I've been studying for an exam and in one of the exams it asks me the parametrise the unit circle in $R^3$ which lines in the plane $$x+y=0$$.
I know how to parametrise a normal circle, but I don't really have any idea how to parametrise it in a plane like that. Any help would be appreciated, thanks
Here’s a general solution:
In $\mathbb R^2$, if $\vec u$ and $\vec v$ are any pair of orthogonal unit vectors, then $\vec c+r\vec u\cos t+r\vec v\sin t$ is a parameterization of a circle of radius $r$ centered at $\vec c$. The same parameterization will work in any Euclidean space, so your problem is reduced to finding a pair of orthogonal unit vectors parallel to the plane of the circle.
In $\mathbb R^3$, a plane can be given by the equation $ax+by+cz+d=0$, which can be written as $(a,b,c)\cdot(x,y,z)+d=0$. The vector $\vec n=(a,b,c)$ is normal to the plane, and it can be easily verified that both $(b,-a,0)$ and $(c,0,-a)$ are perpendicular to this vector and so parallel to the plane. Pick either one of these vectors and normalize it to get $\vec u$. For $\vec v$, compute the cross product of whichever vector you picked with $(a,b,c)$ and normalize that.
For your specific problem, $\vec n=(1,1,0)$, so we can take $\vec u=\frac1{\sqrt2}(1,-1,0)$. For $\vec v$, $(1,-1,0)\times(1,1,0)=(0,0,2)$, so $\vec v=(0,0,1)$. One possible parameterization of the circle is therefore $$(0,0,0)+\frac1{\sqrt2}\cos t\,(1,-1,0)+\sin t\,(0,0,1)=\left(\frac1{\sqrt2}\cos t,-\frac1{\sqrt2}\cos t,\sin t\right).$$
The unit sphere has equation $x^2+y^2+z^2=1$, its intersection with the plane is the circle you are looking for. So you have $$y=-x$$ and $$2x^2+z^2=1$$. You can now use polar coordinates so put $x=-y={1 \over \sqrt 2} \cos t$ and $z=\sin t$ so that $$2{1 \over 2} \cos^2 t+ \sin^2 t=1$$ is identically verified.
You can generate two sets of coordinates. One with the regular $(x, y, z)$ you're used to and another: $(u, v, w)$ These latter coordinates will be a "slanted" version of your $(x, y, z)$ coordinate system.
Since $x+y=0$ is a vertical plane, I'm going to simply go with a rotation of x and y for my transformation between the two systems. Namely,
$$u = x*cos(\pi/4) - y*sin(\pi/4)$$ $$v = x*sin(\pi/4) + y*cos(\pi/4)$$ $$w=z$$
(This is a simple rotation, read up on the 'rotation matrix' for an insight into this transformation)
if we simplify this transformation we get $u = \frac{x-y}{\sqrt2}, v = \frac{x+y}{\sqrt2}, w=z$
We know, additionally, that in this circle $x+y =0$ so $v=0$ in our circle, so we have the plane $(u, w)$ in which the circle is contained.
Here, the unit circle is parametrized by an angle $\alpha$. So we have: $Circle: (u, v, w) = (sin(\alpha), 0, cos(\alpha))$ (I'm sure you recognize this parametrization.)
Now we also know that this can be transformed back into the regular $(x,y,z)$ system we are used to (because rotations are reversible transformations).
If we play around with the transformation equations we can get back to $x = \frac{u+v}{\sqrt2}, y = \frac{-u+v}{\sqrt2}, z=w$
so we can use: $Circle: (x,y,z) = (\frac{sin(\alpha)+0}{\sqrt2}, \frac{-sin(\alpha)-0}{\sqrt2}, cos(\alpha))$ because $(x,y,z) = (\frac{u+v}{\sqrt2}, \frac{-u+v}{\sqrt2}, w)$
And there you have it, a parametrized unit circle in the $x+y=0$ plane (you can check that it is in this plane because $x + y = \frac{sin(\alpha)}{\sqrt2} - \frac{sin(\alpha)}{\sqrt2} = 0$)
The Radius of the circle is given by $R$ in
$$ R^2 = x^2 + y^2 + z^2 = \left(\frac{sin(\alpha)}{\sqrt2}\right)^2 + \left(-\frac{sin(\alpha)}{\sqrt2}\right)^2 + cos(\alpha)^2 $$
$$ = \frac{sin^2(\alpha)}{2} + \frac{sin^2(\alpha)}{2} + cos^2(\alpha) = sin^2(\alpha) + cos^2(\alpha) = 1$$
Here is what I think. The way to derive a circle is to set up the identity of distance, i.e. the square of the distance of a point on the circle to its center (1 for unit circle) is equal to the sum of squares(Pythagorean theorem). If you are given the points on $x+y=0$ plane, then the setting of such points would be limited to $(x,-x,z)$. Do the same thing. Since now you are guaranteed the points in a plane, it is for sure a circle, not a sphere.
