Projections and Planes

Recall (dot product)

let $$\begin{gathered} \vec{a}=[a_1 \\ {a}_2 \\ {a}_3] \end{gathered}$$ , be is the same in ${\mathbb{R}}^3$.

Dot product of $\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$.

$\vec{a}\cdot \vec{a}=||\vec{a}|{|}^2$ so $||\vec{a}||=\sqrt{\vec{a}\cdot \vec{a}}$

Properties of dot products:

1. $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$
2. $\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\vec{b}+\vec{a}\vec{c}$
3. $(\alpha \vec{a})\cdot \vec{b}=\vec{a}\cdot (\alpha \vec{b}),\alpha \in \mathbb{R}$

Fact: $\vec{a}\cdot \vec{b}=||\vec{a}||\cdot ||\vec{b}||cos\theta$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Proof (law of cosine)

$$\begin{array}{r}||\vec{b}-\vec{a}|{|}^2=||a|{|}^2+||b|{|}^2-2||\vec{a}||||\vec{b}||\cos \theta \\ \\ (\vec{b}-\vec{a})(\vec{b}-\vec{a})=\vec{a}\cdot \vec{a}+\vec{b}\cdot \vec{b}-2||\vec{a}||||\vec{b}||\cos \theta \end{array}$$

after reduction..

$$\begin{array}{r}-2\vec{a}\cdot \vec{b}=-2||\vec{a}||||\vec{b}||\cos \theta \\ \\ \vec{a}\cdot \vec{b}=||\vec{a}||\cdot ||\vec{b}||\cos \theta \end{array}$$

Corrolary: $\vec{a}\perp \vec{b}⟺\vec{a}\cdot \vec{b}=0$

Ex: Find the angle between u=[-1\\1\\2] and v[2\\1\\-1]

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{||\vec{a}||\cdot ||\vec{b}||}$

dot product works out to:

-2 + 1 - 2 = -3

$||u||=\sqrt{-1^2+1^2+2^2}=\sqrt{6}$ $||v||=\sqrt{2^2+1^2+-1^2}=\sqrt{6}$

plug into $\cos \theta$

$\frac{-3}{{\sqrt{6}}^2}=\frac{-3}{6}=\frac{-1}{2}$