The [[dot product]] is $\boldsymbol v \cdot \boldsymbol w=0$ when $\boldsymbol v$ is perpendicular to $\boldsymbol w$. #### Proof In the case that 2 vectors $\boldsymbol v$ and $\boldsymbol w$ are perpendicular, their lengths, along with the length of their difference, would be the legs and hypotenuse repectively of a right triangle. By [[Pythagorean Theorem]] $ \begin{aligned} \|\boldsymbol{v}\|^{2}&+\|\boldsymbol{w}\|^{2}&&=&&\|\boldsymbol{v}-\boldsymbol{w}\|^{2}\\ \left({v_1}^{2}+{v_2}^{2}\right)&+\left({w_1}^{2}+{w_2}^{2}\right)&&=&&\left({v_1}-{w_1}\right)^{2}+\left(v_{2}-w_{2}\right)^{2}\\ \left({v_1}^{2}+{v_2}^{2}\right)&+\left({w_1}^{2}+{w_2}^{2}\right)&&=&&({v_1}^2-2v_1 w_1+{w_1}^2)+({v_2}^2-2v_2 w_2+{w_2}^2)\\ \left(\cancel{{v_1}^{2}}+\cancel{{v_2}^{2}}\right)&+\left(\cancel{{w_1}^{2}}+\cancel{{w_2}^{2}}\right)&&=&&(\cancel{{v_1}^2}-2v_1 w_1+\cancel{{w_1}^2})+(\cancel{{v_2}^2}-2v_2 w_2+\cancel{{w_2}^2})\\ &\ 0&&=&&-2v_1 w_1-2v_2 w_2\\ 2v_1 w_1&+2v_2 w_2 &&=&&0\\ v_1 w_1&+v_2 w_2 &&=&&0\\ \boldsymbol v &\ \cdot\ \boldsymbol w &&=&&0\\ \end{aligned} $ The same process can be applied for $n$ dimensional vectors, just canceling $2n$ quadratic terms on both sides.