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I have an exercise where if $\Sigma=(0,E)$ is a orthogonal coordinate system, then the line $r$ is such that: $$r\colon X=A+\lambda\vec{AB}$$ where $A(1,2,1)_{\Sigma},B(2,1,2)_{\Sigma}$ and $C(3,1,2)_{\Sigma}$.

There also a plan $\pi$ such that $A\in\pi$ and the vector normal of $\pi$ is $$\vec{AB}=\vec{AB}\wedge\vec{AC}$$.

Now, I have to find a new coordinate system $\Sigma'=(O',f_{1},f_{2},f_{3})$ with coordinates $(u,v,w)$ such that

  • the line $r$ is given by the parametric equations $u=\lambda,v=w=0,\lambda\in\mathbb{R}$
  • the plan $\pi$ is given by the general equation $w=0,u,v\in\mathbb{R}$

I really have no idea how to find this coordinate system.

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  • $\begingroup$ Surely you must have some ideas. Rename $u$, $v$, $w$ to $x$, $y$, $z$. What line is $r$ then? $\endgroup$ – amd Jul 20 '18 at 6:35
  • $\begingroup$ I have no ideas. I can do this but what do I get? I mean, how do I find $f_{1},f_{2},f_{3}$? $\endgroup$ – mvfs314 Jul 20 '18 at 10:56
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Think about the geometric meaning of the requirements. In the new coordinate system, the coordinates of points on $r$ are of the form $(\lambda,0,0)_{\Sigma'}$. In other words, $r$ is the $u$-axis, and when $\lambda=0$ the corresponding point is the origin. Assuming that this $\lambda$ and the one in the original definition of $r$ are identical, this completely determines $O'$ and $f_1$.

Now look at the second requirement. Points on $\pi$ have $\Sigma'$-coordinates of the form $(a,b,0)_{\Sigma'}$—it is the $u$-$v$ plane. So, for $f_2$ you can pick any vector parallel to $\pi$ that’s not a multiple of the one you chose for $f_1$ and for $f_3$ any vector that’s not parallel to $\pi$. There are obvious choices for both of them.

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