Intersecting Planes and Solving Systems of Equations

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To solve for the intersection of two planes, one must first determine the normal vectors of each plane and then compute their cross product to find a vector parallel to the intersection line. Next, algebraic methods can be used to identify a common point on both planes, which will define the line of intersection. An example was provided using the equations X + Y + Z = 0 and 2X + Y + Z = 0, illustrating the process of setting Z to zero to simplify the system. The discussion also emphasized the importance of posting in a single section to adhere to forum guidelines. Clear communication of mathematical concepts is crucial for effective understanding.
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This is my post that I posted in the calculus discussion area, but I know that there are a few people here in the same class (and using the same book!) so it's probably beneficial to post it here too.

edit: go me! of course I forget to post the link lol! https://www.physicsforums.com/showthread.php?p=252564#post252564

I must have been :zzz: ! =p
 
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You need to find the normal vectors for both planes, and simply take the cross product of them to produce a vector parallel to the line that represents the planes' intersection. Then use some algebra to find a point common to both planes; and therefore on the line, and you have your line. Sorry if this didn't make any sense, it makes perfect sense in my head, but it's hard to determine if the same is true in words. And I think you should try to limit your posts to one section to keep the moderators happy. :wink:
 
Ahh ok.

I thought about it a little, and that makes sense. Like if my planes were X + Y + Z = 0 and 2X + Y + Z = 0 I'd set Z equal to zero, and then I'd get a zyztem of equations and the point I'd use would be (X,Y,0, right? That makes sense, thanks gza!

Anyway, do you have any answers to the other questions?

(I'll only post in one section from now on if it matters! =p)
 
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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