How do you write the components of vectors?

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.
.