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I was wondering how would one go about converting a displacement vector in cartesian coordinates to cylindrical. I am getting a bit confused on how to deal with the unit vectors; converting a point in space is a simple task, but when it's a vector it just confuses me.

I am specifically working on describing a displacement vector along the axis of symmetry (z-axis) of a arbitrary helix i.e. the displacement vector from the z-axis to any point on the helix:

Describing the Helix in Cartesian:

{x=Rcosθ, y=Rsinθ, z=(h/2∏)θ}; h is the pitch of the helix, or height of one revolution of coil

Describing the Helix in Cylindrical:

{r=R, ∅=θ, z=(h/2∏)θ}

In Cartesian I have the displacement vector as...

(x-x')i + (y-y')j + (z-z')k =...

...(x-Rcos[θ])i + (y-Rsin[θ])j + (z-((h/2∏)θ))k=...; apply condition of being on z axis {0,0,z}

...(0-Rcos[θ])i + (0-Rsin[θ])j + (z-((h/2∏)θ))k=

...(-Rcos[θ])i + (-Rsin[θ])j + (z-((h/2∏)θ))k

Now, converting to cylindrical (or so I think):

{(-Rcos[θ])[itex]^{2}[/itex]+(-Rsin[θ])[itex]^{2}[/itex]}[itex]\hat{r}[/itex]=(R)[itex]\hat{r}[/itex]...

ArcSin[y/r][itex]\hat{θ}[/itex]=ArcSin[(([itex]\frac{-Rsinθ}{R}[/itex])]=-θ[itex]\hat{θ}[/itex]

(z-((h/2∏)θ))[itex]\hat{z}[/itex]

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# Converting a DISPLACEMENT vector in Cartesian to Cylindrical Corrdinates!

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