Thread: Solve solution set geometrically

1. Solve solution set geometrically

x1 - x2 - x3 = 7
x1 - x2 - 2x3 = 2

I know how to solve the problem, but I don't know how to show it geometrically so here is the answer:

$\left(\begin{array}{cc}x1\\x2\\x3\end{array}\right )$ = x2 $\left(\begin{array}{cc}1\\1\\0\end{array}\right)$ + $\left(\begin{array}{cc}12\\0\\5\end{array}\right)$

How do you graph this? Help please

Thank You

2. The graphs are planes in 3 space, and their intersection is a line.

To plot a plane, you need 3 points.
Easiest 3 points to find are on the axes (e.g. x1 when x2 = x3 =0, x2 when x1=x3=0, ...)
Connect the 3 points and you will have triangle in the plane you want to sketch.

Your solution can be given as a parameteric line (let x2 on the right hand side be t)
x1 = t + 12
x2 = t
x3 = 5
Since x3 is constant, the line is parallel to the x1-x2 plane.
You already have it in the answer. The starting point is (12,0,5) and the direction vector is (1,1,0)

3. Originally Posted by JohnJames
x1 - x2 - x3 = 7
x1 - x2 - 2x3 = 2

I know how to solve the problem, but I don't know how to show it geometrically so here is the answer:

$\left(\begin{array}{cc}x1\\x2\\x3\end{array}\right )$ = x2 $\left(\begin{array}{cc}1\\1\\0\end{array}\right)$ + $\left(\begin{array}{cc}12\\0\\5\end{array}\right)$
Where did you get that? I don't see how that expression is related to the two equations given. I would write them, rather, as
$\begin{bmatrix}1 & -1 & -1 \\ 1 & -1 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\\ x_3\end{bmatrix}= \begin{bmatrix} 7 \\ 2\end{bmatrix}$
but, as snowtea says, geometrically, a line equation in three variables corresponds to a plane so the solution set is the line of intersection of those planes.

How do you graph this? Help please

Thank You
One way to solve that is to set it up as an "augmented matrix" and row-reduce:
$\begin{bmatrix}1 & -1 & -1 &| & 7 \\ 1 & -1 & 2 & | & 2\end{bmatrix}$
$\begin{bmatrix}1 & -1 & -1 &| & 7 \\ 0 & 0 & -1 & | & -5\end{bmatrix}$
$\begin{bmatrix}1 & -1 & 0 & | & 12 \\ 0 & 0 & 1 & | & 5\end{bmatrix}$
which says that $x_1- x_2= 12$, so that $x_1= x_2+ 12$ and $x_3= 5$. That line can be written as the parametric equations $x_1= t+ 12$, $x_2= t$, $x_3= 5$. Taking t= 0 gives (12, 0, 5) and taking t= 1 gives (13, 1, 5). The graph is the straight line through those two points.

4. Thank you Snowtea and Hallsofivy for responding

Originally Posted by HallsofIvy
Where did you get that? I don't see how that expression is related to the two equations given. I would write them, rather, as
$\begin{bmatrix}1 & -1 & -1 \\ 1 & -1 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\\ x_3\end{bmatrix}= \begin{bmatrix} 7 \\ 2\end{bmatrix}$
The expression I wrote in my original post is exactly what you wrote here:

Originally Posted by HallsofIvy
One way to solve that is to set it up as an "augmented matrix" and row-reduce:
$\begin{bmatrix}1 & -1 & -1 &| & 7 \\ 1 & -1 & 2 & | & 2\end{bmatrix}$
$\begin{bmatrix}1 & -1 & -1 &| & 7 \\ 0 & 0 & -1 & | & -5\end{bmatrix}$
$\begin{bmatrix}1 & -1 & 0 & | & 12 \\ 0 & 0 & 1 & | & 5\end{bmatrix}$
which says that $x_1- x_2= 12$, so that $x_1= x_2+ 12$ and $x_3= 5$. That line can be written as the parametric equations $x_1= t+ 12$, $x_2= t$, $x_3= 5$. Taking t= 0 gives (12, 0, 5) and taking t= 1 gives (13, 1, 5). The graph is the straight line through those two points.
As I stated in my original post, I already knew how to row reduce, so I just posted the answer in that form, instead of showing the whole calculation as you did.

Again thanks for your help, but I still don't understand how to graph this:

Originally Posted by snowtea
You already have it in the answer. The starting point is (12,0,5) and the direction vector is (1,1,0)
Originally Posted by HallsofIvy
Taking t= 0 gives (12, 0, 5) and taking t= 1 gives (13, 1, 5). The graph is the straight line through those two points.
This is where it gets confusing for me, I have no clue how to graph that. Help Please.

5. Do you know how to plot points on a 3d cartesian (x,y,z) grid on a piece of paper?

Something like this:

Plot (12,0,5) and (13,1,5) in the 3d grid and connect them with a straight line. This will be the 2d projection of the line on your paper grid.

6. so you connect 12 to 0, 0 to 5, 5 to 13, 13 to 1, and 1 to 5 in that order?

7. One more thing about a geometric solution.

The normal vectors for the two planes are:
(1, -1, -1)
(1, -1, -2)
just by looking at the coefficients.

The line at the intersection of both of them must be orthogonal to both these vectors.
So the vector parallel to the line should be a scalar multiple of the cross product of (1, -1, -1) and (1, -1, -2).

8. OK I just tried to draw it, here is the image:

Did I do it correctly snowtea?

9. Not quite.
I think once you learn how to plot one point correctly you should be set.
So lets try (12,0,5).
Remember this is only a point, so the lines are only for reference only the final point location matters.

Start at (0,0,0)
Move parallel to x1 12 units (you did this correctly)
Move parallel to x2 0 units (stay where you are)
Move parallel to x3 5 units (directly move up 5 units)

You point should go here.

Look at the diagram I linked to before, and see if this makes sense.

10. OK one more question, and I think I will have it! When I start plotting the (13,1,5) do I start from (0,0,0) also, or do I start from the 5 from the previous plot?

11. You start from (0,0,0)

Just think about how you plot two points in a 2d grid.
E.g. How do you plot (12, 0) and (13, 1)
This is the same thing except in 3d.