Which Hand Points in the Positive Z Direction in a 3D Coordinate System?

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In a 3D Cartesian coordinate system where the origin acts as a source of gravity, a person standing on a tightrope with arms extended forms a cross. With their torso in the xy-plane and facing the y-axis, the right hand points in the positive z direction while the left hand points in the negative z direction. The discussion emphasizes the importance of visualizing the scenario accurately to understand the orientation of the hands. The participant seeks confirmation on their drawing and expresses concern about the simplicity of their question. Overall, the right-hand rule applies in this context, confirming the direction of the hands.
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Hello everyone. I'm suppose to imagine the cartesian product or three-dimensional rectangular coordinate system (R^3 = {(x,y,z)|x,y,z R}) that the orgin is a source of gravity and a tight rope connects the point (10,0,0) to the point (10,0,0) to the point (0,10,0) on the tight rope holding their hand straight out to their sides so their boyd forms a cross. Which hand will be pointing in the postive z direction? Draw pictures describing this situation and make sue to account for the effect of the orgin being the only source of gravity. Did I draw the picture right? Here is my drawing: http://img303.imageshack.us/img303/5346/math2gn.jpg

Thanks!
 
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Yes. Since the origin is a "source of gravity", the person will be standing on the rope with his feet toward the origin and so his torso will be in the xy-plane, his arms stretched out perpendicular to the xy-plane and so parallel to the z-axis. Assuming, as in the picture (you seem to have left out some of the words), that the person is facing y-axis, since this is a right-hand coordinate system, his right hand will be pointing in the direction of the positive z-axis and his left in the direction of the negative z-axis.
 
sorry about the double post, i'll keep it in the college section! Somtimes I don't know if my questions are too simplistic for college level even though it is.
 
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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