Here we learn how to determine the distance of a point from a hyperplane. This is a very practical problem which one can encounter in many different guises. Later, it will also enable us to carry out ‘least squares’ approximations.
So, how do solve the problem of finding the distance of a point \(Q\) from a hyperplane \(H\)? – We do this in two steps: first we find the unique point \(R\) on \(H\) which is closest to \(Q\). This point \(R\) is the only point on \(H\) for which the arrow \(\Arrow{R}{Q}\) is perpendicular to \(H\). Then we take the distance from \(Q\) to \(R\) to be the distance from \(Q\) to \(H\).
To compute the distance of a point \(Q\) from a hyperplane \(H\), we first find the point \(R\) on \(H\) closest to \(Q\).
Let \(H\) be the hyperplane in \(\RNrSpc{n}\) consisting of all \(\Vect{x}=(x_1,\dots ,x_n)\) with
\[a_1x_1+ \cdots + a_nx_n = c,\qquad \Vect{a} = (a_1,\dots ,a_n)\neq (0,\dots ,0)\]Given an arbitrary point \(Q\) in \(\RNrSpc{n}\), there is a unique point \(R\) on \(H\) which is closest to \(Q\). It has the position vector
\[\Vect{r} = \Vect{q}\ -\ \frac{ \DotPr{ \Vect{a} }{ \Vect{q} }\ -\ c }{ \DotPr{\Vect{a}}{\Vect{a}} } \, \cdot\, \Vect{a}\]Then we take the distance from \(Q\) to \(R\) as the distance from \(Q\) to the hyperplane.
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
\[\Dstnc{Q}{H} = \frac{\Abs{ \DotPr{\Vect{a}}{ \Vect{q} } - c}}{ \Norm{ \Vect{a} }}\]