Physics - Vectors Question with Solution | TestHub

Given $|\vec{A}\times \vec{B}|=\sqrt{3}(\vec{A}\cdot \vec{B})$

$AB\sin \theta =\sqrt{3}AB\cos \theta$

$\tan \theta =\sqrt{3}$

$\theta =60^{\circ }$

The magnitude of the resultant vector is given by:

$|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2AB\cos \theta }$

Substituting $\theta =60^{\circ }$ and $\cos 60^{\circ }=\frac{1}{2}$:

$|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2AB\left(\frac{1}{2}\right)}$

$|\vec{A}+\vec{B}|=(A^2+B^2+AB{)}^{\frac{1}{2}}$

Answer: $(A^2+B^2+AB{)}^{\frac{1}{2}}$