=|AB→⋅AD→|AB→||AD→||=∣(2i^+10j^+11k^)⋅(−i^+2j^+2k^)(2)2+(10)2+(11)2(−1)2+(2)2+(2)2[ using (i) ]=|−2+20+222259∣=|4015∗3∣=|89|=89… (iii)Using trigonometric identities, we havesin⁡θ=1−cos2⁡θ=1−6481=1781[ using (iii)]=179Taking cosine on both sides of (ii), we getcos⁡a=cos⁡(π2−θ)=sin⁡θ=179Hence, the correct option is (B).