# For the unit vector #hattheta#, geometrically show that #hattheta = -sinthetahati + costhetahatj#? Essentially, converting from cartesian to polar, how would I determine the unit vector for #vectheta# in terms of #theta#, #hati#, and #hatj#?

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I've been able to show that #hatr = costhetahati + sinthetahatj# :

#costheta = hati/hatr#

#sintheta = hatj/hatr#

#=> ||hatr|| = sqrt(hatrcdothatr(cos^2theta + sin^2theta))#

#= sqrt(hatrcdothatrcos^2theta + hatrcdothatrsin^2theta)#

#= sqrt(hatrcostheta * hati + hatrsintheta hatj)#

#=> hatrcdothatr = ||hatr||^2 = hatrcosthetacdothati + hatrsinthetacdothatj#

#= hatrcdot(costhetahati + sinthetahatj)#

Thus, #hatr = costhetahati + sinthetahatj# . But how would I do it for #hattheta# ? I'm probably just missing something really simple, like where the #hattheta# vector points.

I've been able to show that

Thus,

##### 2 Answers

See below.

#### Explanation:

Considering

We have then

but

Here for convenience, we call

so we have

deriving again

Here

and finally

Concluding

See the design in the explanation and the graph.

#### Explanation:

As a matter of convenience, I use

The unit vector in the direction

The parallel position vector through the origin O ( r = 0 ) is

the unit circle r = 1 , in the direction

P' is at

vector of the unit circle, in the direction

graph{(x^2+y^2-1)(y-x/sqrt3)(y+sqrt3x)=0 [-1, 1, -.05, 1]}

Here

would represent

In brief,