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:

= x2 +

How do you graph this? Help please

Thank You

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- Dec 12th 2010, 05:57 AMJohnJamesSolve 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:

= x2 +

How do you graph this? Help please

Thank You - Dec 12th 2010, 09:51 AMsnowtea
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) - Dec 12th 2010, 10:55 AMHallsofIvy
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

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.

Quote:

How do you graph this? Help please

Thank You

which says that , so that and . That line can be written as the parametric equations , , . 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. - Dec 12th 2010, 12:02 PMJohnJames
Thank you Snowtea and Hallsofivy for responding

The expression I wrote in my original post is exactly what you wrote here:

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:

This is where it gets confusing for me, I have no clue how to graph that. Help Please. - Dec 12th 2010, 12:13 PMsnowtea
Do you know how to plot points on a 3d cartesian (x,y,z) grid on a piece of paper?

Something like this: http://upload.wikimedia.org/wikipedi...ystem_CA_0.svg

http://upload.wikimedia.org/wikipedi...m_CA_0.svg.png

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. - Dec 12th 2010, 12:26 PMJohnJames
so you connect 12 to 0, 0 to 5, 5 to 13, 13 to 1, and 1 to 5 in that order?

- Dec 12th 2010, 12:31 PMsnowtea
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). - Dec 12th 2010, 12:38 PMJohnJames
OK I just tried to draw it, here is the image:

http://oi54.tinypic.com/4dcaw.jpg

Did I do it correctly snowtea? - Dec 12th 2010, 12:47 PMsnowtea
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. - Dec 12th 2010, 12:56 PMJohnJames
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?

- Dec 12th 2010, 02:08 PMsnowtea
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.