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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Vector Algebra

Let α=(λ−2)a+b\alpha = (\lambda - 2)\mathbf{a} + \mathbf{b} and β=(4λ−2)a+3b\beta = (4\lambda - 2)\mathbf{a} + 3\mathbf{b} be two given vectors where vectors a\mathbf{a} and b\mathbf{b} are non-collinear. The value of λ\lambda for which vectors α\alpha and β\beta are collinear, is
  1. A.
    4
  2. B.
    −3-3
  3. C.
    3
  4. D.
    −4-4
Show answer and solution

Correct option: D

Solution

For α and β to be collinear, β = kα for some scalar k.

So (4λ-2)a + 3b = k[(λ-2)a + b]

Since a and b are non-collinear, equate coefficients:
- Coefficient of b: 3 = k → k = 3
- Coefficient of a: 4λ-2 = k(λ-2) = 3(λ-2) = 3λ-6

Solving: 4λ-2 = 3λ-6 → λ = -4.

[JEE Main 2019 · 10 Jan · Shift II]

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