Cross Product Magnitude for Triangle Area

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SUMMARY

The discussion centers on calculating the area of a triangle using the cross product of vectors derived from its vertices. The vertices given are [0, 0, -5], [-2, 1, -4], and [-3, -1, -6]. The user correctly computes the vectors as [-2, 1, 1] and [-3, -1, -1], leading to a cross product of [0, 5, 5] and a magnitude of √50. The area is then accurately expressed as (1/2)√50, which simplifies to (5√2)/2.

PREREQUISITES
  • Understanding of vector operations, specifically cross product
  • Knowledge of calculating vector magnitudes
  • Familiarity with geometric interpretations of vectors in three-dimensional space
  • Basic algebra for simplifying square roots and fractions
NEXT STEPS
  • Study the properties of the cross product in vector algebra
  • Learn about the geometric interpretation of the cross product in determining areas
  • Explore different methods for calculating areas of polygons in three dimensions
  • Practice problems involving the cross product with varying vertex configurations
USEFUL FOR

Students studying vector calculus, geometry enthusiasts, and anyone seeking to understand the application of cross products in calculating areas of triangles in three-dimensional space.

dr721
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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 \sqrt{50}

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

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.
 
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I don't see anything wrong with what you are doing.
 
50= (2)(52) so it might be better to write it as
\frac{\sqrt{50}}{2}= \frac{5\sqrt{2}}{2}
 

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