Find a parametric form of a circle that is in a plane $x-2y+z=0$ with radius 1 and center at origin

In the book they picked $\hat u =(2,1,0)$ (hat for unit vector) , did cross product with the plane normal and called it $\hat v$ (after normalizing it) lastly they got

$\vec r(t)=\frac{1}{\sqrt{5}} \cdot (2,1,0) \cdot \cos(t) + \frac{1}{\sqrt{30}} \cdot (-1,2,5) \cdot \sin(t)$

What I did was take a random vector $(a,b,c)$ and did a cross product with the plane normal and got $(b+2c,c-a,-2a-b)$ since this vector is perpendicular to the plane and the vector I picked I could randomly pick $a,b,c$ so I pick $a=1,b=2,c=3$ so my vector would be $(8,2,-4)=(4,1,-2)$ and after making it also a unit vector I got $\hat v=\frac{(4,1,-20)}{\sqrt{21}}$ according to the $a,b,c$ I picked I could find the second unit vector which is $\hat u=\frac{(1,2,3)}{\sqrt6}$

lastly $\vec r(t)=\frac{(1,2,3)}{\sqrt6} \cdot \cos(t) + \frac{(4,1,-20)}{\sqrt{21}} \cdot \sin(t)$

is what I did correct? , the idea of doing a cross product is according to instruction but why is that? what does the cross product have to do with a circle in a plane (I know that it gives a perpendicular vector but how does it help here)?

sorry for my English hopefully it is understandable


1 Answer 1


The idea is that if $\vec{u}$ and $\vec{v}$ are two perpendicular unit vectors, then the curve $\cos(t)\vec{u} + \sin(t)\vec{v}$ is a unit circle centered at $(0,0)$ in the plane spanned by $\vec{u}$ and $\vec{v}$.

The cross product is used to create such a pair of vectors in the given plane $P$. One can take a unit vector $\vec{u}\in P$ and define $\vec{v} = \vec{u}\times \vec{n}$ where $\vec{n}$ is a unit vector normal to $P$, here $(1,-2,1)$ normalized.

  • $\begingroup$ And vector $u$ has to be in the plane ? a vector that solves the plane equation $\endgroup$
    – Adamrk
    Mar 22, 2023 at 11:23
  • $\begingroup$ Yes because the circle is in the $(\vec{u}, \vec{v})$ plane. $\endgroup$ Mar 22, 2023 at 12:08

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