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AIEEE 2010

JEE Main 2010 Mathematics question on Vector Algebra

Two adjacent sides of a parallelogram ABCDABCD are given by AB⃗=2i^+10j^+11k^\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k} and AD⃗=−i^+2j^+2k^\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}. The side ADAD is rotated by an acute angle α\alpha in the plane of the parallelogram so that ADAD becomes AD′AD'. If AD′AD' makes a right angle with the side ABAB, then the cosine of the angle α\alpha is given by
  1. A.
    89\dfrac{8}{9}
  2. B.
    179\dfrac{\sqrt{17}}{9}
  3. C.
    19\dfrac{1}{9}
  4. D.
    459\dfrac{4\sqrt{5}}{9}
Show answer and solution

Correct option: B

Solution

|AB| = √(4+100+121) = 15, |AD| = √(1+4+4) = 3.

AB·AD = -2+20+22 = 40, so cos(angle DAB) = 40/(15·3) = 8/9.

AD' is in the plane of the parallelogram with |AD'| = |AD| = 3, and AD' ⊥ AB. The angle between AD and AD' is α, so cos α = cos(angle between old and new position).

Since AD' ⊥ AB, the angle between AD' and AB is 90°. The original angle between AD and AB has cos = 8/9, so angle DAB = arccos(8/9). Thus α = 90° - angle DAB, giving cos α = sin(angle DAB) = √(1 - 64/81) = √(17/81) = √17/9.

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