Let $\vec{u}$ and $\vec{v}$ be vectors in $\mathbb{R}^n$ .

If $|\vec{u}+\vec{v}| = 5$ and $|\vec{u}-\vec{v}| = 3$ , find $\vec{u} \cdot \vec{v}$ .

The answer is 4.

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$4~\color{#D61F06}{\vec{u} \cdot \vec{v}}=|\vec{u}+\vec{v}|^2-|\vec{u}-\vec{v}|^2=16$ $\implies \color{#D61F06}{\vec{u} \cdot \vec{v}}=\dfrac{16}{4}=\boxed 4$

$\begin{aligned}|\vec{u}+\vec{v}|^2=&|\vec{u}+\vec{v}|\cdot|\vec{u}+\vec{v}|\\&=|\vec u|^2+\vec{v}\cdot\vec{u}+\vec{u}\cdot\vec{v}+|\vec{v}|^2\\&=|\vec{v}|^2+|\vec{u}|^2+2~\color{#D61F06}{\vec{u} \cdot \vec{v}}\end{aligned}$ Similarly, $|\vec{u}-\vec{v}|^2=|\vec u|^2+|\vec v|^2-2~\vec u\cdot\vec v$ And hence, $4~\color{#D61F06}{\vec{u} \cdot \vec{v}}=|\vec{u}+\vec{v}|^2-|\vec{u}-\vec{v}|^2$