Answer
The proof is below.
Work Step by Step
To check the distributive property of a vector:
u=$\left[ \begin{array}{ c c } u_{1} \\ u_{2}\\.\\. \\u_{n}\end{array} \right]$ v=$\left[ \begin{array}{ c c } v_{1} \\ v_{2}\\.\\. \\v_{n}\end{array} \right]$ w=$\left[ \begin{array}{ c c } w_{1} \\ w_{2}\\.\\. \\w_{n}\end{array} \right]$
We must show:
$u\cdot (v+w)=u\cdot v+u\cdot w$
so $u.(v+w)=\left[ \begin{array}{ c c } u_{1} \\ u_{2}\\.\\. \\u_{n}\end{array} \right].(\left[ \begin{array}{ c c } v_{1} \\ v_{2}\\.\\. \\v_{n}\end{array} \right]+\left[ \begin{array}{ c c } w_{1} \\ w_{2}\\.\\. \\w_{n}\end{array} \right])$
$u\cdot (v+w)=u_{1}(v_{1}+w_{1})+u_{2}(v_{2}+w_{2}).....+u_{n}(v_{n}+w_{n})$
$u\cdot (v+w)=u_{1}v_{1}+u_{1}w_{1}+u_{2}v_{2}+u_{2}w_{2}.......+u_{n}v_{n}+u_{n}w_{n}$
$u\cdot(v+w)=(u_{1}v_{1}+u_{2}v_{2}.....+u_{n}v_{n})+(u_{1}w_{1}+u_{2}w_{2}......+u_{n}w_{n})$
Thus:
$u\cdot (v+w)=u\cdot v+u\cdot w$
