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Question
If \[\vec{a} + \vec{b} + \vec{c} = \vec{0} ,\] show that the angle θ between the vectors \[\vec{b} \text{ and } \vec{c}\] is given by \[\frac{\left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 - \left| \vec{c} \right|^2}{2\left| \vec{b} \right| \left| \vec{c} \right|} .\]
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Solution
\[\text{ Given },\]
\[ \vec{a} + \vec{b} + \vec{c} = 0\]
\[ \Rightarrow \vec{b} + \vec{c} = - \vec{a} \]
\[ \Rightarrow \left| \vec{b} + \vec{c} \right|^2 = \left| - \vec{a} \right|^2 \]
\[ \Rightarrow \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2 + 2 \vec{b} . \vec{c} = \left| \vec{a} \right|^2 \]
\[ \Rightarrow 2 \vec{b} . \vec{c} = \left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 - \left| \vec{c} \right|^2 \]
\[ \Rightarrow 2 \left| \vec{b} \right| \left| \vec{c} \right| \cos \theta = \left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 - \left| \vec{c} \right|^2 \]
\[ \therefore \cos \theta = \frac{\left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 - \left| \vec{c} \right|^2}{2 \left| \vec{b} \right| \left| \vec{c} \right|}\]
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