vectors

**Quick** · July 18th 2006, 06:20 AM

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?

**CaptainBlack** · July 18th 2006, 06:49 AM

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

**ThePerfectHacker** · July 18th 2006, 09:58 AM

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 $2\bold{i}+\bold{j}-\bold{k}$

**Quick** · July 18th 2006, 04:41 PM

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

**ThePerfectHacker** · July 18th 2006, 07:30 PM

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.

**Quick** · July 18th 2006, 08:13 PM

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 $0\bold{i}+0\bold{j}+1\bold{k}$

**CaptainBlack** · July 18th 2006, 09:06 PM

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

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