Calculating Triangle Area in 4-Space: Can Cross Products Be Used?

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SUMMARY

The area of a triangle in 4-dimensional space can be calculated using displacement vectors and the dot product, rather than relying on the cross product, which is not defined in 4-space. To find the area, form displacement vectors between the triangle's vertices: (-2,-2,2,2), (0,0,1,-1), and (-1,-2,1,1). Calculate the magnitudes of these vectors and use the dot product to determine the cosine of the angle between two sides. This method effectively allows for the calculation of the triangle's area in R4.

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  • Familiarity with dot products and their geometric interpretations
  • Basic knowledge of trigonometry
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Homework Statement


Find the area of triangle with vertices (-2,-2,2,2), (0,0,1,-1), (-1,-2,1,1)


Homework Equations


The Attempt at a Solution


The only way I know how to find the area of a triangle is by finding half the parallelogram. I.e. A = (1/2)||u x v||

But this requires cross product and you can't find the cross product in 4-space, can you?

Any help/explanation would be greatly appreciated.
 
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CheesePlease said:

Homework Statement


Find the area of triangle with vertices (-2,-2,2,2), (0,0,1,-1), (-1,-2,1,1)


Homework Equations


The Attempt at a Solution


The only way I know how to find the area of a triangle is by finding half the parallelogram. I.e. A = (1/2)||u x v||

But this requires cross product and you can't find the cross product in 4-space, can you?

Any help/explanation would be greatly appreciated.

You'll need to form displacement vectors between pairs of vertices. You can then find the magnitudes of those displacement vectors. After that you can use the dot product to find the cosine of the angle between two adjacent sides of the triangle.

All three points lie in a plane (it happens to be in R4 but that's not important), so I would advise drawing a rough sketch of the three points. With a bit of trig you should be able to get the area of the triangle.
 
Great, got it. Thanks you!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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