Parallelogram Law

Vector u
\(u_1\) = 3.20
\(u_2\) = 0.40
\(u_3\) = 0.00

Vector v
\(v_1\) = 0.70
\(v_2\) = 1.66
\(v_3\) = 0.00
Rotate the graph and you can watch the parallelogram with sides constructed by $u$ and $v$ take shape in three dimensions, with the sum $w = u + v$ drawing the diagonal (and $u-v$ drawing the other one). This is completely analogous to the two-dimensional picture, just embedded in 3-D, showing this image exists in higher dimensions. \begin{remark} This geometric picture has a clean algebraic counterpart, written entirely in terms of norms. Herein, we prove a geometric fact completely algebraically. The diagonals of the parallelogram are $u+v$ and $u-v$, and adding their squared lengths gives the \textbf{parallelogram law} $$\|u+v\|^2 + \|u-v\|^2 = 2\|u\|^2 + 2\|v\|^2.$$ This holds in $\R^n$ for any $n$, and the cross terms $\langle u,v\rangle$ and $-\langle u,v\rangle$ appearing when you expand $\|u+v\|^2$ and $\|u-v\|^2$ cancel exactly. \end{remark}