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Let p ^ × q → × p ^ + p ^ ⋅ q → q → = x 2 + y 2 q → + 14 - 4 x - 6 y p ^ , where p ^ and q → are non-collinear vectors ( p ^ is a unit vector) and x ,   y are scalars, then the value of x 2 + y 2 is equal to

Options

  1. A10
  2. B11
  3. C12
  4. D13

Correct answer

D. 13

Step-by-step solution

p ^ × q → × p ^ + p ^ ⋅ q → q → = x 2 + y 2 q → + 14 - 4 x - 6 y p ^ p ^ 2 q → - p ^ ⋅ q → p ^ + p ^ ⋅ q → q → = x 2 + y 2 q → + 14 - 4 x - 6 y p ^ Comparing both sides, we get, 14 - 4 x - 6 y = - p ^ ⋅ q → x 2 + y 2 = 1 + p ^ ⋅ q → x 2 + y 2 - 4 x - 6 y + 14 = 1 x 2 + y 2 - 4 x - 6 y + 13 = 0 x - 2 2 + y - 3 2 = 0 ⇒ x = 2 ,   y = 3

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