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Question: If \(|\overrightarrow{A} \times \overrightarrow{B}| = \sqrt{3}\overrightarrow{A}.\overrightarrow{B},...

If ∣A→×B→∣=3A→.B→,|\overrightarrow{A} \times \overrightarrow{B}| = \sqrt{3}\overrightarrow{A}.\overrightarrow{B}, then the value of∣A→+B→∣|\overrightarrow{A} + \overrightarrow{B}| is

A

(A2+B2+AB3)1/2\left( A^{2} + B^{2} + \frac{AB}{\sqrt{3}} \right)^{1/2}

B

A+BA + B

C

(A2+B2+3AB)1/2(A^{2} + B^{2} + \sqrt{3}AB)^{1/2}

D

(A2+B2+AB)1/2(A^{2} + B^{2} + AB)^{1/2}

Answer

(A2+B2+AB)1/2(A^{2} + B^{2} + AB)^{1/2}

Explanation

Solution

∣A→×B→∣=3(A→.B→)|\overset{\rightarrow}{A} \times \overset{\rightarrow}{B}| = \sqrt{3}(\overset{\rightarrow}{A}.\overset{\rightarrow}{B})

ABsin⁡θ=3ABcos⁡θ⇒AB\sin\theta = \sqrt{3}AB\cos\theta \Rightarrow tan⁡θ=3\tan\theta = \sqrt{3}∴θ=60∘\theta = 60{^\circ}

Now ∣R→∣=∣A→+B→∣=A2+B2+2ABcos⁡θ|\overset{\rightarrow}{R}| = |\overset{\rightarrow}{A} + \overset{\rightarrow}{B}| = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}

=A2+B2+2AB(12)=(A2+B2+AB)1/2= \sqrt{A^{2} + B^{2} + 2AB\left( \frac{1}{2} \right)} = (A^{2} + B^{2} + AB)^{1/2}