Find the rectangular equation for the surface given by $\displaystyle r(u,v)=\frac{v}{2}i+uj+vk$ by eliminating the paranmeter.
I'm having a bad day. I can see that $\displaystyle 2x=z$. But, that's as far as I get. Any Help?
As it turns out, $\displaystyle 2x=z$ is the equation of the plane.
The position vector, $\displaystyle \displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k,$ gives the coordinates of a point at $\displaystyle \displaystyle (x,\ y,\ z)$$\displaystyle \displaystyle =\left(\frac{v}{2},\ u,\ v\right)\,.$
$\displaystyle \displaystyle \vec r(0,0)=0\hat i+0\hat j+0\hat k,$ so the surface passes through the origin, (0, 0, 0).
Define vector, $\displaystyle \displaystyle \vec A$$\displaystyle \displaystyle =\vec{r}(2,\ 0)-\vec{r}(0,\ 0)=1\hat i+0\hat j+2\hat k$.
Define vector, $\displaystyle \displaystyle \vec B$$\displaystyle \displaystyle =\vec{r}(0,\ 1)-\vec{r}(0,\ 0)=0\hat i+1\hat j+0\hat k$.
Notice that any vector of the form $\displaystyle \displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k,$ may be written as a linear
combination of the constant vectors, $\displaystyle \displaystyle \vec A$ and $\displaystyle \displaystyle \vec B$.
Therefore, the surface specified by $\displaystyle \displaystyle \vec r(u,v)$ is a plane.
Vectors, $\displaystyle \displaystyle \vec A$ and $\displaystyle \displaystyle \vec B$ lie in the plane, so their vector product (cross product) is perpendicular to the plane.
$\displaystyle \displaystyle \vec A\times\vec B = \begin{vmatrix}
\ \hat i&\hat j&\hat k\ \\
\ 1&0&2\ \\
\ 0&1&0\ \end{vmatrix}=-2\,\hat i+0\,\hat j+1\,\hat k$
So, the equation of the plane is $\displaystyle \displaystyle -2(x-0)+0(y-0)+1(z-0)=0,$
which may be written as $\displaystyle \displaystyle 2x-z=0\,.$