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1983

JEE Main 1983 Mathematics question on Vector Algebra

The points with position vectors 60i^+3j^60\hat{i} + 3\hat{j}, 40i^−8j^40\hat{i} - 8\hat{j}, ai^−52j^a\hat{i} - 52\hat{j} are collinear, if
  1. A.
    a=−40a = -40
  2. B.
    a=40a = 40
  3. C.
    a=20a = 20
  4. D.
    None of these
Show answer and solution

Correct option: A

Solution

For three points A, B, C to be collinear, vector AB and AC must be parallel.

AB = (40-60)î + (-8-3)ĵ = -20î - 11ĵ
AC = (a-60)î + (-52-3)ĵ = (a-60)î - 55ĵ

For collinearity: (-20)/( a-60) = (-11)/(-55)

-11/(-55) = 1/5, so:
-20/(a-60) = 1/5
a - 60 = -100
a = -40

[1983]

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