Parametrization of the intersection between a sphere and a plane

Question

I can't find a way to get the parametric equation $\gamma(t)=(x(t),y(t),z(t))$ of a curve that is the intersection of a sphere and a plane (not parallel to any coordinate planes). That is

$$\begin{cases} x^2+y^2+z^2=r^2 \\ ax+by+cz=d \end{cases}$$

I don't know how to move the variables in order to get something easily parameterizable.\

Can anyone let me know the main steps to get the parameterization of this type of curve?

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Example

$$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 \end{cases}$$

Answer: $\gamma(t)=(\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,-\frac{\sqrt{2}}{2}cost +\frac{\sqrt{6}}{6}sint,\frac{\sqrt{6}}{3}sint) , \,\,\,\, t \in [0,2\pi]$

Answer 1

Just to present a vectorial approach to the problem

Consider the plane equation written as: $$ \frac{{a\,x + b\,y + c\,z}}{{\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } }} = \frac{d}{{\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } }} $$ That means that the points ${\bf p} = \left( {x,y,z} \right)$ on the plane shall project onto the unit vector ${\bf n} = \left( {a,b,c} \right)/\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } $ at a constant distance $\delta = d/\sqrt {a^{\,2} + b^{\,2} + c^{\,2} } $, i.e. $$ {\bf p} \cdot {\bf n} = \delta $$ so that the plane is distant $\delta$ from the origin. At the same time we shall have $$ \left| {\,{\bf p}\,} \right| = r $$ So with reference to the sketch, we can put $$ {\bf p} = \delta \,{\bf n} + \sqrt {r^{\,2} - \delta ^{\,2} } \,{\bf t} = \delta \,{\bf n} + \rho \,{\bf t} $$ where ${\bf t}$ is a generic unit vector parallel to the plane, that is normal to ${\bf n}$. We can express ${\bf t}$ by taking two unit vectors ${\bf u}$ and ${\bf v}$, normal to ${\bf n}$ and to each other, and then putting $$ {\bf t} = \cos \theta \,{\bf u} + \sin \theta \,{\bf v} $$ To determine ${\bf u}$ and ${\bf v}$ we can take (if, e.g. $c \ne 0$) $$ {\bf u} = \left( {0, - c,b} \right)/\sqrt {b^{\,2} + c^{\,2} } \quad \quad {\bf v} = {\bf n} \times {\bf u} $$ example

with $r=4 \; a=1 \; b=2 \; c=3 \; d=7$ we get

$$ {\bf n} = \,\sqrt {14} /14\,\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right)\;\quad \delta = \sqrt {14} /2\quad \rho = 5\sqrt 2 /2 $$ $$ {\bf u} = \sqrt {13} /13\;\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right)\quad {\bf v} = \sqrt {182} /182\;\left( {\begin{array}{*{20}c} {13} \\ { - 2} \\ { - 3} \\ \end{array}} \right) $$

and thus $$ \begin{array}{l} {\bf p} = \delta \,{\bf n} + \rho \,\left( {\cos \theta \,{\bf u} + \sin \theta \,{\bf v}} \right)\quad \Rightarrow \\ \Rightarrow \quad \left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right) = \frac{1}{2}\,\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right) + \frac{{5\sqrt {364} }}{{364}}\,\left( {\sqrt {14} \cos \theta \;\left( {\begin{array}{*{20}c} 1 \\ 2 \\ 3 \\ \end{array}} \right) + \sin \theta \,\;\left( {\begin{array}{*{20}c} {13} \\ { - 2} \\ { - 3} \\ \end{array}} \right)} \right) \\ \end{array} $$ and you can verify that: $$ {\bf p} \cdot {\bf n} = \delta \,\quad {\bf p} \cdot {\bf p} = r^{\,2} $$

Answer 2

If the plane passes through the origin (so $d = 0)$, the intersection will be a circle of radius $1$ that lies on the plane. Thus, you can choose a pair $(e_1,e_2)$ of orthogonal vectors of unit length that lie on your plane and then the intersection will be parametrized as

$$ \gamma(t) = \cos(t) e_1 + \sin(t) e_2. $$

To calculate this explicitly, you can choose $e_1$ to be any unit length vector that lies on the plane and set $e_2 = e_1 \times \frac{(a,b,c)}{\|(a,b,c)\|}$ (as $(a,b,c)$ is the normal vector to the plane). For example, for $x + y + z = 0$ we can take

$$ e_1 = \frac{(1,-1,0)}{\|(1,-1,0)\|} = \frac{1}{\sqrt{2}}(1,-1,0), \\ e_2 = \frac{1}{\sqrt{2}} (1,-1,0) \times \frac{1}{\sqrt{3}} (1,1,1) = \frac{1}{\sqrt{6}} (-1,-1,2) $$

and so

$$ \gamma(t) = \cos(t) \frac{1}{\sqrt{2}} (1,-1,0) + \sin(t) \frac{1}{\sqrt{6}} (-1,-1,2) = \left( \frac{\cos t}{\sqrt{2}} - \frac{\sin t}{\sqrt{6}}, -\frac{\cos t}{\sqrt{2}} - \frac{\sin t}{\sqrt{6}}, \frac{2}{\sqrt{6}} \sin t\right) $$

which is almost the parametrization up wrote but note that you have a sign error in your parametrization.

Answer 3

Hint

Let $D=\frac{|d|}{\sqrt{a^2+b^2+c^2}}$ be the distance between the plane and the center of the sphere.

There are three cases

$D>a \implies $ no interesection

$D=a \implies $the plane is tangent.

$D<a \implies $ the interesection is a circle of radius$=\sqrt{a^2-D^2}$.