# How do you write the components of vectors?

Jul 31, 2018

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

#### Explanation:

For location of a point, use

$r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$.

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

$\vec{u} = \frac{\vec{r}}{r} = \frac{1}{\sqrt{{x}^{2} + {y}^{2}}} < x , y >$

In other words,

any direction can be represented by

unit vector $\vec{u} = \frac{\vec{r}}{r}$, in the direction of $\vec{v}$.

Note that length of the vector $\vec{r} , r = \sqrt{{x}^{2} + {y}^{2}} \ge 0$.

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 $- \vec{j} = 1 < 0 , - 1 >$ 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

$x \vec{i} + y \vec{j} = r \left(\cos \theta \vec{i} + \sin \theta \vec{j}\right)$

Graph of the like-parallel and equal vector $\vec{r}$ at $O \left(0 , 0\right)$,

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.
.