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AIEEE 2012

JEE Main 2012 Mathematics question on Vector Algebra

If a⃗\vec{a} and b⃗\vec{b} are vectors such that ∣a⃗+b⃗∣=29|\vec{a} + \vec{b}| = \sqrt{29} and a⃗×(2i^+3j^+4k^)=(2i^+3j^+4k^)×b⃗\vec{a} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{b}, then a possible value of (a⃗+b⃗)⋅(−7i^+2j^+3k^)(\vec{a} + \vec{b}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) is
  1. A.
    0
  2. B.
    3
  3. C.
    4
  4. D.
    8
Show answer and solution

Correct option: C

Solution

The condition a × (2i + 3j + 4k) = (2i + 3j + 4k) × b can be rewritten as a × (2i + 3j + 4k) + b × (2i + 3j + 4k) = 0, i.e., (a + b) × (2i + 3j + 4k) = 0. This means (a + b) is parallel to (2i + 3j + 4k).

So a + b = λ(2i + 3j + 4k) for some scalar λ. Since |a + b| = √29 and |2i + 3j + 4k| = √(4+9+16) = √29, we get λ = ±1.

Then (a + b)·(-7i + 2j + 3k) = λ(2i + 3j + 4k)·(-7i + 2j + 3k) = λ(-14 + 6 + 12) = λ(4).

For λ = 1, the value is 4.

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