~~REVEAL~~


====== Chapter 3: Vectors and geometry ======

==== Vectors ====

  * $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a $2\times 1$ column vector
  * i.e., a pair of numbers written in a column
  * We also use pairs of numbers to write points in the plane $\mathbb R^2$
  * e.g., $(4,3)$ is a point
    * you get there by starting from the origin, moving $4$ units to the right and $3$ units up.
  * We think $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up. 
  * This movement is called "translation by $\vec v$". 

==== Translation by $\vec v$ ====

The vector $\vec v=\m{4\\3}$ moves:

  * $(0,0)$ to $(4,3)$
  * $(-2,6)$ to $(2,9)$
  * $(x,y)$ to $(x+4,y+3)$.

  * We're not too fussy about the difference between points like $(4,3)$ and vectors like $\m{4\\3}$.
  * If we write points as column vectors, we can perform algebra (addition, subtraction, scalar multiplication) using points and column vectors.

==== ====

For example, we could rewrite the examples above by saying that $\vec v=\m{4\\3}$ moves:

  * $\m{0\\0}$ to $\m{0\\0}+\m{4\\3}=\m{4\\3}$
  * $\m{-2\\6}$ to $\m{-2\\6}+\m{4\\3}=\m{2\\9}$
  * $\m{x\\y}$ to $\m{x\\y}+\m{4\\3}=\m{x+4\\y+3}$.

  * More generally: a column vector $\vec v$ moves a point $\vec x$ to $\vec x+\vec v$.

==== Example ====

Which vector $\vec v$ moves the point $A=(-1,3)$ to $B=(5,-4)$?

  * We need a vector $\vec v$ with $A+\vec v=B$
  * So $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$. 

  * We write $\vec{AB}=\m{6\\-7}$, since this is the vector which moves $A$ to $B$.

==== Definition of $\vec{AB}$ ====

If $A$ and $B$ are any points in $\mathbb{R}^n$, then the vector $\vec{AB}$ is defined by 
\[ \vec{AB}=B-A\]
(on the right, we interpret points as column vectors so we can subtract them to get a column vector).

  * $\vec{AB}$ is the vector which moves $A$ to $B$.

==== Example ====

In $\mathbb{R}^3$, 
  * if $A=(3,-4,5)$ 
  * and $B=(11,6,-2)$ 
  * then $\vec{AB}=\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\def\m#1{\mat{#1}}\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.

==== The uses of vectors ====

Vectors are used in geometry and science to represent quantities with both a **magnitude** (size/length) and a **direction**. For example:

  * displacements (in geometry)
  * velocities
  * forces

Recall that a column vector moves points. Its magnitude, or length, is how far it moves points.

==== Definition: the length of a vector ====

If $\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ is a column vector in $\mathbb{R}^n$, then its **magnitude**, or **length**, or **norm**, is the number
\[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\]

==== Examples ====

  * $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

  * \begin{align*}\left\|\m{1\\0\\-2\\3}\right\|&=\sqrt{1^2+0^2+(-2)^2+3^2}\\&=\sqrt{1+0+4+9}=\sqrt{14}\end{align*}

==== Exercise ====

Prove that if $c\in \mathbb{R}$ is a scalar and $\vec v$ is a vector in $\mathbb{R}^n$, then 
\[ \|c\vec v\|=|c|\,\|\vec v\|.\]
That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.

  * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$ whenever $x,y\ge0$, and $\sqrt{c^2}=|c|$ for any $c\in \mathbb R$.

==== Distance between two points ====

$\|\vec{AB}\|$ is the distance from point $A$ to point $B$
  * since this is the length of vector which takes point $A$ to point $B$.

  * e.g. how far from from $A=(1,2)$ to $B=(-3,4)$?
  * $\small\|\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}$.

  * e.g. what's the length of the main diagonal of the unit cube in $\mathbb{R}^3$?
  * = distance from $0=(0,0,0)$ to $A=(1,1,1)$
  * $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.


==== Scalar multiplication and direction ====

Multiplying a vector by a scalar changes its length, but doesn't change its direction.

==== Definition: unit vectors ====

{{page>unit vector}}

==== Proposition: finding a unit vector in the same direction as a given vector ====

If $\vec v$ is a non-zero vector, then $\vec w=\frac1{\|\vec v\|}\vec v$ is a unit vector (in the same direction as $\vec v$).

=== Proof ===

  * Use the formula $\|c\vec v\|=|c|\,\|\vec v\|$ and the fact that $\|\vec v\|>0$:
  * $\|\vec w\|=\left\|\frac1{\|\vec v\|}\vec v\right\|=\left|\frac1{\|\vec v\|}\right|\,\|\vec v\|=\frac1{\|\vec v\|}\,\|\vec v\| = 1$.
  * So $\vec w$ is a unit vector.
  * It's a scalar multiple of $\vec v$, so is in the same direction as $\vec v$. ■ 

==== Example ====

What is unit vector in the same direction as $\vec v=\m{1\\2}$? 

  * By the Proposition, $\frac{1}{\|\vec v\|}$ is a unit vector in the same direction as $\vec v$.
  * $\|\vec v\|=\sqrt{1^2+2^2}=\sqrt5$
  * So the unit vector in the same direction as $\vec v$ is:
    *  $\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$.

===== Addition of vectors =====

If $\vec v=\vec{AB}$, then $\vec v$ moves $A$ to $B$, so $A+\vec v=B$.

If $\vec w=\vec {BC}$, then $\vec w$ moves $B$ to $C$, so $B+\vec w=C$.

What about $\vec v+\vec w$? We have $A+\vec v+\vec w=B+\vec w=C$. So $\vec v+\vec w=\vec{AC}$.

==== The triangle law for vector addition ====

{{ :tri.png?nolink&400 |}}

====  The parallelogram law for vector addition ====

{{ :par.png?nolink&400 |}}

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

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

{{page>dot product}}

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

===== 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: the relationship between angle and the dot product ====

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 $\vec w$.

  * The proof will be given soon, but for now let's work out an example.

==== Example ====

What's the angle between $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$?
  * $\dp vw=1(-2)+2(1)=-2+2=0$.
  * $\|\vec v\|=\sqrt5=\|\vec w\|$
  * So the angle $\theta$ between $\vec v$ and $\vec w$ has $ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta$
  * So $5\cos\theta=0$, 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&400 |}}


==== Proof of the Theorem ====

We wish to show that $\def\vv{\vec v}
\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$. 

  * Recall the [[wp>cosine rule]]:{{ :t1.png?nolink&300 |}}

==== Proof of the Theorem, slide 2====

We wish to show that $\def\vv{\vec v}
\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$. 

  * Consider a triangle with two sides $\vv$ and $\ww$. 
  * By the triangle rule for vector addition, the third side $\vec x$ has $\ww+\vec x=\vv$, so $\vec x=\vv-\ww$:{{ :t2.png?nolink&300 |}}
  * Apply the cosine rule: $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta.$

==== Proof of the Theorem, slide 3====

We wish to show that $\def\vv{\vec v}
\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$. 

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

==== Corollary 1 ====

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

==== Corollary 2 ====
If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.

==== Example 1 ====

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

==== 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 is also orthogonal to $\vv$). 
  * So $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.