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Table of Contents
Remark
$\|\vec{AB}\|$ is the distance from point $A$ to point $B$, since this is the length of vector which takes point $A$ to point $B$.
Examples
- The distance from $A=(1,2)$ to $B=(-3,4)$ is $\|\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\vec{AB}\|=\left\|\m{-3\\4}-\m{1\\2}\right\|=\left\|\m{-4\\2}\right\|=\sqrt{(-4)^2+2^2}=\sqrt{20}=2\sqrt{5}$.
- The length of the main diagonal of the unit cube in $\mathbb{R}^3$ is the distance between $0=(0,0,0)$ and $A=(1,1,1)$, which is $\|\vec{0A}\|=\left\|\m{1\\1\\1}\right\|=\sqrt{1^2+1^2+1^2}=\sqrt3$.
Scalar multiplication and direction
Multiplying a vector by a scalar changes its length, but doesn't change its direction.
Definition: unit vectors
A unit vector is a vector $\vec v$ with $\|\vec v\|=1$.
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
Using the formula $\|c\vec v\|=|c|\,\|\vec v\|$ and the fact that $\|\vec v\|>0$, we have \[ \|\vec w\|=\left\|\frac1{\|\vec v\|}\vec v\right\|=\left|\frac1{\|\vec v\|}\right|\,\|\vec v\|=\frac1{\|\vec v\|}\,\|\vec v\| = 1.\] So $\vec w$ is a unit vector, and since it's scalar multiple of $\vec v$, it's in the same direction as $\vec v$. ■
Example
What is unit vector in the same direction as $\vec v=\m{1\\2}$?
We have $\|\vec v\|=\sqrt{1^2+2^2}=\sqrt5$, so the proposition tells us that is $\vec w=\frac1{\|\vec v\|}\vec v = \frac1{\sqrt 5}\vec v=\frac1{\sqrt5}\m{1\\2}=\m{1/\sqrt{5}\\2/\sqrt5}$ is a unit vector in the same direction as $\vec v$.
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 $\vec v+\vec w=\vec{AC}$.
This gives us the triangle law for vector addition: $\vec v$, $\vec w$ and $\vec v+\vec w$ may be arranged to form a triangle:
We get another triangle by starting at $A$ and translating first by $\vec w$ and then by $\vec v$; the other side of this triangle is $\vec w+\vec v$ But we know that $\vec v+\vec w=\vec w+\vec v$, so we can put these two triangles together to get the parallelogram law for vector addition:
The dot product
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.
Example
If $\vec v=\m{3\\5}$ and $\vec w=\m{4\\-7}$, then $\vec v\cdot \vec w=\m{3\\5}\cdot \m{4\\-7} = 3(4)+5(-7)=-23$.
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 here is an example.
Example
If $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$, then $\dp vw=1(-2)+2(1)=-2+2=0$. On the other hand, we have $\|\vec v\|=\sqrt5=\|\vec w\|$, so the angle $\theta$ between $\vec v$ and $\vec w$ satisfies \[ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta\] so $5\cos\theta=0$, so $\cos\theta=0$, so $\theta=\pi/2$ or $\theta=3\pi/2$ (measuring angles in radians). This tells us that the angle between $\vec v$ and $\vec w$ is a right angle. We say that these vectors are orthogonal. We can draw a convincing picture which indicates that these vectors are indeed at right angles:
