3D glasses

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

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

If the system of equations $u + v = 5,\ u - v = 3$ has a solution in positive real numbers, then the vectors $\vec u = u\vec n,\ \vec v = v\vec n$ satisfy the equation, with $\vec n$ a unit vector; in that case $\vec u\cdot \vec v = uv$ .

So I solved the equations $u + v = 5,\ u - v = 3$ for real numbers and found $u = 4$ and $v = 1$ . Therefore the answer is $u\cdot v = 4$ .