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Cross Product Magnitude for Triangle Area

  • Thread starter dr721
  • Start date
  • #1
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Homework Statement



So I have several problems on my homework which deal with the application of cross product and its magnitude to find areas. I know how to do cross product, and I know how to find the magnitude, however, no matter how many times I try to calculate it, I get the wrong answer.

Take for example:

Find the area of the triangle with the three vertices [0, 0, -5], [-2, 1, -4], and [-3, -1, -6].


2. The attempt at a solution

This is where I'm not sure if I'm doing something incorrectly. I have generally been trying to create the two vectors for the cross product by subtracting one vertex from the other two.

So,

[-2, 1, -4] - [0, 0, -5] = [-2, 1, 1]

and

[-3, -1, -6] - [0, 0, -5] = [-3, -1, -1]

Then, computing the cross product to be [0, 5, 5]

So the magnitude comes out to be [itex]\sqrt{50}[/itex]

And then the area of the triangle would be [itex]\frac{1}{2}[/itex][itex]\sqrt{50}[/itex]

Is this right? If not, where did I go wrong? Is there some proper approach to deciding which vertex to use as the starting point?

Thanks for your help.
 

Answers and Replies

  • #2
Dick
Science Advisor
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I don't see anything wrong with what you are doing.
 
  • #3
HallsofIvy
Science Advisor
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50= (2)(52) so it might be better to write it as
[tex]\frac{\sqrt{50}}{2}= \frac{5\sqrt{2}}{2}[/tex]
 

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