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JEE Main 2023 · 25 Jan · Shift 2

JEE Main 2023 Mathematics question on Vector Algebra

Let a⃗=−i^−j^+k^\vec{a} = -\hat{i} - \hat{j} + \hat{k}, a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1 and a⃗×b⃗=i^−j^\vec{a} \times \vec{b} = \hat{i} - \hat{j}.

Then a⃗−6b⃗\vec{a} - 6\vec{b} is equal to
  1. A.
    3(i^+j^−k^)3(\hat{i} + \hat{j} - \hat{k})
  2. B.
    3(i^−j^−k^)3(\hat{i} - \hat{j} - \hat{k})
  3. C.
    3(i^+j^+k^)3(\hat{i} + \hat{j} + \hat{k})
  4. D.
    3(i^−j^+k^)3(\hat{i} - \hat{j} + \hat{k})
Show answer and solution

Correct option: C

Solution

We need to find b given a·b = 1 and a×b = î - ĵ.

Let b = xî + yĵ + zk̂, with a = -î - ĵ + k̂.

From a·b = 1: -x - y + z = 1.

From a×b = î - ĵ + 0k̂:
a×b = |î ĵ k̂; -1 -1 1; x y z| = (-z - y)î - (-z - x)ĵ + (-y + x)k̂

So: -z - y = 1, z + x = -1, x - y = 0.

From x - y = 0: x = y. From -z - y = 1: z = -1 - y. From z + x = -1: (-1 - y) + y = -1 ✓ (consistent).

From a·b: -x - y + z = 1 → -2y + (-1 - y) = 1 → -3y = 2 → y = -2/3.

So x = -2/3, z = -1 + 2/3 = -1/3.

Thus b = (-2/3)î + (-2/3)ĵ + (-1/3)k̂.

Then 6b = -4î - 4ĵ - 2k̂.

a - 6b = (-1+4)î + (-1+4)ĵ + (1+2)k̂ = 3î + 3

[JEE Main 2023 · 25 Jan · Shift 2]

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