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Chapter 3: Vectors and geometry

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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 $\vec v$”.

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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 $\vec x+\vec v$.

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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 $B$.

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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 $B$.

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Example

In $\mathbb{R}^3$,

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

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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.

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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 \[ \|\vec v\|=\sqrt{v_1^2+v_2^2+\dots+v_n^2}.\]

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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*}

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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 $c\in \mathbb R$.

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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 $A=(1,1,1)$
$\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.

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Scalar multiplication and direction

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

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Definition: unit vectors

A unit vector is a vector $\vec v$ with $\|\vec v\|=1$.

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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 $\vec v$. ■

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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:
- $\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$.

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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 $\vec v+\vec w=\vec{AC}$.

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The triangle law for vector addition

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The parallelogram law for vector addition

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The dot product

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Definition of the dot product

Let $\def\m#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\vec v=\m{v_1\\v_2\\\vdots\\v_n}$ and $\vec w=\m{w_1\\w_2\\\vdots\\w_n}$ be two vectors in $\mathbb{R}^n$.

The dot product of $\vec v$ and $\vec w$ is the number $\vec v\cdot \vec w$ given by \[ \color{red}{\vec v\cdot\vec w=v_1w_1+v_2w_2+\dots+v_nw_n}.\]

Note that while $\vec v$ and $\vec w$ are vectors, their dot product $\vec v\cdot \vec w$ is a scalar.

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Example

Let $\vec v=\m{3\\5}$ and $\vec w=\m{4\\-7}$.

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

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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.

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Angles and the dot product

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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.

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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.

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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:

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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 cosine rule:

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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 $\vec x=\vv-\ww$:
Apply the cosine rule: $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta.$

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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 $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■

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Corollary 1

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

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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.

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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 $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.

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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.

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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 $\vv=\c12$.