Question

Find a point with z=2 on the intersection line of the planes x+y+3z=6 and x-y+z=4.Find the point with z=0. Find a third point halfway between.

Can someone tell how i create x+y+3z=6 and x-y+z=4 as a row picture?

Answer 1

I mean the plotting

Answer 2

For the system of linear equations x+y+3z=6 and x-y+z=4 is equivalent to
\[\left[\begin{matrix}1 & 1 & 3 \\ 1 & -1 & 1\end{matrix}\right]\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}6 \\ 4\end{matrix}\right]\] Using row elimination (subtracting one of row 1 from row 2) takes us from A to U...
\[\left[\begin{matrix}1 & 1 & 3 \\ 0 & -2 & -2\end{matrix}\right]\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}6 \\ -2\end{matrix}\right]\], which is the same as the system of equations x + y + 3z = 6 and -2y - 2z = -2.

By back substitution, we can then find the point where z = 0:
x = 5
y = 1
z = 0.
And similarly where z = 2:
x = 1
y = -1
z = 2.

To find the point midway between those two, you simply take the average for each coordinate.

x = (5 + 1)/2 = 3
y = (1 + (-1))/ 2 = 0
z = (0 + 2)/2 = 1