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NDA 1 2025 Maths, question 69

Topic: Vectors

Q69

The position vectors of three points AA, BB and CC are a→\overrightarrow{a}, b→\overrightarrow{b} and c→\overrightarrow{c} respectively, where

c→=(cos⁡2θ)a→+(sin⁡2θ)b→\overrightarrow{c} = \left( \cos^{2}\theta \right)\overrightarrow{a} + \left( \sin^{2}\theta \right)\overrightarrow{b}. What is (a→×b→)+(b→×c→)+(c→×a→)\left( \overrightarrow{a} \times \overrightarrow{b} \right) + \left( \overrightarrow{b} \times \overrightarrow{c} \right) + \left( \overrightarrow{c} \times \overrightarrow{a} \right) equal to?

  1. (a) 0→\overrightarrow{0}Correct answer
  2. (b) 2c→2\overrightarrow{c}
  3. (c) 3c→3\overrightarrow{c}
  4. (d) Unit vector

Solution.

Answer: (a) 0→\overrightarrow{0}

  1. Put u=cos⁡2θu = \cos^{2}\theta and v=sin⁡2θv = \sin^{2}\theta, so u+v=1u + v = 1.
  2. Then b→×c→=−u(a→×b→)\overrightarrow{b} \times \overrightarrow{c} = - u\left( \overrightarrow{a} \times \overrightarrow{b} \right) and c→×a→=−v(a→×b→)\overrightarrow{c} \times \overrightarrow{a} = - v\left( \overrightarrow{a} \times \overrightarrow{b} \right).
  3. The required sum is (1−u−v)(a→×b→)=0→(1 - u - v)\left( \overrightarrow{a} \times \overrightarrow{b} \right) = \overrightarrow{0}.