1. Parametric Surfaces

Find the rectangular equation for the surface given by $r(u,v)=\frac{v}{2}i+uj+vk$ by eliminating the paranmeter.

I'm having a bad day. I can see that $2x=z$. But, that's as far as I get. Any Help?

2. Originally Posted by VonNemo19
Find the rectangular equation for the surface given by $r(u,v)=\frac{v}{2}i+uj+vk$ by eliminating the parameter.

I'm having a bad day. I can see that $2x=z$. But, that's as far as I get. Any Help?

As it turns out, $2x=z$ is the equation of the plane.

The position vector, $\displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k,$ gives the coordinates of a point at $\displaystyle (x,\ y,\ z)$ $\displaystyle =\left(\frac{v}{2},\ u,\ v\right)\,.$

$\displaystyle \vec r(0,0)=0\hat i+0\hat j+0\hat k,$ so the surface passes through the origin, (0, 0, 0).

Define vector, $\displaystyle \vec A$ $\displaystyle =\vec{r}(2,\ 0)-\vec{r}(0,\ 0)=1\hat i+0\hat j+2\hat k$.

Define vector, $\displaystyle \vec B$ $\displaystyle =\vec{r}(0,\ 1)-\vec{r}(0,\ 0)=0\hat i+1\hat j+0\hat k$.

Notice that any vector of the form $\displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k,$ may be written as a linear

combination of the constant vectors, $\displaystyle \vec A$ and $\displaystyle \vec B$.

Therefore, the surface specified by $\displaystyle \vec r(u,v)$ is a plane.

Vectors, $\displaystyle \vec A$ and $\displaystyle \vec B$ lie in the plane, so their vector product (cross product) is perpendicular to the plane.

$\displaystyle \vec A\times\vec B = \begin{vmatrix}
\ \hat i&\hat j&\hat k\ \\
\ 1&0&2\ \\
\ 0&1&0\ \end{vmatrix}=-2\,\hat i+0\,\hat j+1\,\hat k$

So, the equation of the plane is $\displaystyle -2(x-0)+0(y-0)+1(z-0)=0,$

which may be written as $\displaystyle 2x-z=0\,.$

3. Hey man, thanks a lot.