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)

||\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{aligned} $$ after reduction.. $$ \begin{aligned} -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{aligned} $$ Corrolary: $\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$ 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 $\cos\theta$ $\frac{-3}{\sqrt{6}^2} = \frac{-3}{6} = \frac{-1}{2}$