We have $\gamma:I\to \Bbb R^3$ a regular and simple curve which lies in a plane that does not contain the origin. Then we have a unit vector $c$ and a nonzero scalar d s.t. $\gamma \cdot c \equiv d$.
Prove that $\sigma :I\times (0,\infty) \to R^3$ given by $(x,y)\mapsto y\gamma(x)$ is an injective and regular surface patch.
So the curve is injective because it's simple?
And to show it's a regular surface patch, I have to show it's smooth, the vectors $\sigma_x$ and $\sigma_y$ are linearly independent and the vector product between vectors are nonzero at every point of $I$. Though I'm not sure how to show it's regular.
A curve $\gamma\colon [a,b] \to \Bbb R^3$ is simple if $\gamma(a) = \gamma(b)$ and $\gamma\big|_{]a,b[}$ is injective.
Now, the surface patch $\sigma$ is injective because $\gamma$ is injective and $y > 0$ always. If $y_1\gamma(x_1) = y_2\gamma(x_2)$, dotting it with $c$ gives $y_1d=y_2d$. Since $d \neq 0$ we have $y_1 = y_2$. Then $y_1\gamma(x_1) = y_1\gamma(x_2)$ gives $\gamma(x_1) = \gamma(x_2)$ and so $x_1 = x_2$, as wanted.
Now we want to see that $\sigma$ is regular. That is, we want to see that the derivatives $\sigma_x(x,y) = y\gamma'(x)$ and $\sigma_y(x,y) = \gamma(x)$ are always linearly independent. One checks this as follows: differentiating $\gamma(x) \cdot c = d$ one obtains $\gamma'(x) \cdot c = 0$. Write the linear combination $$ay\gamma'(x) + b\gamma(x) = 0.$$Dot that with $c$, so $$0=ay\gamma'(x)\cdot c + b \gamma(x) \cdot c = bd,$$and so $b = 0$. Since $y>0$ and $\gamma$ is regular, $ay\gamma'(x) = 0$ implies $a=0$.