1. ## vectors

I would like to know what the $i,j,k$ stand for in things like:
$v=2i+j-k$

my guess: they stand for north, east, and ?forward?

2. Originally Posted by Quick
I would like to know what the $i,j,k$ stand for in things like:
$v=2i+j-k$

my guess: they stand for north, east, and ?forward?
They are unit vectors parallel to the coordinate axes, usualy $\bold{i}=[1,0,0],\ \bold{j}=[0,1,0],\ \bold{k}=[0,0,1]$

RonL

3. Originally Posted by Quick
I would like to know what the $i,j,k$ stand for in things like:
$v=2i+j-k$

my guess: they stand for north, east, and ?forward?
Firstly,
it is $\bold{i}$ not $i$, if you are writing on paper and do not want to draw a bold line you write $\vec{i}$ (with an arrow on top).

Let me explain it differently.
This is a vector,
$\bold{v}=2\bold{i}+\bold{j}-\bold{k}$
where, $\bold{i},\bold{j},\bold{k}$ are called vector-components.

Basically you are in 3 dimensions. Instead of 2 (xy graph) you know have an xyz graph. So think of a point located at $(2,1,-1)$ now from that point draw a line to the orogin $(0,0,0)$ the resulting line would be called a vector defined by $2\bold{i}+\bold{j}-\bold{k}$

4. Originally Posted by ThePerfectHacker
Firstly,
it is $\bold{i}$ not $i$, if you are writing on paper and do not want to draw a bold line you write $\vec{i}$ (with an arrow on top).

Let me explain it differently.
This is a vector,
$\bold{v}=2\bold{i}+\bold{j}-\bold{k}$
where, $\bold{i},\bold{j},\bold{k}$ are called vector-components.

Basically you are in 3 dimensions. Instead of 2 (xy graph) you know have an xyz graph. So think of a point located at $(2,1,-1)$ now from that point draw a line to the orogin $(0,0,0)$ the resulting line would be called a vector defined by $2\bold{i}+\bold{j}-\bold{k}$
so then $\bold{i}$ would be the x-coordinate, $\bold{j}$ would be y, and $\bold{k}$ would be z

and the distance and direction between that point and the origin is the vector? seems logical to me

5. Originally Posted by Quick
so then $\bold{i}$ would be the x-coordinate, $\bold{j}$ would be y, and $\bold{k}$ would be z

and the distance and direction between that point and the origin is the vector? seems logical to me
Look what CaptainBlank said. Yes, if that is what you mean.

6. just a question about directions...

would I consider the y axis as elevation, like if I said the ball was thrown upward at 3 ft/s would be $0\bold{i}+3\bold{j}+0\bold{k}$

would I consider the z axis moving forward, like If I said that a man walks forward at a pace of 1 ft/s would be $0\bold{i}+0\bold{j}+1\bold{k}$

7. Originally Posted by Quick
so then $\bold{i}$ would be the x-coordinate, $\bold{j}$ would be y, and $\bold{k}$ would be z

and the distance and direction between that point and the origin is the vector? seems logical to me
The coefficients of $\bold{i}$ would be the x-coordinate (etc) for
a position vector, $\bold{i}$ itself is the unit vector parallel to the
x-axis.

RonL