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Question

If three points A, B and C have position vectors i^+xj^+3k^, 3i^+4j^+7k^ and yi^-2j^-5k^ respectively are collinear, then (x, y) =
(a) (2, −3)
(b) (−2, 3)
(c) (−2, −3)
(d) (2, 3)

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Solution

(a) (2, −3)
Given position vectors of A, B and C are i^+xj^+3k^, 3i^+4j^+7k^ and yi^-2j^-5k^. Then,
AB= 3i^+4j^+7k^-i^-xj^-3k^ = 2i^+4-xj^+4k^BC = yi^-2j^-5k^-3i^-4j^-7k^ = y-3i^-6j^-12k^
Since, the given vectors are collinear.
AB=λBC2i^ +4-xj^+ 4k^ = λ y-3i^ -6λ j^-12λ k^2=λ y-3 , 4-x = -6λ, 4=-12λ 2=λ y-3 , 4-x = -6λ, λ=-132=-13y-3 , 4-x = -6×-13-6=y-3 , 4-x = 2 y=-3 , x= 2

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