Like what you saw?
Create FREE Account and:
- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics
Area With the Cross Product - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Remember that, any two vectors will form or determine a parallelogram. Here I have an example; find the area of a parallelogram determined by vectors u equals <0 -1 2> and v equals <3 -1 0>. And I’ve drawn that parallelogram down here. This is v <3 -1 0> and this is u <0 -1 2>. Then I filled up the parallelogram that they determine. I want to find the area of that parallelogram.
So just remember, that we use the formula; area of the parallelogram equals the magnitude of u cross v. You remember that if you cross two vectors in the opposite or you get the opposite vector. But since we are taking the magnitude, we'll always get the right answer. So it actually doesn’t matter what order you cross these vectors in, because the magnitude will take care of that. But let’s go ahead and calculate this.
We’ve got i j k and I need the components of u first <0 -1 2> and then <3 -1 0>. So this is going to be i times the minor -1 2, -1 0, minus j times the minor 0 2, 3 0. Plus k times the minor 0 -1, 3 -1 and then I just calculate that. I have 0 minus -2, is plus 2i. 0 minus 6 times -1 is plus 6j, and this is 0 minus -3 plus 3, k.
The only problem here is I’ve calculated u cross v. This is u cross v, it’s not the magnitude of u cross v, maybe I should leave that off until later. So now I’ll do the area. The area is the magnitude of u cross v.
So the magnitude of this vector <2 6 3>. So that’s the square root of 2 square which is 4, plus 6 square root 36. Plus 3² which is 9, 13 plus 36 is 49. So that’s going to be 7 and that’s the area of our parallelogram. The area of this guy is 7.
Please enter your name.
Are you sure you want to delete this comment?