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NDA 2 2024 Maths, question 51

Topic: Vectors

Q51

What is 3α+2β3\alpha+2\beta equal to if (2i^+6j^+27k^)×(i^+αj^+βk^)(2\hat{i}+6\hat{j}+27\hat{k})\times (\hat{i}+\alpha\hat{j}+\beta\hat{k}) is a null vector?

  1. (a) 3636Correct answer
  2. (b) 3333
  3. (c) 3030
  4. (d) 2727

Solution.

Answer: (a) 3636

  1. A zero cross product means the two nonzero vectors are parallel.
  2. Their first components give multiplier 2, so 2α=6 and 2β=27.
  3. Thus α=3, β=27/2, and 3α+2β=9+27=36.