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Find the parametric equation

Hi guys i'm stuck on this problem, Find the parametric equation of the curve generated by the intersection of: \(x^{2} + y^{2} + z^{2} = a^{2}\)

and

\(x + y + z = 1\)

\(a\) is constant

I know that is a circunference on the plane \(x+y+z=1\) but i can't find the equation, since i don't know how to make a 3D rotation. I'm thinking on how to use the simmetry of the problem.

About parametric equation:

https://en.wikipedia.org/wiki/Differentialgeometryof_curves

http://en.wikipedia.org/wiki/Parametric_equation

Note by Felipe Sousa
3 years, 11 months ago

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First, what is a parametrization of the intersection of the sphere \( x^2 + y^2 + z^2 = a^2 \) and the plane \( z = 1/\sqrt{3} \)? This is easy: \[ (x(\theta), y(\theta), z(\theta)) = \left( \sqrt{a^2 - 1/3} \cos \theta, \sqrt{a^2 - 1/3} \sin \theta, 1/\sqrt{3} \right), \] which, as we expect, is defined if and only if \( a \ge 1/\sqrt{3} \). Now all we need to do is rotate this parametrization in the plane \( x = y \) by a suitable angle \( \alpha = \tan^{-1} \sqrt{2} \). But it's not quite that simple; we multiply by an initial rotation matrix \[ R_{xz}(\alpha) = \begin{bmatrix} \cos\alpha & 0 & \sin\alpha \\ 0 & 1 & 0 \\ -\sin\alpha & 0 & \cos \alpha \end{bmatrix}, \] followed by a (counterclockwise) rotation of \( \beta = \pi/4 \) in the \(xy\)-plane: \[ R_{xy}(\beta) = \begin{bmatrix} \cos \beta & -\sin\beta & 0 \\ \sin\beta & \cos\beta & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] Therefore the transformed parametrization is \[ (x'(\theta), y'(\theta), z'(\theta)) = R_{xy}(\pi/4) R_{xy}(\tan^{-1}\sqrt{2}) \cdot (x(\theta), y(\theta), z(\theta)), \] which gives \[ \begin{align*} x'(\theta) &= \frac{1}{3}\left( 1 + \sqrt{6a^2-2} \cos(\theta+\frac{\pi}{3}) \right), \\ y'(\theta) &= \frac{1}{3} \left(1 + \sqrt{6a^2-2} \cos(\theta-\frac{\pi}{3}) \right), \\ z'(\theta) &= \frac{1}{3} \left( 1 - \sqrt{6a^2-2} \cos\theta \right), \end{align*} \] for \( \theta \in [0, 2\pi) \). It is then easy to verify that \( x' + y' + z' = 1 \) and \( x'^2 + y'^2 + z'^2 = a^2 \). Note that any such parametrization is not unique: we can find any number of equivalent parametrizations for this curve. Hero P. · 3 years, 11 months ago

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