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Describe the motion of the path $$ \mathbf{r}=3\cos{t}\mathbf{\hat{i}}+4\cos{t}\mathbf{\hat{j}}+5\sin{t}{\mathbf{\hat{k}}} $$ The answers is:

Path: the circle of intersection of the sphere $x^2+y^2+z^2=25$ and the plane $4x=3y$.

I understand that the sphere has radius 5 since: $$ \left|\mathbf{r}\right|= \sqrt{(3\cos{t})^2+(4\cos{t})^2+(5\sin{t})^2)}= \sqrt{25\cos^2{t}+25\sin^2{t}}=5 $$ But I don't understand the connection to the plane or how to find it...

Thanks!

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    $\begingroup$ Notice that the coordinates of the point $(3\cos t,\;4\cos t,\;5\sin t)$ satisfy the equation $4x=3y$ for any real number $t$. $\endgroup$ – Ángel Mario Gallegos Jun 7 '16 at 6:40
  • $\begingroup$ Is it because $z(t)=5\sin{t}$ always is zero? So we have $(3\cos t,\;4\cos t,\;0)$ which satify $4x-3y=0$. Is it right? $\endgroup$ – JDoeDoe Jun 7 '16 at 9:50
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The way to find a plane in your question is as follows:

You must understand that the equation of the plane will be devoid of $t$, i.e. would be something like a locus..

It will consist of single powers of $x,y,\& z$

So you see, if you want to make a linear relation of $z$ with $x$ or $y$, you can't avoid $\tan(t)$ or something like that. The only way to do is to find a linear relation of $x$ and $y$. Since they have the same $\cos(t)$ which cancels out in a linear relation.

Hence the answer

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