Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$

Question

I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?

Answer 1

When you substitute $z=-(x+y)$ into $x^2+y^2+z^2=6$ we get \begin{eqnarray*} x^2+y^2+(x+y)^2=6 \\ x^2+xy+y^2 =3. \end{eqnarray*} Now multiply this by $4$ & complete the square \begin{eqnarray*} (2x+y)^2+3y^2=12. \end{eqnarray*} This is easily parameterised by $2x+y = \sqrt{12} \cos ( \theta)$ and $y = 2 \sin ( \theta)$. \begin{eqnarray*} x &=& \frac{\sqrt{12} \cos ( \theta)-2 \sin ( \theta)}{2} \\ y &=& 2 \sin ( \theta) \\ z &=& \frac{-\sqrt{12} \cos ( \theta)- 2\sin ( \theta)}{2} \end{eqnarray*}

Answer 2

The equation $$ x^2 + y^2 + z^2 = 6 $$ gives a sphere around the origin with radius $R=\sqrt{6}$. $$ 0 = x + y + z = (1,1,1) \cdot (x,y,z) $$ gives a plane $H$ with normal vector $(1,1,1)^\top$ including the origin.

The intersection is a circle of radius $R$ on that plane $H$, with the origin as midpoint.

GeoGebra seems to give: $$ u(t) = (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t))^\top $$ Checking: $$ (-\sqrt{3} \cos(t) +\sin(t))^2 + (\sqrt{3} \cos(t) + \sin(t))^2 + (-2 \sin(t))^2 = \\ 3 \cos^2(t) + \sin^2(t) - 2\sqrt{3} \cos(t)\sin(t) + \\ 3 \cos^2(t) + \sin^2(t) + 2\sqrt{3} \cos(t)\sin(t) + \\ 4 \sin^2(t) = \\ 6 \cos^2(t) + 6 \sin^2(t) = 6 $$ so the curve lies on the sphere. $$ (1,1,1) \cdot (-\sqrt{3} \cos(t) +\sin(t), \sqrt{3} \cos(t) + \sin(t), -2 \sin(t)) = 0 $$ so that curve is on the plane $H$.

How to get there? One might try a coordinate transformation of a circle in the $x$-$y$-plane with radius $r$ onto that plane by two rotations:

We would need $$ T\,(0,0,1)^\top = (1,1,1)/\sqrt{3} $$

Answer 3

$z=(-x-y)$ leads to $$ x^2+y^2+(x+y)^2 = 6 $$ that is equivalent to $x^2+xy+y^2=3$, i.e. the equation of an ellipse in the $xy$-plane. Such ellipse goes through $(1,1)$, hence by considering the intersections between such ellipse and the lines through $(1,1)$, i.e. by solving $x^2+xy+y^2=3$ and $y=m(x-1)+1$, we get that $$ (x,y)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1}\right) $$ for $m\in\overline{\mathbb{R}}$ is a parametrization of the ellipse in the $xy$-plane and $$ (x,y,z)=\left(\frac{m^2-2m-2}{m^2+m+1},\frac{-2m^2-2m+1}{m^2+m+1},\frac{m^2+4m+1}{m^2+m+1}\right) $$ is a parametrization of the wanted section (circle) in the Euclidean space.

Answer 4

With spherical coordinate system, $r=\sqrt{6}$ and $$z=-x-y=-\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi$$ so your parametric equation will be $$ \begin{aligned}x&=\sqrt{6}\,\sin \theta \,\cos \varphi \\y&=\sqrt{6}\,\sin \theta \,\sin \varphi \\z&=\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi \end{aligned}$$

where $ θ ∈ [0, π], φ ∈ [0, 2π)$.