Thursday, 15 September 2005, 14:01:43 EDT

Before I get into this post I would like to say that the mathematics herein are the type I have been wanting to learn for a long time. Until recently, I didn't realize what would be required to learn and practice such math. I think if someone had actually *shown* me and explained it a long time ago then I wouldn't be where I am today. Instead, they just say "this is required for graduation" and "if you don't do it like this you won't pass." That is not how math should be taught and I find myself despising some of my previous teachers now even more than I did before. If, by some strange happenstance, you are a math teacher and reading this then take this advice into consideration -- show your students some of the cool aspects of what you are trying to teach them. Even if what you show them is beyond what you will teach them. It will give the student a reason to learn what you are teaching.

Ready for some math now? First, the problem: Find an equation for the plane that passes through the point (-1, 2, 1) and contains the line of intersection of the planes x + y - z = 2 and 2x - y + 3z = 1.

Let's consider the given information. We have two planes which form a line where the two planes meet and we have a point outside of those two planes but parallel to the line formed by them. We need to find a plane, on which the given point lies, that passes through the line formed by the two given planes. Let's call the plane "x + y - z = 2" P1 and the plane "2x - y + 3z = 1" P2. We will call the given point, (-1, 2, 1), p1. Don't get confused; the capital 'P' stands for "plane" while the lower case 'p' stands for "point." Let's also call the line of intersection L.

Now let's visualize the information given to us. Figure 1 is a representation of P1, figure 2 is a representation of P2, and figure 3 is a representation of both planes, P1 and P2, showing the line of intersection, L. Finally, figure 4 shows p1 in relation to P1 and P2.

What must we do to find a plane that contains p1 and passes through L? First, we need to determine a point on L. We can do this by setting z = 0 in each equation and solving the resulting system of equations. Solving P1 for y results in y = 2 - x which we then substitute into P2 to determine that x = 1 yielding the point p2 = (1, 1, 0). Now we have two points that are in the plane we wish to find an equation for. We can use these two points to determine a vector, a line with direction in space, on our plane, which we will call **v**. This vector is simply <(-1 - 1), (2 - 1), (1 - 0)> = <-2, 1, 1> represented by the blue line in figure 5.

That gives us one-third of the information we need to construct our equation. Next we need to determine another vector on our plane. We can do this by crossing the normal vectors of the two given planes. This gives us a new vector, **u** = <2, -5, -3>. This vector **u** is represented by the orange line in figure 6.

The standard equation of a plane is a*(x - x1) + b*(y - y1) + c*(z - z1) = 0 where a, b, and c are the components of the normal vector to the plane and x1, y1, z1 are the components of a point on the plane. Since we already have two points on the plane all we have left to find is a normal vector to the plane. This is done by simply crossing the vectors **v** and **u** — **v** x **u** = <-2, 4, -8>. We can scale this vector by -(1/2) to give us **u** = <1, -2, 4>. This vector can be seen as the yellow line in figure 7.

We now have all the information necessary to construct the equation for the plane we are trying to find. By substituting into the standard equation of a plane we get 1(x - 1) - 2(y - 1) + 4(z - 0) = 0 or x - 2y + 4z = -1 which we will call P3 and is represented by the red plane in figure 8. Substituting p1 or p2 into this equation results in a valid answer so our points are on P3.

There you have it. The solution to a problem that I spent a few hours on. This one problem helped me realize that, while a point may be a vector, a vector constructed from a point is not necessarily in a plane. That one thing had me tripped up for quite some time on this problem. I knew what I needed to do but I kept using the wrong vectors.

If you have Mac OS 10.4 (and hopefully later) you can download the Grapher file I used to create the figures in this post. Grapher is a really awesome program and is helping me quite a bit in Calculus III. It is very easy to visualize problems with Grapher. I use it all the time to check my answers visually.

Categories:

- mathematics

thank you

Posted by jones on Monday, 24 March 2008, 7:18:26 EDT.

thank you

Posted by jones on Monday, 24 March 2008, 7:19:14 EDT.

This has been a great help! Thank you!

Posted by pauline on Wednesday, 14 May 2008, 13:52:36 EDT.