Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $

Question

Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation:

\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j , \ \ 0 \leq t \leq 2 \pi \end{align} But I am not sure . Any help is there ?

Answer 1

$\mathbf u = (\frac {\sqrt 2}{2} \mathbf i - \frac {\sqrt 2}{2} \mathbf j)\\ \mathbf v = (\frac {\sqrt 6}{6} \mathbf i + \frac {\sqrt 6}{6} \mathbf j -\frac {\sqrt 6}{3} \mathbf k)$

$(x,y,z) = (1,1,-1) + 3\mathbf u \cos t + 3\mathbf v \sin t\\ x = 1 + 3\frac {\sqrt 2}{2} \cos t + \frac{\sqrt {6}}{2} \sin t\\ y = 1 - 3\frac {\sqrt 2}{2} \cos t + \frac {\sqrt {6}}{2} \sin t\\ z = -1 - \sqrt {6} \sin t$

Answer 2

The intersection of the sphere $$(x-1)^2+(y-1)^2+(z+1)^2=9$$ and the plane $$x+y+z=1$$ can be parametrised by spherical coordinates : $$x=1+3\sin (\phi)\cos (\theta) $$ $$y=1+3\sin (\phi)\sin (\theta) $$ $$z=-1-3\sin (\phi)(\cos (\theta)+\sin (\theta)) $$.

with $0\le \theta \le 2\pi $ and $0\le \phi \le \pi$.

Answer 3

The plane can be parameterised as follows \begin{eqnarray*} x=1+t \\ y=1+s \\ z=-1-t-s. \end{eqnarray*} Now substitute this into the equation for the sphere $(x-1)^2+(y-1)^2+(z+1)^2=9$. We have \begin{eqnarray*} t^2+s^2+(t+s)^2=9 \\ \end{eqnarray*} Rearrange this to

\begin{eqnarray*} (2t+s)^2+3s^2=18 \end{eqnarray*} This can be paramterised

\begin{eqnarray*} s=\sqrt{6} \sin(\theta) \\ 2t+s= \sqrt{18} \cos(\theta) \end{eqnarray*} Now substitute back into the parameterisation of the plane

\begin{eqnarray*} x=1+ \frac{\sqrt{18} \cos(\theta) - \sqrt{6} \sin(\theta)}{2} \\ y=1+\sqrt{6} \sin(\theta) \\ z=-1-\frac{\sqrt{18} \cos(\theta)+\sqrt{6} \sin(\theta)}{2} \end{eqnarray*}