3D vector projection

Question

If I have an arbitrary vector in the 3D coordinate plane with given azimuth angle and polar angle with radius r=1, how could I use this information to project this vector onto the three 2D planes?(xy,yz,xz) There are no specific number available because I need to represent these "component" vectors symbolically rather than numerically.

Answer 1

If your vector is $$\vec{p} = \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}r \cos\theta \cos\varphi \\ r \cos\theta \sin\varphi \\ r \sin\theta \end{matrix}\right]$$ where $-180° \le \varphi \le +180°$ is the azimuth angle, and $-90° \le \theta \le +90°$ the polar angle, then its projection on the $xy$, $xz$, and $yz$ planes are $$\vec{p}_{xy} = \left[\begin{matrix}x\\y\\0\end{matrix}\right], \quad \vec{p}_{xz} = \left[\begin{matrix}x\\0\\z\end{matrix}\right], \quad \vec{p}_{yz} = \left[\begin{matrix}0\\y\\z\end{matrix}\right]$$ respectively.

In spherical coordinates, $$\left\lbrace\begin{aligned} r_{xy} &= r \cos\theta \\ \varphi_{xy} &= \varphi \\ \theta_{xy} &= 0 \\ \end{aligned}\right.$$ $$\left\lbrace\begin{aligned} r_{xz} &= r \sqrt{1 - (\cos\theta)^2(\cos\varphi)^2} \\ \varphi_{xz} &= \begin{cases} 0°, & \cos\varphi \ge 0 \\ 180°, & \cos\varphi \lt 0 \\ \end{cases} \\ \theta_{xz} &= \arcsin\left(\frac{\sin\theta}{\sqrt{1 - (\cos\theta)^2(\sin\varphi)^2}}\right) \\ \end{aligned}\right.$$ $$\left\lbrace\begin{aligned} r_{yz} &= r \sqrt{1 - (\cos\theta)^2(\sin\varphi)^2} \\ \varphi_{yz} &= \begin{cases} +90°, & \cos\varphi \ge 0 \\ -90°, & \cos\varphi \lt 0 \\ \end{cases} \\ \theta_{yz} &= \arcsin\left(\frac{\sin\theta}{\sqrt{1 - (\cos\theta)^2(\cos\varphi)^2}}\right) \end{aligned}\right.$$

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In general, if you have a plane through origin with unit normal vector $\hat{n}$, $\lVert\hat{n}\rVert = 1$, then the projection of $\vec{p}$ on that plane in Cartesian coordinates is $\vec{p}_n$, $$\vec{p}_n = \vec{p} - \vec{p}\cdot\hat{n}$$