Parametric equation of a curve in $R^3$

Question

How to find the parametric equation of the curve in $R^3$, which is the intersection of the sphere of radius $a>0$ centred at the origin, and the plane $x+y+z=0$?

I've tried to start looking for $x$ and $y$, but I had no sucess.

Thanks in advance!

Answer 1

First, find an orthonormal basis $\{ \vec{u}_1, \vec{u}_2 \}$ for the plane $x + y + z = 0$. Start with any old vector in the plane: $$ \vec{v}_1 = \begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix} $$ will do. Notice that you just have to make the coordinates sum to zero. To find an orthogonal vector $\vec{v}_2$ in the plane, engineer it so that $\vec{v}_1 \cdot \vec{v}_2 = 0$. This isn't too hard: $$ \vec{v}_2 = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} $$ works. (You could have found $\vec{v}_2$ up to a scalar by computing the vector product $\vec{n} \times \vec{v}_1$, where $$ \vec{n} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$ is the normal vector to the plane.)

Second, normalize the vectors (make them have unit length): $$ \vec{u}_1 = \frac{\vec{v}_1}{\| \vec{v}_1 \|} = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix} $$ and $$ \vec{u}_2 = \frac{\vec{v}_2}{\| \vec{v}_2 \|} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} $$

Finally, parametrize the circle of radius $a$ in terms of the basis $\{ \vec{u}_1, \vec{u}_2 \}$: $$ \begin{align} \vec{\gamma}(t) &= (a \cos t) \vec{u}_1 + (a \sin t) \vec{u}_2 \\ &= \frac{a}{\sqrt{6}} \begin{bmatrix} \cos t - \sqrt{3} \sin t \\ \cos t + \sqrt{3} \sin t \\ -2 \cos t \end{bmatrix} \end{align} $$

Answer 2

This is a circle that projects to an ellipse in the $x$-$y$ plane. Once you parametrize the projection as $(x(t),y(t))$ you can parametrize the original circle as $(x(t),y(t),-x(t)-y(t))$.

The equation of the sphere is $x^2+y^2+z^2=a^2$, substituting $z=-x-y$ into it gives $x^2+xy+y^2=a^2/2$. This is the projected ellipse. To eliminate the $xy$ term we change variables $x=x'+y'$, $y=x'-y'$, this is the usual rotation of axes for ellipses up to a factor. This brings us to $3x'^2+y'^2=a^2/2$, which is easy to parametrize since $\cos^2 t+\sin^2 t=1$. Namely, $x'(t)=\frac{a}{\sqrt{6}}\cos t$ and $y'(t)=\frac{a}{\sqrt{2}}\sin t$, so for the circle: $$\Big(\frac{a}{\sqrt{6}}\cos t+\frac{a}{\sqrt{2}}\sin t,\, \frac{a}{\sqrt{6}}\cos t-\frac{a}{\sqrt{2}}\sin t,\, -\frac{2a}{\sqrt{6}}\cos t\Big).$$