How to parametrize this region surface

Question

$S$ is the portion of the plane $$x+2y-3z=3$$ in the octan bounded by the positive direction of the $x$ and $y$ axis and the negative direction of the $z$ axis.

How can I parametrize this crazy region??

Answer 1

We can let x = t, y = s, and s > 0, t > 0. Then we need that z < 0. So:

x + 2y - 3 = 3z < 0 ==> x + 2y < 3 ==> t + 2s < 3. Thus we can have the parametrization for

the region S:

S = {(t, s, t/3 + 2s/3 - 1): s > 0, t > 0, and t + 2s < 3}

Note that if we use the condition 0 < s < 3/2 and 0 < t < 3 we will still violate the condition that z < 0. Take for example s = 1.499, and t = 2.999, then t + 2s = 2.999 + 2(1.499) = 5.997 > 3. Thus z = (5.997 - 3)/3 = 0.999 > 0 but we require that z < 0.

Answer 2

This equation has the intercept form $$\frac{x}{3}+\frac{y}{3/2}+\frac{z}{-1}=1$$ So the plane cuts the axes at $(3,0,0),(0,3/2,0),(0,0,-1)$. So this portion is a triangle. Let $x=t,y=s$ then $$r(t,s)=(t,s,\frac{t+2s-3}{3}),\ \ 0<t<3,\ \ 0<s<3/2$$