Unit
Systems of Linear Equations and Matrices
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.
So let’s compute some cross products. I have two vectors u equals 4 1 0 and v equals -5 0 1. And I want to compute both u cross v, and v cross u. I often do this just to see if the operations commutative, whenever I’m dealing with a new operation.
So let’s calculate both of these. Remember that, in order to calculate a cross product, I need the vectors in i j k form. So the first thing I should probably do is convert each of these to i j k form. So this will be 4i plus 1j which is plus j plus 0k, which I don’t have to write. And then v is going to be -5i plus 0j, which I don’t have to write, plus k, 1k.
So this is going to be 4i, that means I have a 4 here, 1 here and a 0 here. U comes first and then v comes next, -5 0 1. You just copy down the components. Let’s expand across the top. I have i times this minor here; 1 0, 0 1, minus j times this minor; 4 0, -5 1. Plus k times its minor, 4 1. -5 0. So let’s compute those.
It’s going to be i times 1 minus 0 times 1, minus j times 4 minus 0, times 4, plus k times 0 minus -5, 0 plus 5, so 5. And so my answer is i minus 4j plus 5k, or you can write it in component form 1 -4 5.
Let’s try v cross u. Again we have the determinant i j k or we start that and then you put. Since v comes first in this product, I’m going to put the components of v across the top here. Minus 5 0 1 and then u; 4 1 0. So I have i times this minor 0 1, 1 0, minus j times this minor -5 1, 4 0 plus k times the minor -5 0 4 1.
This two by two determinant is 0 minus 1. So I have i times negative 1 minus j times 0 minus 4 plus k times -5 minus 0. And that gives me -i plus 4j minus 5k. And that’s the same as in component form -1, 4 -5.
So let’s compare. This is v cross u, this is u cross v. You can see that they are different vectors, so the cross product is not commutative. But not only are they different, but they are very specifically different. This is exactly the opposite of this vector.
So what that tells me is that, u cross v is the opposite of v cross u. And that’s always true. The two vectors have exactly the same length and they point in opposite directions.