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Question:

Let A be a point on the line r = (1+μ)i + (μ)j + (2+5μ)k and B(3,2,6) be a point in the space. Then the value of μ for which the vector AB is parallel to the plane x-y+3z=1 is?

12

-4

14

18

Solution:

Let point A be (1+μ)i + (μ)j + (2+5μ)k and point B be (3,2,6) then AB = (2+3μ)i + (2-μ)j + (4-5μ)k which is parallel to the plane x-y+3z=1
∴ 2+3μ - (2-μ) + 3(4-5μ) = 0
8μ = -12
μ = -14