The two lines x = ay + b , z = cy + d and x = a'y + b' , z = c'y + d' will be perpendicular, if and only if: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The two lines x = a y + b, z = cy + d and x = a^{'} y + b^{'}, z = c^{'} y + d^{'} will be perpendicular, if and only if:

Solution & Step-by-step Explanation

Rewrite the lines in symmetric form: Line 1: \frac{x - b}{a} = \frac{y - 0}{1} = \frac{z - d}{c} . Direction vector v_{1} = (a, 1, c) . Line 2: \frac{x - b ^{'}}{a ^{'}} = \frac{y - 0}{1} = \frac{z - d ^{'}}{c ^{'}} . Direction vector v_{2} = (a^{'}, 1, c^{'}) . For the lines to be perpendicular, v_{1} \cdot v_{2} = 0 : a a^{'} + (1) (1) + c c^{'} = 0 ⟹ a a^{'} + c c^{'} + 1 = 0

Try it yourself before checking the explanation above.

The two lines x = a y + b, z = cy + d and x = a^{'} y + b^{'}, z = c^{'} y + d^{'} will be perpendicular, if and only if:

a a^{'} + b b^{'} + c c^{'} + 1 = 0

a a^{'} + b b^{'} + c c^{'} = 0

(a + a^{'}) (b + b^{'}) + (c + c^{'}) = 0

a a^{'} + c c^{'} + 1 = 0

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