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

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

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