Answer
$u \cdot (v+w)=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w} $
Work Step by Step
Recall the dot product property for n-dimensional vectors, which states that if $\vec{u}=(u_1,u_2.....u_n)$ and $\vec{v}=(v_1,v_2.....v_n)$, then their dot product is:
$\vec{u}\cdot \vec{v}=u_1v_1+u_2v_2+........u_nv_n$
We find: $u \cdot (v+w)=(u_1+u_2) \cdot (v_1+w_1, v_2+w_2)\\=u_1v_1+u_1w_1+u_2v_2+u_2w_2 \\=(u_1v_1+u_2v_2)+(v_1w_1+v_2w_2)\\=(u_1, u_2)\cdot (v_1, v_2)+(u_1, u_2)\cdot (w_1, w_2)\\=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w} ~~~~~\bf{(Proved)}$
