Let \(H\) be the hyperplane in \(\RNrSpc{n}\) consisting of all solutions of the linear equation
\[a_1x_1 + \cdots + a_nx_n = c\]If \(\Vect{a}\DefEq (a_1,\dots ,a_n)\neq \Vect{0}\), then the distance from a point \(Q\), with position vector \(\Vect{q}=(q_1,\dots ,q_n)\), in \(\RNrSpc{n}\) from \(H\) is
\[\Dstnc{Q}{H} = \frac{\Abs{ \DotPr{\Vect{a}}{ \Vect{q} } - c}}{ \Norm{ \Vect{a} }}\]Let \(\Vect{p}\) be the position vector of an arbitrary point on \(H\). Then we know that
\[\DotPr{ \Vect{a} }{ \Vect{p} } = c\]Now already found that there is a unique point \(R\) on \(H\) which is closest to \(Q\). Moreover, \(R\) has position vector
\[\Vect{r} = \Vect{q}\ -\ \frac{ \DotPr{ \Vect{a} }{ (\Vect{q}-\Vect{p}) }}{ \DotPr{\Vect{a}}{\Vect{a}} } \, \cdot\, \Vect{a}\]Consequently,
| \(\Dstnc{Q}{H}\) | \(=\) | \(\Dstnc{Q}{R}\) |
| \(\) | \(=\) | \(\Norm{ \Vect{q} - \Vect{r} }\) |
| \(\) | \(=\) | \(\Norm{ \frac{ \DotPr{ \Vect{a} }{ (\Vect{q}-\Vect{p}) }}{ \DotPr{\Vect{a}}{\Vect{a}} } \, \cdot\, \Vect{a} }\) |
| \(\) | \(=\) | \(\frac{\Abs{ \DotPr{ \Vect{a} }{ (\Vect{q}-\Vect{p}) } }}{ \Norm{ \Vect{a} }^2 } \cdot \Norm{\Vect{a}}\) |
| \(\) | \(=\) | \(\frac{\Abs{ \DotPr{ \Vect{a} }{ (\Vect{q}-\Vect{p}) } }}{ \Norm{ \Vect{a} } }\) |
| \(\) | \(=\) | \(\frac{\Abs{ \DotPr{ \Vect{a} }{ \Vect{q} }\ -\ \DotPr{ \Vect{a} }{ \Vect{p} } }}{ \Norm{ \Vect{a} } }\) |
| \(\) | \(=\) | \(\frac{\Abs{ \DotPr{ \Vect{a} }{ \Vect{q} }\ -\ c }}{ \Norm{ \Vect{a} } }\) |
This is exactly what we wanted to show.