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For example, let the surface $S$ in $\mathbb R^3$ be formed by taking the union of the straight lines joining pairs of points on the lines $$\{x=t,y=0,z=1\},\qquad \{x=0,y=1,z=t\}$$ with the same parameter $t$.

Then this surface can be given in the parametric form, say $\gamma$: $$x=ut,\quad y=1-u,\quad z=u+t-ut.$$

I was wondering how to understand $S$ can be given in(?) $\gamma$.

I notice that if we take $u=1$ and $u=0$ in $\gamma$, then respectively, we could get the equations of two original lines. But, does this suffice to show that "$S$ can be given in $\gamma$"? I guess this only shows that the $S$ contains in the curve described by $\gamma$. If so, how to show its converse? Thanks.

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1 Answer 1

For any $t_0$ the equation of the straight line joining the pair of points $(t_0,0,1)$ and $(0,1,t_0)$ is given, parametrically, by $u\cdot (t_0,0,1)+(1-u)\cdot (0,1,t_0)=(ut_0, 1-u, u+t_0-ut_0).$ Hence $\gamma$ can be given in parametric form by the desired equation.

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