0

I need to find that intersection and I followed every step in the answer here: Parametrization for intersection of sphere and plane. $x^2+y^2+z^2=81$, $x+y+z=15$.

(the first method), I've plugged $z = -3x-2y$ in the sphere equation and got $$10\cdot\left(x+\frac{3}{10}y\right)^{2}+\frac{41}{10}y^{2}=1$$ So parametrization for that ellipse is $$\left(\frac{1}{\sqrt{10}}\cdot\cos(t)-\frac{3}{10}\cdot\sqrt{\frac{10}{41}}\cdot\sin(t),\sqrt{\frac{10}{41}}\cdot\sin(t)\right)$$ this is correct, but now if I set $$z(t) = -3\cdot\left(\frac{1}{\sqrt{10}}\cdot\cos(t)-\frac{3}{10}\cdot\sqrt{\frac{10}{41}}\cdot\sin(t)\right)-2\sqrt{\frac{10}{41}}\cdot\sin(t)$$ the parametrization $$(x(t),y(t),\underbrace{-3x(t)-2y(t)}_{z(t)})$$ is not on the sphere, I don't get what I did wrong, I'll be glad for some explanation why it's not correct.

Thank you all in advance!

chirico
  • 1,339
oneneedsanswers
  • 161

1 Answers1

2

In case that you are interested in a different approach: The plane contains the origin, and the center of the sphere is the origin. The sphere is a unit sphere. This means that the intersection is a circle with radius $1$ which has its center also at the origin. All we need are two orthogonal normalized vectors $v$ and $w$ which are perpendicular to the plane's normal $(3\;2\;1)^T.$ Then $$ \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} = v\,\cos(t)+w\,\sin(t) $$ Those vectors can easily be found. The first one (without normalization) can be found by switching two coordinates, set one value to the negative value, and set the third one to $0$. Using the first and the third coordinate, we get $\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}.$ The other vector must be orthogonal to this one and to the plane's normal, so we can use the cross product and get $2\begin{pmatrix} 3 \\ -5 \\ 1 \end{pmatrix}.$ After normalization, we get $$ v = \frac{1}{\sqrt{10}}\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} \;\;\;\;,\;\;\;\; w = \frac{1}{\sqrt{35}}\begin{pmatrix} 3 \\ -5 \\ 1 \end{pmatrix} $$

Reinhard Meier
  • 8,302