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99 Percentile Qs Bank for JEE MainMathematicsVector Algebra

A unit vector perpendicular to the plane of a → = 2 i ^ - 6 j ^ - 3 k ^ and b → = 4 i ^ + 3 j ^ - k ^ is

Options

  1. A4 i ^ + 3 j ^ - k ^ 26
  2. B2 i ^ - 6 j ^ - 3 k ^ 7
  3. C3 i ^ - 2 j ^ + 6 k ^ 7
  4. D2 i ^ - 3 j ^ - 6 k ^ 7

Correct answer

C. 3 i ^ - 2 j ^ + 6 k ^ 7

Step-by-step solution

A unit perpendicular to the plane a → and b → = a → × b → a → × b → Now, a → × b → = i ^ j ^ k ^ 2 - 6 - 3 4 3 - 1 = i ^ 6 + 9 - j ^ - 2 + 12 + k ^ 6 + 24 = 15 i ^ - 10 j ^ + 30 k ^ and a → × b → = 15 2 + - 10 2 + 30 2 = 1225 = 35 ∴ Required vector = 15 i ^ - 10 j ^ + 30 k ^ 35 = 3 i ^ - 2 j ^ + 6 k ^ 7

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