Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Let’s try a slightly harder problem. Here I have triangle CAT, find the measure of angle C. All I’m given are the coordinates of the 3 vertices of the triangle. I want to find the measure of this angle here. I can do that using the angle between the vectors, if I define 2 vectors to find the angle between. I’ll define vectors CA and CT.
Let’s look at CA first. In components, CA would be 1 minus -5 or 6 and 8 minus 0, 8. And CT is going to be 10 minus -5, 15 and 8 minus 0, 8. In the formula for the angle between vectors, I need the magnitudes of both of these. So magnitude of CA is the square root of 6² plus 8², 36 plus 64. That’s 100, so the square root of 100 is 10. For the magnitude of CT, I need the square root of 15² or 225 plus 8² or 64. 225 plus 64, that’s 289. 289 is a perfect square. It’s 17², so the square root is 17.
I have everything I need to find the measure of angle C. The cosine of C is going to be CA.CT. The dot product of these two vectors, over the product of the magnitudes; CA and CT. The dot product is 6 times 15, that’s 60 plus 30, 90. Plus 8 times 8, 64. Over CA which is 10, times CT, which is 17. That’s going to be 154 over 170. Angle C is going to be the inverse cosine of 154 divided by 170. Let’s use our calculators to approximate that quickly. Inverse cosine, 154 divided by 170 and I get 25.1 degrees. That’s the measure of angle C; 25.1 degrees. You could find the measures of angle A, and angle T the same way.