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MHT CET 2026 (15 May, Shift 1) · Vector (Cross) Product

Vector (Cross) Product question from MHT CET 2026 (15 May, Shift 1)

Q2MHT CET 2026 · 15 MAY · SHIFT 1Medium

Vector (Cross) Product

If ∣A⃗×B⃗∣=3(A⃗⋅B⃗)|\vec{A} \times \vec{B}| = \sqrt{3}(\vec{A} \cdot \vec{B}) then the value of ∣A⃗+B⃗∣|\vec{A} + \vec{B}| is (tan⁡60∘=3cos⁡60∘=12\tan 60^\circ = \sqrt{3} \quad \cos 60^\circ = \frac{1}{2})

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Answer: (d)

  1. ABsin⁡θ=3 ABcos⁡θ⇒tan⁡θ=3⇒θ=60∘AB\sin\theta = \sqrt{3}\,AB\cos\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ.
  2. ∣A⃗+B⃗∣2=A2+B2+2ABcos⁡60∘=A2+B2+AB|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos 60^\circ = A^2 + B^2 + AB
  3. So the answer is (d) (A2+B2+AB)1/2(A^2 + B^2 + AB)^{1/2}.
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