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.
I want to use a cross product to find the area of another triangle. This time the triangle is going to be formed by the points AB and C. A is (2 1 0) B is (3 1 4) and C is (5 0 4). The first thing I need to do is form two vectors. Vectors that represent two sides of the triangle.
So I?m going to have vector AB and vector AC. The components of vector AB will be 3 minus 2, 1, 1 minus 1, 0 and 4 minus 0, 4. The components of AC will be 5 minus 2, 3, 0 minus 1, -1 and 4 minus 0, 4. First, I want to take the cross product of these two sides. That?s the first step, so AB cross AC. And that?s i j k and then I?m going to write the components of each of these vectors. <1 0 4>, <3 -1 4>.
Now expanding along the top row, I get i times the minor 0 4, -1 4 minus j times this minor 1 4, 3 4. Plus k times this minor 1 0, 3 -1. So we complete these determinants. We have 0 minus -4, 0 plus 4, 4. We have 4 minus 12, -8 times -1 is plus 8j. -1 minus 0 I get minus k. In component form this would be <4 8 -1>.
This isn?t representing anything by itself. I need to take the magnitude of this and multiply by a 1/2 to get the area of my triangle. So the area of the triangle, and that?s triangle ABC, is ½ the magnitude of my cross product.
AB cross AC that's ½ and the cross product was 4 8 -1, the magnitude of 4 8 -1. The magnitude of this vector is going to be the square root of the sum of the squares of the components. 16 plus 64 plus 1. And this gives me 80 plus 1, 81. And root 81 is 9, so ½ times 9, 4.5 that?s my answer. That?s the area of the triangle.