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3.3 Orthogonality: Class Notes

Definition

We write $\overrightarrow{u}⟂\overrightarrow{v}$ for “$\overrightarrow{u}$ is perpendicular to $\overrightarrow{v}$.”

Definition

A line $L$ through $(x_0,y_0)$ with normal vector $(a,b)$. For any other point $(x,y)$ on $L$, the vector $(x-x_0,y-y_0)$ lies along the line, so it is perpendicular to the normal.

$$(a,b)·(x-x_0,y-y_0)=0\Rightarrow a(x-x_0)+b(y-y_0)=0$$

#4

Find a point-normal form of the equation of the plane passing through $P$ and having $𝐧$ as a normal.

$P(1,1,4)$, $𝐧=(1,9,8)$

Solution

A plane through $P(1,1,4)$ with normal $\overrightarrow{n}=(1,9,8)$ drawn from $P$. The vector $\overrightarrow{v}$ goes from $P$ to any point $(x,y,z)$ in the plane.

$\overrightarrow{v}=(x-1,y-1,z-4)$ is a vector in the plane for a point $(x,y,z)$ in the plane. Then

$$\overrightarrow{n}⟂\overrightarrow{v}\Rightarrow \overrightarrow{n}·\overrightarrow{v}=0\Rightarrow (1,9,8)·(x-1,y-1,z-4)=0$$

$$\Rightarrow (x-1)+9(y-1)+8(z-4)=0$$

Expanding gives the general form $x+9y+8z-42=0$.

Theorem 3.3.1

Theorem 3.4.3 (not a typo)

Consider the homogeneous system

$$\begin{array}{cc}\hfill {\displaystyle a_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle ⋮}& {\displaystyle ⋮}\hfill \\ \hfill {\displaystyle a_{m1}{x}_1+a_{m2}{x}_2+\cdots +a_{mn}{x}_n}& {\displaystyle =0}\hfill \end{array}$$

Each equation is a dot product of a row of $A$ with $\overrightarrow{x}$. For example, the first equation is

$$(a_{11},a_{12},\dots ,a_{1n})·(x_1,x_2,\dots ,x_n)=0.$$

Shifting gears a bit,

Theorem 3.3.2 Projection Theorem

Proof

${\overrightarrow{w}}_1$ parallel to $\overrightarrow{a}$ means ${\overrightarrow{w}}_1=k\overrightarrow{a}$. Then

$$\begin{array}{cc}\hfill {\displaystyle \overrightarrow{u}·\overrightarrow{a}}& {\displaystyle =({\overrightarrow{w}}_1+{\overrightarrow{w}}_2)·\overrightarrow{a}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(k\overrightarrow{a}+{\overrightarrow{w}}_2)·\overrightarrow{a}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k\overrightarrow{a}·\overrightarrow{a}+{\overrightarrow{w}}_2·\overrightarrow{a}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k{\Vert \overrightarrow{a}\Vert }^2,}\hfill \end{array}$$

since ${\overrightarrow{w}}_2·\overrightarrow{a}=0$ (${\overrightarrow{w}}_2$ is orthogonal to $\overrightarrow{a}$). So

$${\displaystyle \overrightarrow{u}·\overrightarrow{a}=k{\Vert \overrightarrow{a}\Vert }^2\Rightarrow k=\frac{\overrightarrow{u}·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}.}$$

$${\displaystyle \text{So }{\overrightarrow{w}}_1=\frac{\overrightarrow{u}·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}\overrightarrow{a}=\frac{\overrightarrow{u}·\overrightarrow{a}}{\Vert \overrightarrow{a}\Vert }\frac{\overrightarrow{a}}{\Vert \overrightarrow{a}\Vert }.}$$

$${\displaystyle \text{Then }{\overrightarrow{w}}_2=\overrightarrow{u}-\frac{\overrightarrow{u}·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}\overrightarrow{a}.}$$

Definition

#20

Find the vector component of $𝐮$ along $𝐚$ and the vector component of $𝐮$ orthogonal to $𝐚$.

$𝐮=(5,0,-3,7)$, $𝐚=(2,1,-1,-1)$

Solution

The component along $\overrightarrow{a}$ is ${\overrightarrow{w}}_1$ and the component orthogonal to $\overrightarrow{a}$ is ${\overrightarrow{w}}_2$.

$${\displaystyle {\overrightarrow{w}}_1={proj}_{\overrightarrow{a}}\overrightarrow{u}=\frac{\overrightarrow{u}·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}\overrightarrow{a}=\frac{10+0+3-7}{4+1+1+1}\overrightarrow{a}=\frac{6}{7}\overrightarrow{a}}$$

$${\displaystyle {\overrightarrow{w}}_2=\overrightarrow{u}-{\overrightarrow{w}}_1=(5,0,-3,7)-(\frac{12}{7},\frac{6}{7},-\frac{6}{7},-\frac{6}{7})}$$

Check: ${\overrightarrow{w}}_1+{\overrightarrow{w}}_2=(5,0,-3,7)=\overrightarrow{u}$, and

$${\displaystyle \overrightarrow{a}·{\overrightarrow{w}}_2=(2,1,-1,-1)·(\frac{23}{7},-\frac{6}{7},-\frac{15}{7},\frac{55}{7})=\frac{46}{7}-\frac{6}{7}+\frac{15}{7}-\frac{55}{7}=0.}$$

Standard matrix for a transformation: its columns are the images of the standard basis vectors,

$$[T_A]=\left[\begin{array}{cccc}\hfill {\displaystyle T_A({\overrightarrow{e}}_1)}\hfill & \hfill {\displaystyle T_A({\overrightarrow{e}}_2)}\hfill & \hfill {\displaystyle \cdots }\hfill & \hfill {\displaystyle T_A({\overrightarrow{e}}_n)}\hfill \end{array}\right].$$

#38

Find the standard matrix for the orthogonal projection of $R^2$ onto the stated line, and then use that matrix to find the orthogonal projection of the given point onto that line.

