I have a skew coordinate system with axes x, y, z’, where x and y are orthogonal to each other. In the third dimension, the z-axis would be the orthogonal axis to x and y, but I don’t have z. Instead, I have axis z’, which is tilted by $\beta$ from yz-plane and by $\alpha$ from xz-plane. Here is plot for better clarity (the blue axes are the skew coordinate system that I have):
My problem: In this skew coordinate system, I have a vector or point ($x_0$, $y_0$, $z’_0$). I now want to tilt the xy-plane by $\alpha$ and $\beta$, so that it is perpendicular to the z’-axis, i.e. I want to create an orthogonal system with axes x’,y’,z’. (Yes, the task is not to change z’ to z, but x and y to x’ and y’).
My question: How do coordinates of ($x_0$, $y_0$, $z’_0$) change when expressed in this new coordinate system, i.e. I want ($x’_0$, $y’_0$, $z’_0$)?
My approach: For sure, the xy-plane would need to be first tilted by $\beta$ to the right and then by $\alpha$ upwards to be perpendicular to z’. However, the point ($x_0$, $y_0$, $z’_0$) is not rotated, but its coordinates need to be expressed in the new system. The first step, x to x’, would change the $x_0$ coordinate to $x’_0$, and I think it is simply
$$x’_0 = x_0 \cdot cos(\beta)$$
correct? I then need to tilt the y-axis by $\alpha$, but is it as simple as
$$y’_0 = y_0 \cdot cos(\alpha)$$
or do the two transformations affect each other (x and y are ortogonal and so are x' and y'…)? In other words, is my point ($x_0$, $y_0$, $z’_0$) in the new coordinates simply $(x_0 \cdot cos(\beta), y_0 \cdot cos(\alpha), z’_0)$?
And is there a more formal way of transformation? I was reading about transformation matrices, but failed to construct one for this problem…
Best Answer
This answer assumes that $0\lt\alpha\lt\frac{\pi}{2}$ and $0\lt\beta\lt\frac{\pi}{2}$.
Let $P(a,b,c)$ be a point on $xyz$ coordinate system.
Since $z'$ axis is a line whose equation is $$\dfrac{x}{\cos\alpha\sin\beta}=\dfrac{y}{\sin\alpha}=\dfrac{z}{\cos\alpha\cos\beta}$$ where $\vec v=(\cos\alpha\sin\beta,\sin\alpha,\cos\alpha\cos\beta)$ is an unit vector, a point on $z'$ axis is represented as $$Q_t(t\cos\alpha\sin\beta,t\sin\alpha,t\cos\alpha\cos\beta)$$ So, we see that the $z'$ coordinate of $P$ on $xyz'$ coordinate sysytem is given by $t$ such that $$\vec v\cdot \vec{PQ_t}=0$$ i.e. $$(\cos\alpha\sin\beta,\sin\alpha,\cos\alpha\cos\beta)(t\cos\alpha\sin\beta-a,t\sin\alpha-b,t\cos\alpha\cos\beta-c)=0$$ i.e. $$t=a\cos\alpha\sin\beta+b\sin\alpha+c\cos\alpha\cos\beta$$
Since $x'$ axis is a line whose equation is $$\dfrac{x}{\cos\beta}=\dfrac{z}{-\sin\beta}\quad\text{and}\quad y=0$$ the $x'$ coordinate of $P$ on $x'y'z'$ coordinate system is given by $t$ such that $$(\cos\beta,0,-\sin\beta)(t\cos\beta-a,-b,-t\sin\beta-c)=0$$ i.e. $$t=a\cos\beta-c\sin\beta$$
Since $y'$ axis is a line whose equation is $$\dfrac{x}{-\sin\alpha\sin\beta}=\dfrac{y}{\cos\alpha}=\dfrac{z}{-\sin\alpha\cos\beta}$$ the $y'$ coordinate of $P$ on $x'y'z'$ coordinate system is given by $t$ such that $$(-\sin\alpha\sin\beta,\cos\alpha,-\sin\alpha\cos\beta)(-t\sin\alpha\sin\beta-a,t\cos\alpha-b,-t\sin\alpha\cos\beta-c)=0$$ i.e. $$t=-a\sin\alpha\sin\beta+b\cos\alpha-c\sin\alpha\cos\beta$$
From what we've got, we see that a point $P(a,b,c)$ on $xyz$ coordinate system is represented as $$\bigg(a,b,a\cos\alpha\sin\beta+b\sin\alpha+c\cos\alpha\cos\beta\bigg)$$ on $xyz'$ coordinate system, and is represented as $$\bigg(a\cos\beta-c\sin\beta,-a\sin\alpha\sin\beta+b\cos\alpha-c\sin\alpha\cos\beta,a\cos\alpha\sin\beta+b\sin\alpha+c\cos\alpha\cos\beta\bigg)$$ on $x'y'z'$ coordinate system.
Therefore, we have a system $$\begin{cases}x_0=a \\y_0=b \\z_0'=a\cos\alpha\sin\beta+b\sin\alpha+c\cos\alpha\cos\beta \\x_0'=a\cos\beta-c\sin\beta \\y_0'=-a\sin\alpha\sin\beta+b\cos\alpha-c\sin\alpha\cos\beta\end{cases}$$ Eliminating $a,b$ and $c$, we finally get $$\color{red}{\begin{align}x_0'&=\frac{x_0}{\cos\beta}+y_0\tan\alpha\tan\beta-z_0'\frac{\tan\beta}{\cos\alpha} \\\\y_0'&=\frac{y_0}{\cos\alpha}-z_0'\tan\alpha\end{align}}$$