AIEEE 2004

JEE Main 2004 Mathematics question on Vector Algebra

The unit vector which is orthogonal to the vector 3i^+2j^+6k^3\hat{i} + 2\hat{j} + 6\hat{k} and is coplanar with the vectors 2i^+j^+k^2\hat{i} + \hat{j} + \hat{k} and i^−j^+k^\hat{i} - \hat{j} + \hat{k} is
  1. A.
    2i^−6j^+k^41\dfrac{2\hat{i} - 6\hat{j} + \hat{k}}{\sqrt{41}}
  2. B.
    2i^−3j^13\dfrac{2\hat{i} - 3\hat{j}}{\sqrt{13}}
  3. C.
    3j^−k^10\dfrac{3\hat{j} - \hat{k}}{\sqrt{10}}
  4. D.
    4i^+3j^−3k^34\dfrac{4\hat{i} + 3\hat{j} - 3\hat{k}}{\sqrt{34}}
Show answer and solution

Correct option: C

Solution

A vector coplanar with a = 2î + ĵ + k̂ and b = î - ĵ + k̂ can be written as v = λa + μb = (2λ+μ)î + (λ-μ)ĵ + (λ+μ)k̂.

For orthogonality with n = 3î + 2ĵ + 6k̂:
v·n = 3(2λ+μ) + 2(λ-μ) + 6(λ+μ) = 6λ+3μ + 2λ-2μ + 6λ+6μ = 14λ + 7μ = 0, giving μ = -2λ.

Substituting μ = -2λ: v = (2λ-2λ)î + (λ+2λ)ĵ + (λ-2λ)k̂ = 0î + 3λĵ - λk̂ = λ(3ĵ - k̂).

The magnitude is |λ|√(9+1) = |λ|√10, so the unit vector is (3ĵ - k̂)/√10.

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