Question

How do you write the components of vectors?

Answer 1

See explanation that is almost exhaustive. I would review and edit my answer myself, if necessary..

Explanation:

For location of a point, use

`r ( cos theta, sin theta ) = ( x, y )`.

Let any `vec v = vec r`

`= r < cos theta, sin theta >`

`<` x-component x, y-component y `>`

`= < x, y >`.

`vec r` from the ( origin ) pole, r = 0 rArr x = y =0,

is the like-parallel and equal to `vec v`.

Length r = length v.

`theta` given by `vec v`.

Now, the unit vector

in the direction `vec r` ( for that matter `vec v` ) is

`vec u = vec r/r = 1/sqrt ( x^2 +y^2) < x, y >`

In other words,

any direction can be represented by

unit vector `vec u = vec r/r`, in the direction of `vec v`.

Note that length of the vector `vec r, r = sqrt ( x^2 + y^2 ) >= 0`.

If r = 1, `vec r = vec u`.

Unit vector `vec i = 1 < 1, 0 >` for #theta = 2kpi, gives x-positive x-axis.

Unit vector `- vec i` = 1 < -1, 0 >for #theta = ( 2kp + 1 )pi,

gives x-negative x-axis.

Unit vector `vec j = 1 < 0, 1 >` for #theta = ( 2k +1/2 )pi, gives y-positive y-axis.

Unit vector `- vecj = 1 < 0, - 1 >` for #theta = ( 2k + 3/2 )pi, gives y-negative y-axis.

It is evident that `vec v = < x, y >`

is a short form for the sum of the component vectors

`x veci + y vecj = r ( cos theta veci + sin theta vecj)`

Graph of the like-parallel and equal vector `vecr` at `O ( 0, 0 )`,

for `vec v = < 3, -4 >`, positioned elsewhere, in the direction

#theta = arctan( -4/3 ) .
graph{3y+4x=0[0 8 -4 0]}

The direction is away from O.
.