Mathematics in Physics

JEE Main level, single correct, medium

28 of 121

Question 28 of 121

If A⃗\vec{A} and B⃗\vec{B} are two vectors satisfying the relation A⃗⋅B⃗=∣A⃗×B⃗∣\vec{A}\cdot \vec{B}=|\vec{A}\times \vec{B}|. Then the value of ∣A⃗−B⃗∣|\vec{A}-\vec{B}| will be:

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Answer: (D) A2+B2−2AB\sqrt{{A}^{2}+{B}^{2}-\sqrt{2}AB}

A⃗⋅B⃗=∣A⃗×B⃗∣\vec{A}\cdot \vec{B}=|\vec{A}\times \vec{B}|
ABcosθ=ABsinθ⇒θ=45∘ABcos\theta =ABsin\theta \Rightarrow \theta =45^{\circ}
∣A⃗−B⃗∣=A2+B2−2ABcos⁡45∘|\vec{A}-\vec{B}|=\sqrt{{A}^{2}+{B}^{2}-2AB\cos 45^{\circ}}
=A2+B2−2AB=\sqrt{{A}^{2}+{B}^{2}-\sqrt{2}AB}
Hence option (4).(4).

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