The orthogonal projection of $(1,2)$ onto the line that makes an angle of $\pi /4(={45}^{\circ })$ with the positive $x$-axis.

Solution

The line $y=x$ makes angle $\theta =\pi /4$ with the positive $x$-axis. The unit vector $\overrightarrow{a}$ points along the line. The point $(1,2)$ drops perpendicularly (dashed) to the point $(\frac{3}{2},\frac{3}{2})$ on the line.

Find the image of $T_A$ on ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$. A unit vector on the line is $\overrightarrow{a}=(\cos \theta ,\sin \theta )$, so $\Vert \overrightarrow{a}\Vert =1$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {proj}_{\overrightarrow{a}}{\overrightarrow{e}}_1}& {\displaystyle =\frac{{\overrightarrow{e}}_1·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}\overrightarrow{a}=\cos \theta (\cos \theta ,\sin \theta )=({\cos }^2\theta ,\cos \theta \sin \theta )}\hfill \\ \hfill {\displaystyle {proj}_{\overrightarrow{a}}{\overrightarrow{e}}_2}& {\displaystyle =\frac{{\overrightarrow{e}}_2·\overrightarrow{a}}{{\Vert \overrightarrow{a}\Vert }^2}\overrightarrow{a}=\sin \theta (\cos \theta ,\sin \theta )=(\cos \theta \sin \theta ,{\sin }^2\theta )}\hfill \end{array}}$$

These are the first and second columns of the standard matrix:

$$[T_A]=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right].\text{For }\theta ={45}^{\circ },[T_A]=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right].$$

$$[T_A]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}\frac{3}{2}\\ \frac{3}{2}\end{array}\right]$$

In general, the matrix for orthogonal projection onto the line at angle $\theta$ is denoted $P_{\theta }$:

$$P_{\theta }=\left[\begin{array}{cc}{\cos }^2\theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & {\sin }^2\theta \end{array}\right]=\left[\begin{array}{cc}{\cos }^2\theta & \frac{1}{2}\sin 2\theta \\ \frac{1}{2}\sin 2\theta & {\sin }^2\theta \end{array}\right]$$

#36

Find the standard matrix for the reflection of $R^2$ about the stated line, and then use that matrix to find the reflection of the given point about that line.

The reflection of $(1,2)$ about the line that makes an angle of $\pi /4(={45}^{\circ })$ with the positive $x$-axis.

Solution

The given vector $\overrightarrow{x}$ ends at $(1,2)$. Its projection $P_{\theta }\overrightarrow{x}$ ends on the line $y=x$, halfway between $\overrightarrow{x}$ and the desired vector $H_{\theta }\overrightarrow{x}$, the reflection, which ends at $(2,1)$. The dotted segment from $(1,2)$ to $(2,1)$ is perpendicular to the line.

The distance between $\overrightarrow{x}$ and $P_{\theta }\overrightarrow{x}$ is half the distance between $\overrightarrow{x}$ and $H_{\theta }\overrightarrow{x}$, and they point the same way. Thus

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle P_{\theta }\overrightarrow{x}-\overrightarrow{x}}& {\displaystyle =\frac{1}{2}(H_{\theta }\overrightarrow{x}-\overrightarrow{x})}\hfill \\ \hfill {\displaystyle H_{\theta }\overrightarrow{x}}& {\displaystyle =2P_{\theta }\overrightarrow{x}-\overrightarrow{x}=(2P_{\theta }-I)\overrightarrow{x}}\hfill \end{array}}$$

where $I$ is the identity matrix. So

$$\begin{array}{cc}\hfill {\displaystyle H_{\theta }=2P_{\theta }-I}& {\displaystyle =\left[\begin{array}{cc}2{\cos }^2\theta -1& \sin 2\theta \\ \sin 2\theta & 2{\sin }^2\theta -1\end{array}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right].}\hfill \end{array}$$

(Using $\cos 2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$.) Here $2\theta =\pi /2$, so

$$H_{\pi /4}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right].$$

Theorem 3.3.3 Theorem of Pythagoras in Rⁿ

The Euclidean inner product is the dot product.

Proof

$$\begin{array}{cc}\hfill {\displaystyle {\Vert \overrightarrow{u}+\overrightarrow{v}\Vert }^2}& {\displaystyle =(\overrightarrow{u}+\overrightarrow{v})·(\overrightarrow{u}+\overrightarrow{v})={\Vert \overrightarrow{u}\Vert }^2+2\overrightarrow{u}·\overrightarrow{v}+{\Vert \overrightarrow{v}\Vert }^2}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\Vert \overrightarrow{u}\Vert }^2+{\Vert \overrightarrow{v}\Vert }^2}\hfill \end{array}$$

since $\overrightarrow{u}⟂\overrightarrow{v}\Rightarrow \overrightarrow{u}·\overrightarrow{v}=0$.