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If $|\vec{a}|=1,|\vec{b}|=2,|\vec{a}-\vec{b}|^2+|\vec{a}+2 \vec{b}|^2=20$ then $(\vec{a}, \vec{b})=$
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The correct answer is:
$\frac {2\pi}{3}$
$|\vec{a}-\vec{b}|^2+|\vec{a}+2 \vec{b}|^2=20$
$\begin{aligned} & \Rightarrow \quad(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})+(\vec{a}+2 \vec{b}) \cdot(\vec{a}+2 \vec{b})=20 \\ & \Rightarrow \quad|\vec{a}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2+|\vec{a}|^2+4 \vec{a} \cdot \vec{b}+4|\vec{b}|^2=20 \\ & \Rightarrow \quad 2|\vec{a}|^2+5|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=20 \\ & \Rightarrow \quad 2(1)^2+5(2)^2+2(1)(2) \cos \theta=20 \\ & \Rightarrow \quad 22+4 \cos \theta=20 \Rightarrow \cos \theta=\frac{-1}{2} \\ & \therefore \quad \theta=\frac{2 \pi}{3} .\end{aligned}$
$\begin{aligned} & \Rightarrow \quad(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})+(\vec{a}+2 \vec{b}) \cdot(\vec{a}+2 \vec{b})=20 \\ & \Rightarrow \quad|\vec{a}|^2-2 \vec{a} \cdot \vec{b}+|\vec{b}|^2+|\vec{a}|^2+4 \vec{a} \cdot \vec{b}+4|\vec{b}|^2=20 \\ & \Rightarrow \quad 2|\vec{a}|^2+5|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=20 \\ & \Rightarrow \quad 2(1)^2+5(2)^2+2(1)(2) \cos \theta=20 \\ & \Rightarrow \quad 22+4 \cos \theta=20 \Rightarrow \cos \theta=\frac{-1}{2} \\ & \therefore \quad \theta=\frac{2 \pi}{3} .\end{aligned}$
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