Parametrization of the intersection of two surfaces in $\mathbb R^3$ [closed]

Solution 1:

Plotted above are your two equations, the first in purple, and the second in blue. Observe that,

\begin{align*} 2&=1+1\\\\ &=(x^2+3y^2+z^2)+(x^2-3y^2+z^2)\\\\ &=2x^2+2z^2\\\\ 2&=2(x^2+z^2)\\\\ 1&=x^2+z^2 \end{align*}

Note that this is a circle of radius 1 in the $xz$-plane.

The parametrization of a circle is given by

\begin{align*} \begin{bmatrix}x\\z\end{bmatrix}=\begin{bmatrix}R\cos t\\ R\sin t\end{bmatrix} \end{align*}

Since our circle is of radius 1, $R=1$. Then we have

\begin{align*} \begin{bmatrix}x\\z\end{bmatrix}=\begin{bmatrix}\cos t\\ \sin t\end{bmatrix} \end{align*}

Extending to $\mathbb{R}^3$, we have,

\begin{align*} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\cos t\\0\\ \sin t\end{bmatrix}=r(t) \end{align*}

Equivalently written, $r(t)=(\cos t,0,\sin t)$.