parametrization of plane in $\mathbb R^3$

Question

Parametrize the plane in $\mathbb R^3$ with direction vectors $\hat u$ and $\hat v$ and through the point $p$ as in representation as the range of a $C^1$ function $f:\mathbb R^2\to\mathbb R^3$. (Use parameters $s$ and $t$ and use $f(s,t)$).

Then write the equation of the plane as the locus of a linear function $F(x,y,z)=0$. What is the relation of gradient of $F$ to $d_sf \times d_tf$?

So for the first part, I come up with $f(s,t)=(p_1+tu_1+sv_1,p_2+tu_2+sv_2,p_3+tu_3+sv_3)$, is it correct?.

But then I am confused on how to write the equation as $F(x,y,z)=0$

Answer 1

If you write your parametrized plane in vector form $\boldsymbol\pi = \boldsymbol p + r \boldsymbol{u} + {s} \boldsymbol v$ where $\boldsymbol\pi = (x, y, z)$ is an arbitrary point in your plane and use this to rewrite your plane equation as a dot product $ (\boldsymbol\pi-\boldsymbol p) \cdot (\boldsymbol u \times \boldsymbol v) = 0$ you can just expand that and you'll have an equation in $x, y, z$.

Answer 2

If $\boldsymbol c=d_sf\times d_tf$ then $F(w)=\boldsymbol c\bullet (w-p)$ will work, that is
$$F(x,y,z)=c_1(x-p_1)+c_2(y-p_2)+c_3(z-p_3),$$ where $\boldsymbol c=c_1e_1+c_2e_2+c_3e_3$, is the function what you are seeking. From this expression of $F$ you can see that the ${\rm grad}F=d_sf\times d_tf$.