~~REVEAL~~

==== Last time ====

  * the vector $\vec{AB}$ joining $A$ to $B$
  * length (norm) of a vector
  * unit vectors
  * triangle law and parallelogram law for vector addition

====== The dot product ======

==== Definition of the dot product ====

{{page>dot product}}

  * In other words, $\vec v\cdot \vec w$ is the [[row-column product]] $(\vec v)^T\vec w$ of the [[transpose]] of $\vec v$ with $\vec w$.
  * Note that while $\vec v$ and $\vec w$ are vectors, their dot product $\vec v\cdot \vec w$ is a scalar.

==== Example ====

Let $\vec v=\m{3\\5}$ and $\vec w=\m{4\\-7}$.
  * $\vec v\cdot \vec w=\m{3\\5}\cdot \m{4\\-7} = 3(4)+5(-7)=12-35=-23$.

===== Properties of the dot product =====

For any vectors $\vec v$, $\vec w$ and $\vec u$ in $\mathbb{R}^n$, and any scalar $c\in \mathbb{R}$:

  - $\def\dp#1#2{\vec #1\cdot \vec #2}\dp vw=\dp wv$ (the dot product is commutative)
  - $\vec u\cdot(\vec v+\vec w)=\dp uv+\dp uw$
  - $(c\vec v)\cdot \vec w=c(\dp vw)$
  - $\dp vv=\|\vec v\|^2\ge 0$, and $\dp vv=0 \iff \vec v=0_{n\times 1}$

The proofs of these properties are exercises.

===== Angles and the dot product =====

==== Theorem: geometric dot product formula ====

If $\vec v$ and $\vec w$ are non-zero vectors in $\mathbb{R}^n$, then
\[ \dp vw=\|\vec v\|\,\|\vec w\|\,\cos\theta\]
where $\theta$ is the angle between $\vec v$ and $\vec w$.

  * Why's it called a geometric formula?
    * It expresses the dot product using only geometric quantities: lengths and angle
  * The proof will be given soon, but for now let's work out an example.

==== Example ====

What's the angle $\theta$ between $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$?
  * $\|\vec v\|\,\|\vec w\|\cos\theta=\dp vw$
  * $=1(-2)+2(1)=-2+2=0$.
  * So $\|\vec v\|\,\|\vec w\|\cos\theta=0$
  * We have $\|\vec v\|\,\|\vec w\|\ne 0$ (why?) so we can cancel this from both sides
  * So $\cos\theta=0$
  * So $\theta=\pi/2$ or $\theta=3\pi/2$ (measuring angles in radians). 
  * The angle between $\vec v$ and $\vec w$ is a right angle. 
  * We say $\vec v$ and $\vec w$ are **orthogonal**. 
 
==== Picture of $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$====
We can draw a convincing picture which indicates that these vectors are indeed at right angles:

{{ :orth1.png?nolink&500 |}}


==== Reminder: the cosine rule ====

{{ :t1.png?nolink&500 |}}

  * This is the key to proving the geometric dot product formula.

==== Proof of the geometric dot product formula ====

Aim: $\def\vv{\vec v}\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. 

  * Consider a triangle with two sides $\vv$ and $\ww$. By the triangle rule for vector addition, the third side is $\vv-\ww$:{{ :t2.png?nolink&400|}}
  * Use the cosine rule with $A=\theta$, $a=\|\vv-\ww\|$, $b=\|\vv\|$, $c=\|\ww\|$ 
  * $ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta$

==== Proof of the geometric dot product formula, continued====

Aim: $\def\vv{\vec v}
\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$.

  * $\|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta$
  * So $2\|\vv\|\,\|\ww\|\cos\theta=\|\vv\|^2+\|\ww\|^2-\|\vv-\ww\|^2$
    * $=\vv\cdot\vv+\ww\cdot\ww-(\vv-\ww)\cdot(\vv-\ww)$
    * $=\vv\cdot\vv+\ww\cdot\ww-(\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww)$
    * $=2\vv\cdot\ww$.
  * So $2\vv\cdot\ww=2\|\vv\|\,\|\ww\|\,\cos\theta$
  * So $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■ 

==== Corollary 1: angle formula====

If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$.

  * Proof: since $\vv$ and $\ww$ are non-zero, we have $\|\vv\|\,\|\ww\|\ne0$ so we can divide both sides of the geometric dot product formula by $\|\vv\|\,\|\ww\|$ to get the angle formula.
==== Example 1 ====

What is the angle $\theta$ between $\def\c#1#2{\left[\begin{smallmatrix}{#1}\\{#2}\end{smallmatrix}\right]}\c12$ and $\c3{-4}$?
  * Use the angle formula $\cos\theta=\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$:
  * $ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=\frac{-5}{5\sqrt 5}=-\frac1{\sqrt5}$
  * So $\theta=\cos^{-1}(-\tfrac1{\sqrt5}) \approx 2.03\,\text{radians}\approx 116.57^\circ$.

==== Corollary 2: orthogonal vectors ====
If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.
  *Proof: the angle formula gives $\cos \theta=\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}=0$
  *So $\cos\theta=0$
  *So $\theta$ is a right-angle.
==== Example 2 ====
Prove that $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle.
  * $\vec{AB}=\c36-\c23=\c13$ 
  * $\vec{AC}=\c{-4}5-\c23=\c{-6}2$
  * So $\vec{AB}\cdot\vec{AC}=\c13\cdot\c{-6}2=1(-6)+3(2)=0$
  * So the sides $AB$ and $AC$ are at right-angles.
  * So $ABC$ is a right-angled triangle.

==== Example 3 ====

Find a unit vector orthogonal to the vector $\vv=\c12$.
  * Observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$
  * So $\vv$ and $\ww$ are orthogonal
  * $\vec u=\frac1{\|\ww\|}\ww$ is a unit vector in the same direction as $\ww$
    * so it is also orthogonal to $\vv$.
  * So $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.