Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I want to use what we’ve learnt about cross products to find the equation of a plane. Now let’s remember that, in order to find the equation of a plane, you need to find a vector that’s normal to the plane.
That means that a vector normal to the plane means that, it's perpendicular to any vector that’s contained within the plane. So what I should do here, is find 2 vectors contained within the plane, and then their cross product will be perpendicular to each of them. And that's normal to the plane, and that’s exactly what I want to do. I have three points a b c to make vectors out of.
So let’s consider vector AB and vector AC. These two vectors will lie on the plane, and I’ll create a normal vector by computing their cross product.
So the vector from A to B is 2 minus 2, 0 1 minus 3 - 2 and 0 minus - 4, +4. The vector from A to C is 3 minus 2, 1, 3 minus 3, 0, and -2 minus -4, -2 plus 4, 2. So my normal vector, and I’ll call that n, is going to be the cross product AB cross AC. The cross product is i j k. And then I need the components of AB first 0 - 2, 4 and then 1 0 ,2 the components of AC. So let me just expand this. I get i times this minor; - 2 4 0 2, minus j times this minor; 0 4 1 2, plus k times this minor; 0 - 2 1 0. Now what does that give me?
-4 minus 0 -4i. 0 minus 4, -4 times -j plus 4j. 0 minus -2 is plus 2, k. And so this is my normal vector -4i plus 4j plus 2k, that’s the same as -4, in component form <-4 4 2>.
Now remember, from our previous result, we know that if we have a normal vector to a plane, in component form, then these end up being the coefficients of x, y and z. So I can make any equation -4x plus 4y plus 2z equals some value d. I just need to find the d value and it’s really easy to do that. All you have to do is pick one of the three points that you know is in the plane, and plug that point into the equation. I want to pick this one because there is a 0 and it's going to make this a little easier.
So 2 1 0 lies in the plane. So plugging that in I get -4 times 2, plus 4 times 1, plus 2 times 0 equals d. -8 plus 4, -4, that’s what d is. D is -4 and so my plane is -4x plus 4y plus 2z equals -4.
Now of course you can get another equation for the same plane by multiplying this by another concept. You'll probably notice that I could simplify this. I need to cancel out the factor of 2. It doesn’t really matter, it’s still an equation for the plane.
Let’s just go over the process one more time. This is really important because one of the things about the problems we did before is, we were giving you the normal vector for the plane. That’s a pretty big gift. Suppose all you know is that these three points are in the plane, 3 points determine a plane. So you can come up with two vectors AB and AC that lie in the plane, take their cross product and that gives you a normal. And once you have vector normal of the plane, it's really easy to turn that into an equation. Just plug in one of the points and solve for the value of d.
Unit
Systems of Linear Equations and Matrices