If $a+b+c=0,|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is

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Aicharya · Accepted Answer

Let $a+b+c=0 \Rightarrow(a+b)=-c$ $\Rightarrow(a+b)^2=c^2$ $ \begin{aligned} & \Rightarrow a^2+b^2+2 a \cdot b=c^2 \ & \Rightarrow 9+25+2.3 .5 \cos 	heta=49 \ & (\because|\vec{a}|=3,|\vec{b}|=5 	ext { and }|\vec{c}|=7) \ & 	herefore \cos 	heta=\frac{1}{2} \Rightarrow 	heta=\frac{\pi}{3} \end{aligned} $

If $a+b+c=0,|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is

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