Graphing a plane

Question

For $x\geq 0, y\leq 0, z\geq 0$

I understand how to graph each trace, and that the traces intercept each other on the coordinate axis but I do not understand why the plane is represented by the shaded region, i.e How do I know that a point in the region satisfies the equation of the plane?

Answer 1

First trace (red) $x-2 y=6$ has parametrization like this $(2 t+6,t,0) $ for $-5<t<-1$ Straight-Line is complete in x-y- plane

Second trace (green) $x+3 z=6$ has parametrization like this $(6 - 3 t, 0, t) $ for $-1<t<1$ Straight-Line is complete in x-z- plane

Third trace (magenta) $-2 y + 3 z = 6$ has parametrization like this $(0,t,\frac{2 t}{3}+2) $ for $-1<t<1$ Straight-Line is complete in y-z- plane.

Here's the picture:

Here is another one:

Answer 2

The extent to which these remarks explain anything will depend on how comfortable you are with the geometry of vector addition and scalar multiplication. :)

If $p_{1}$, $p_{2}$, and $p_{3}$ are non-collinear points of $\mathbf{R}^{n}$, and if $t_{1}$, $t_{2}$, and $t_{3}$ are real numbers whose sum is unity, then the general point on the plane through the $p_{i}$ is the convex linear combination \begin{align*} t_{1} p_{1} + t_{2} p_{2} + t_{3} p_{3} &= (1 - t_{2} - t_{3}) p_{1} + t_{2} p_{2} + t_{3} p_{3} \\ &= p_{1} + t_{2} (p_{2} - p_{1}) + t_{3} (p_{3} - p_{1}). \tag{1} \end{align*} Geometrically, this expression may be interpreted as the set of positions that can be reached by starting at $p_{1}$ and traveling by arbitrary amounts parallel to $p_{2} - p_{1}$ (the direction from $p_{1}$ to $p_{2}$) and $p_{3} - p_{1}$ (the direction from $p_{1}$ to $p_{3}$).

In your example, the axis intercepts of the plane are $$ p_{1} = (6, 0, 0), \qquad p_{2} = (0, -3, 0),\qquad p_{3} = (0, 0, 2), $$ and it's straightforward to check every point of the form (1) satisfies $2x - y + 3z = 6$. (One "smart" approach is to check that if $F(x, y, z) = 2x - y + 3z$, then $$ F(p_{1}) = 6,\qquad F(p_{2} - p_{1}) = 0 = F(p_{3} - p_{1}), $$ and that "$F$ respects linear combinations" in the sense that $F(tp + q) = tF(p) + F(q)$ for all points $p$ and $q$, and all real numbers $t$.)

In case it helps, the green region (the triangle with the $p_{i}$ as vertices) is the convex hull of the $p_{i}$, namely, the set of convex linear combinations with all three "weights" non-negative: $t_{i} \geq 0$.