The magnitude of vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are 5, 12 and 13 units respectively. If $\

Question

The magnitude of vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are 5, 12 and 13 units respectively. If $\vec{A}+\vec{B}=\vec{C}$, the angle between $\vec{A}$ and $\vec{B}$ is

Professor PC Thomas Classes · Answer

(a) : Given $A=5, B=12, C=13$. \ We know $|\vec{A}+\vec{B}| = |\vec{C}| \implies A^2 + B^2 + 2AB\cos	heta = C^2$. \ Substituting the values: $5^2 + 12^2 + 2(5)(12)\cos	heta = 13^2 \implies 25 + 144 + 120\cos	heta = 169$. \ $169 + 120\cos	heta = 169 \implies 120\cos	heta = 0 \implies 	heta = 90^{\circ} = \pi/2$.