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# Vector Projection; Space Coordinate System

Why is it that we have to provide a reference point $$P(x, y, z)$$ to convert a Cartesian vector to a Cylindrical vector? Also, how would you interpret the vector $$A_\rho a_\rho + A_\phi a_\phi + A_z a_z$$, geometrically?

Note by JohnDonnie Celestre
4 years, 6 months ago

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Cylindrical polar coordinates are defined in terms of an axis (the axis of the cylinder). The relationship between the fundamental Cartesian coordinate vectors $$a_x,a_y,a_z$$ (which are unit vectors pointing along the $$x$$-, $$y$$- and $$z$$-axes) and the fundamental cylindrical polar coordinate vectors $$a_\rho,a_\phi,a_z$$ varies depending on the point where these vectors are being considered. While, for example, $$a_x$$ always points in the same direction, the vector $$a_\rho$$ points in different directions at different points in space.

To be specific, if $$a_x=(1,0,0)$$, $$a_y = (0,1,0)$$ and $$a_z = (0,0,1)$$. Suppose we are at the point $$P$$ with coordinates $$(\rho\cos\phi,\rho\sin\phi,z)$$.

• $$a_\rho$$ is the unit vector in the "direction of increasing $$\rho$$", namely the unit vector which is perpendicular to the $$z$$-axis and radially outwards, so $$a_\rho = (\cos\phi,\sin\phi,0)$$.

• $$a_\phi$$ is the unit vector in the "direction of increasing $$\phi$$", namely the direction you would go if $$\phi$$ increases, while keeping $$\rho$$ and $$z$$ constant. If you keep $$\rho$$ and $$z$$ constant, you rotate about the $$z$$-axis, and so $$a_\phi = (-\sin\phi,\cos\phi,0)$$.

• $$a_z$$ is still the unit direction in the "direction of increasing $$z$$", namely $$(0,0,1)$$.

If we are converting between the Cartesian and cylindrical polar systems, we are in the business of describing the same vector $$a$$ in two different ways, and so $a \; = \; A_xa_x + A_ya_y + A_za_z \; = \; A_\rho a_\rho + A_\phi a_\phi + A_z a_z$ Taking scalar products: $\begin{array}{rcl} A_\rho & = & a \cdot a_\rho \; = \; A_x \cos\phi + A_y \sin\phi \\ A_\phi & = & a \cdot a_\phi \; = \; -A_x \sin\phi + A_y \cos\phi \\ A_z & = & a \cdot a_z \; = \; A_z \end{array}$ The last equation is trivial, since the $$z$$-coordinate is the same coordinate in both coordinate systems. These are precisely the equations that are being used in the example you provided.

- 4 years, 6 months ago

Thank you Sir Mark :))

- 4 years, 6 months ago