The magnitude of vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are 3,4 and 5 units respectively. If $\vec

Question

The magnitude of vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are 3,4 and 5 units respectively. If $\vec{A}+\vec{B}=\vec{C}$, the angle between $\vec{A}$ and $\vec{B}$ is

Professor PC Thomas Classes · Answer

(a) : Let $	heta$ be angle between $\vec{A}$ and $\vec{B}$. Given: $A=|\vec{A}|=3$ units, $B=|\vec{B}|=4$ units, $C=|\vec{C}|=5$ units\ $|\vec{A}+\vec{B}|=|\vec{C}| \Rightarrow A^{2}+2 A B \cos 	heta+B^{2}=C^{2}$\ $9+2 A B \cos 	heta+16=25$ or $2 A B \cos 	heta=0$ or $\cos 	heta=0 	herefore 	heta=90^{\circ}$.\