Intersection of $x+y+z=0$ and $x^2+y^2+z^2=1$ I got a problem which is to show that the circle that is the intersection of the plane $x+y+z=0 $ and the sphere $x^2+y^2+z^2=1$ can be expressed as
$$x(t)=\frac{\cos(t)-\sqrt{3}\sin(t)}{\sqrt{6}},$$
$$y(t)=\frac{\cos(t)+\sqrt{3}\sin(t)}{\sqrt{6}}$$
and
$$z(t)= \frac{-2\cos(t)}{\sqrt{6}}.$$
The circle turns out like a $45$ degree circle in three dimensions. I know if the circle is in two dimensions like $x$ and $y$, the parametrization will be $(x(t),y(t))=(r\cos(t),r\sin(t))$. But in this question , what should I do?
Here is how I solved for the intersection of the plane $x+y+z=1$ and the sphere $x^2+y^2+z^2 = 1$, which is a little different with your question but the steps are typically the same.
Here is how I solved for the intersection of the plane $x+y+z=1$ and the sphere $x^2+y^2+z^2 = 1$, which is a little different with your question but the steps are typically the same.