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JEE MainMathematicsVector Algebra

Let a = i + j - k and b = 2 i - j + k . A vector c satisfies c a = b a and c a = 3 . Let d = i + j + k and e = 2 i - 2 j + k . If the projection of the vector c d on the vector e is 11 , then the value of | c |^2 is equal to

Options

  1. A18
  2. B33
  3. C54
  4. D81

Correct answer

C. 54

Step-by-step solution

Given c a = b a ( c - b ) a = 0 This implies that c - b is parallel to a , so c = b + t a for some scalar t . Taking the dot product with a on both sides: c a = b a + t| a |^2 We calculate b a = (2)(1) + (-1)(1) + (1)(-1) = 2 - 1 - 1 = 0 . Also, | a |^2 = 1^2 + 1^2 + (-1)^2 = 3 . Substituting these values into the dot product equation: 3 = 0 + 3t t = Thus, c = b + a = (2 + ) i + ( - 1) j + (1 - ) k . Now, we find c d : c d = vmatrix i & j & k 2+ & -1 & 1- 1 & 1 & 1 vmatrix c d = i ( - 1 - (1 - )) - j (2 + - (1 - ))

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