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JEE Main202430 Jan 2024Evening ShiftMathematicsThree Dimensional GeometryActual

Let L 1 : r → = i ^ - j ^ + 2 k ^ + λ i ^ - j ^ + 2 k ^ , λ ∈ R , L 2 : r → = j ^ - k ^ + μ 3 i ^ + j ^ + p k ^ , μ ∈ R and L 3 : r → = δ ( l i ^ + m j ^ + n k ^ ) , δ ∈ R be three lines such that L 1 is perpendicular to L 2 and L 3 is perpendicular to both L 1 and L 2 . Then the point which lies on L 3 is

Options

  1. A( - 1 , 7 , 4 )
  2. B( - 1 , - 7 , 4 )
  3. C( 1 , 7 , - 4 )
  4. D( 1 , - 7 , 4 )

Correct answer

A. ( - 1 , 7 , 4 )

Step-by-step solution

Given: L 1 ⊥ L 2 ⇒ i ^ - j ^ + 2 k ^ . 3 i ^ + j ^ + p k ^ = 0 ⇒ 3 - 1 + 2 p = 0 ⇒ p = - 1 Also, L 3 ⊥ L 1 , L 2 So, L 3 ∥ L 1 × L 2 ⇒ L 1 × L 2 = i ^ j ^ k ^ 1 - 1 2 3 1 - 1 ⇒ L 1 × L 2 = - i ^ + 7 j ^ + 4 k ^ On comparing with L 3 : r → = δ ( l i ^ + m j ^ + n k ^ ) , we get that ( - δ , 7 δ , 4 δ ) will lie on L 3 . Now, for δ = 1 the point will be ( - 1 , 7 , 4 ) .

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