NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

The shortest distance between the lines x - 2 2 = y - 3 2 = z - 0 1 and x + 4 - 1 = y - 7 8 = z - 5 4 lies in the interval

Options

A0,1
B1,2
C2,3
D3,4

Correct answer

C. 2,3

Step-by-step solution

Let, a → = 2 i ^ + 3 j ^ b → = - 4 i ^ + 7 j ^ + 5 k ^ ⇒ b → - a → = - 6 i ^ + 4 j ^ + 5 k ^ c → = 2 i ^ + 2 j ^ + k ^ d → = - i ^ + 8 j ^ + 4 k ^ ⇒ c → × d → = - 9 j ^ + 18 k ^ Hence, shortest distance = b → - a → ⋅ c → × d → c → × d → = - 36 + 90 9 5 = 6 5 ∈ 2,3

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