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I have been looking into this question : we have two surfaces :

$$\big\{(x,y,z)\in \mathbb{R}^3 \mid\;\; S_1\colon\;\; x+z=1 ,\;\; S_2\colon\;\; x^2+y^2=1 \big\}$$

we need to draw or describe the "shape" that we get . I tried to solve it by drawing the two surfaces and imagining the intersection which is an ellipse in $\mathbb{R}^3$.

but in the solution they told us that we can play with equations and get an equation that resembles an ellipse , so I tried doing this but I don't know how to continue :

I have now : $$x^2+y^2=x+z$$

somehow I think we need to uncover the ellipse equation .

So how can we do this ?

M. Strochyk
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Waseem Francis
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3 Answers3

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The approach is wrong. An equation in $\mathbb{R}^3$ is a surface, but an intersection of two surfaces usually is something with lower dimension. $$x^2+y^2=1$$ $$x+z=1$$ are your equations, the right way to go would be to get the parametric equations for the plane $x+z=1$ and put them inside the other. $$(a,b,1-a)$$ describes your plane for all $a,b\in\mathbb{R}$. Now $$a^2+b^2=1$$ are the desired points for the parameters which you know how to solve: $$a=\cos{t}$$ $$b=\sin{t}$$ with $t\in[0,2\pi]$. So your surface is described by: $$(\cos{t},\sin{t},1-\cos{t}).$$

Lolman
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The surface $S_2$ can be described by $(x,\,y,\,z)=(\cos u,\sin u,v)$, $u\in[0,2\pi]$ and $v\in\mathbb{R}$. Then, the curve we are looking is $(x,y,z)=(\cos u, \sin u, 1-\cos u)$, $u\in[0,2\pi]$.

In order to prove that the curve is an ellipse let $A$ be the point $(\frac{1}{\sqrt{2}},0,1-\frac{1}{\sqrt{2}})$ and $B$ the point $(-\frac{1}{\sqrt{2}},0,1+\frac{1}{\sqrt{2}})$, so

\begin{align*} d(A,P)+d(B,P)&=\sqrt{\left(\cos u-\frac{1}{\sqrt{2}}\right)^2+\sin^2 u+\left(-\frac{1}{\sqrt{2}}+\cos u\right)^2}+\sqrt{\left(\cos u+\frac{1}{\sqrt{2}}\right)^2+\sin^2 u+\left(-\frac{1}{\sqrt{2}}-\cos u\right)^2}\\ &=\sqrt{2\cos^2 u-2\sqrt{2}\cos u+1+\sin^2 u}+\sqrt{2\cos^2 u+2\sqrt{2}\cos u+1+\sin^2 u}\\ &=\sqrt{\cos^2 u-2\sqrt{2}\cos u+2}+\sqrt{\cos^2 u+2\sqrt{2}\cos u+2}\\ &=\sqrt{2}-\cos u +\sqrt{2}+\cos u\\ &=2\sqrt{2} \end{align*} Where $P$ is a point of the curve. So that curve is an ellipse.

Ángel Mario Gallegos
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The surfaces are a vertical cylinder and a oblique plane. The intersection is obviously a ellipse. Do you want a parametrization? Project on the plane $XY$, parametrize the projection and use the other equqtion: $$x=\cos t,$$ $$y=\sin t,$$ $$z= 1 - \cos t,$$ $(t\in [0,2\pi])$.

Martín-Blas Pérez Pinilla
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