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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
3 years, 12 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. Mark Hennings · 3 years, 12 months ago

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@Mark Hennings Thank you Sir Mark :)) JohnDonnie Celestre · 3 years, 12 months ago

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