Distance between point and hyperplane given basis

Question

I used the Gram-Schmidt process to find the orthonormal basis for some hyperplane, $V$, in $\mathbb{R}^4$. The vectors are $$ u = \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix}, v = \begin{bmatrix}1\\ 1\\ 1\\ 0\end{bmatrix}, w = \begin{bmatrix}1\\ 2\\ 0\\ 1\end{bmatrix} $$

I end up getting the basis $$ \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix}, \begin{bmatrix}0\\ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{bmatrix}, \begin{bmatrix}0\\ \frac{1}{\sqrt{3}}\\ \frac{-1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} \end{bmatrix} $$

The next part of the question is Calculate the distance from the point $p_0 = (1, 0, 1, 0)$ in $\mathbb{R}^4$ to $V$.

Since this is $\mathbb{R}^4$, I can no longer use the cross product, so I don't know where to go from here. Something about this tells me it should be rather easy.

Answer 1

First find all vectors $r$ orthogonal to $u,v,w.$ I found $r=r_2(0,1,-1,-2).$ Then consider the line orthogonal to $V$ passing through $p=(1,0,-1,0).$ Its points are parameterized by $r_2$ and have the form $p_{r_2}=(1,r_2,1-r_2,-2r_2).$ Next find $r_2=t$ such that $p_{t}$ belongs to $V$, that is find $a,b,c$ and $t$ such that $p_{t}=au+bv+cw.$ Solve this linear system with 4 equations and four unknown $a,b,c,t.$ Finally compute $\|p-p_t\|^2.$

Answer 2

Let $P$ be composed of the orthonormal vectors you found, i.e. \begin{equation*} P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{3} \\ 0 & 1/\sqrt{2} & -1/\sqrt{3} \\ 0 & 0 & 1/\sqrt{3} \end{pmatrix} \end{equation*} Then the projection matrix on V is given by $\operatorname{Proj}_V = PP^T$. The projection of $p_0$ on $V$ is then given by $p_{\parallel} = \operatorname{Proj}_V p_0$, then $\|p_0 - p_{\parallel}\|$ is the desired answer.

Motivation behind projection matrix:

The projection of $\vec{p}_0$ on $V$ is given by $\vec{p}_{\parallel} = (\vec{p}_0 \cdot \vec{v}_1)\vec{v}_1 + (\vec{p}_0 \cdot \vec{v}_2)\vec{v}_2 + (\vec{p}_0 \cdot \vec{v}_3)\vec{v}_3$, where $\vec{v}_i$ are the orthonormal vectors you found. You can verify that the $\operatorname{Proj}_V \vec{p}_0$ above gives exactly the same formula:

\begin{equation*} PP^T \vec{p}_0 = \begin{pmatrix} \vert & \vert & \vert \\ \vec{v}_1 & \vec{v}_2 & \vec{v}_3 \\ \vert & \vert & \vert \end{pmatrix} \begin{pmatrix} -\;\ \vec{v}_1\ - \\ -\;\ \vec{v}_2\ - \\ -\;\ \vec{v}_3\ - \\ \end{pmatrix} \vec{p}_0 = \begin{pmatrix} \vert & \vert & \vert \\ \vec{v}_1 & \vec{v}_2 & \vec{v}_3 \\ \vert & \vert & \vert \end{pmatrix} \begin{pmatrix} \vec{p}_0 \cdot \vec{v}_1 \\ \vec{p}_0 \cdot \vec{v}_2 \\ \vec{p}_0 \cdot \vec{v}_3 \\ \end{pmatrix} \end{equation*} Upon expanding gives exactly what we are looking for, i.e., $(\vec{p}_0 \cdot \vec{v}_1)\vec{v}_1 + (\vec{p}_0 \cdot \vec{v}_2)\vec{v}_2 + (\vec{p}_0 \cdot \vec{v}_3)\vec{v}_3$