If the magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}}$, then

Question

If the magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}}$, then the angle between $\vec{A}$ and $\vec{B}$ will be

Professor PC Thomas Classes · Answer

(a) : We know $|\vec{A}+\vec{B}| = \sqrt{A^{2}+B^{2}+2AB\cos	heta}$. \ Given $|\vec{A}+\vec{B}| = \sqrt{A^{2}+B^{2}}$, this means $2AB\cos	heta = 0$. \ Since $A$ and $B$ are non-zero magnitudes, $\cos	heta = 0 \implies 	heta = 90^{\circ}$.