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Consider the set: $$ A \equiv \left\{ (x,y,z) \in \mathbb{R}^{3} | x + y + z = 0\ \mathrm{and}\ x^{2} + y^{2} + z^{2} = 1 \right\} $$

This is the intersection of the unit sphere with the above plane.

Is there a way to parametrize this set with a curve $\gamma : [0,1] \to A$ going from the point $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$?

I simply can't figure it out since we have the extra constraint of being in the plane $x+y+z=0$

QuantumEyedea
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2 Answers2

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Hint:

The locus of points formed by intersecting a plane with a sphere is a circle (assuming the plane isn't simply tangent to the sphere). One can show that every circle in $\mathbb{R}^3$ centered at the origin can be parametrized as follows:

$$\alpha(t) = \mathbf{v}_1 \cos(t) + \mathbf{v}_2 \sin(t)$$

where $\mathbf{v}_1$ and $\mathbf{v}_2$ are orthogonal vectors that lie in the same plane as the circle with magnitude equal to the radius of the circle.

Fortunately, this plane cuts the origin so we don't have to worry about translating anything. The task at hand is therefore reduced to finding two vectors that satisfy the above. Even better, you've already been given one possible vector.

Kaj Hansen
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  • @Kaj_Hansen I see how $\alpha(0) = \mathbf{v}{1}$ and $\alpha(2\pi) = \mathbf{v}{2}$, so that the path connects these two points. What if I have $\mathbf{v}{1}$ and $\mathbf{v}{2}$ on the surface of the sphere, but that are $not$ orthogonal? For simplicity, setting $\mathbf{v}{1} = (1,0,0)$ and keeping $\mathbf{v}{2} = (x,y,z)$ (normalized to 1, of course). If you can do this in general, can one use this construction for higher-dimensional spheres too? – QuantumEyedea Apr 08 '17 at 00:20
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When a plane intersects a sphere the curve of intersection will be a circle.

the origin is at the center of this cricle (because the plane goes through the origin, and the sphere is centered at the origin). Now you just need to parametarize the circle.

We have two points on this circle, but what we really need are orthogonal vectors. We can use one of the points, and find a vector in the plane that is orthogonal.

$(-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)$ is a vector orthogonal to $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ that lies in the plane.

$(x,y,z) = (-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cos t + (-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)\sin t$

Will trace out the circle.

And for what bounds of $t$ will will that equation trace to the endpoints we seek?

$(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cdot(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}) = 0.5$

$t = \frac{\pi}{3}$ or $\frac{-\pi}{3}$

$t\in[0,\frac {\pi}{3}]$

Doug M
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