Physics - Vectors Question with Solution | TestHub

The magnitude of the resultant is $|\vec{R}|=\sqrt{A^2+B^2+2AB\cos \theta }=\sqrt{3^2+4^2+2(3)(4)\cos 60^{\circ }}=\sqrt{37}$. The angle with $\vec{A}$ is $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }=\frac{4\sin 60^{\circ }}{3+4\cos 60^{\circ }}=\frac{2\sqrt{3}}{5}$. The magnitude of the difference is $|\vec{A}-\vec{B}|=\sqrt{A^2+B^2-2AB\cos \theta }=\sqrt{13}$. The magnitude of the cross product is $|\vec{A}\times \vec{B}|=AB\sin \theta =(3)(4)\sin 60^{\circ }=6\sqrt{3}$.