How do I find the area of a parallelogram using cross product?

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To find the area of a parallelogram using the cross product, two vectors must be established from a common point among the vertices. The vectors can be calculated by taking the difference between the coordinates of the points, such as v1 = p2 - p1 and v2 = p3 - p1. Visualization is not necessary; simply select any three points as p1, p2, and p3. The area can then be determined by calculating the magnitude of the cross product of these two vectors. The final area is confirmed to be 5*sqrt(30).
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11. Calculate the area of the parallelogram having the vertices
(1,2,3),(4,-2,1),(-3,1,0), and (0,-3,-2).

To solve this problem I need to find two vectors that share a common point? Then I can take the magnitude of the cross product of those two vectors to find the area of the parallelogram...no problem.

The problem is how do I know what two vectors will have a common point? Am I just suppose to think in 3d and see it? :bugeye:
 
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Just find two such vectors by saying that one goes from point A to point B while the other goes from point A to point C. The components of these vectors can be computed by (B-A) and (C-A).
 
No visualization is necessary. Pick any three of the four given points. We will call them p1, p2, and p3. Let one of those three be your base point (the common point). Now form the difference v1 = p2 - p1 and v2 = p3 - p1 where p1 is your base point. The two vectors v1 and v2 are the vectors you are looking for, they are the sides of the parallelogram.
 
LeonhardEuler said:
Just find two such vectors by saying that one goes from point A to point B while the other goes from point A to point C. The components of these vectors can be computed by (B-A) and (C-A).

Ah! I see... no problem...so it's what 5*sqrt(30) then...

thanks
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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