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Table of Contents
Example
Which vector moves the point $A=(-1,3)$ to $B=(5,-4)$?
Answer: we need a vector $\vec v$ with $A+\vec v=B$, so $\vec v=B-A = \def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{5\\-4}-\m{-1\\3}=\m{6\\-7}$. 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 \[ \vec{AB}=B-A\] (where on the right hand side, we interpret the points as column vectors so we can subtract them to get a column vector).
Thus $\vec{AB}$ is the vector which moves the point $A$ to the point $B$.
Example
In $\mathbb{R}^3$, the points $A=(3,-4,5)$ and $B=(11,6,-2)$ have $\vec{AB}=\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.
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
- $\left\|\m{4\\3}\right\|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
- $\left\|\m{1\\0\\-2\\3}\right\|=\sqrt{1^2+0^2+(-2)^2+3^2}=\sqrt{1+0+4+9}=\sqrt{14}$
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 $c$.
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:
