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Calc III, confused on (x,y,z);

  1. Aug 31, 2005 #1
    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: Picture

    Thanks!
     
  2. jcsd
  3. Aug 31, 2005 #2

    HallsofIvy

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    You also seemed to be confused as to whether you are in college or not! Please do not post the same thing repeatedly!
     
  4. Aug 31, 2005 #3
    Sorry, I didn't know which category this falls into, somtimes my posts get moved to k-12 even though they are college questions. You said,
    I understand how his arms are perpendicular to the xy-lane, but how is it that they are parallel to the z-axis? Wouldn't his left hand be pointing right at the origin and right hand pointing away from the origin? ALso I see that he is facing the y-axis, but why is his right hand pointing in the direction of the positive z-axis? Is it the right hand rule that tells u this? Thank you for your repsonce and sorry about the double post!
     
  5. Aug 31, 2005 #4

    lightgrav

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    yea, you left out some IMPORTANT words.
    But if the guy's feet are on the rope between
    (10,0,0) and (0,10,0), then his body is in the xy plane.
    You drew a right-hand coordinate system, as is traditional,
    told by right-hand-rule (hand=x, fingers=y, thumb=z).
    If he's facing the y-axis, then his LEFT hand points in +z,
    while his RIGHT hand points in negative-z direction;
    told by looking at your (new) drawing.

    If the guy turns around and faces the x-axis,
    his right hand goes "up" in your picture, along +z.
     
  6. Aug 31, 2005 #5
    If you picture a guy with his arms perendicular to the XY plane, what tells you his left hand is in the +z and his right is in the -z? How does the right hand rule show this? Thanks for the responce
     
  7. Aug 31, 2005 #6
    [tex] \hat x \times \hat y = \hat z [/tex] meaning the crossing the (positive) x and y unit vectors gives you the (positive) z unit vector.
     
  8. Sep 1, 2005 #7

    lightgrav

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    The right-hand rule DOES NOT show this!
    The RHR tells you which direction is called +z ,
    given the directions of +x and +y .

    You need to know WHICH WAY the guy is FACING
    before you can tell which of his hands is along +z .
    You sketch the person in the coordinate system.

    Do you not know which of your hands is the Right-hand?
    The system (upward,backward,rightward) is right-handed,
    upward: from feet to head
    backward: from nose to neck
    rightward: chest to "Right" hand (<= definition of right hand)
     
  9. Sep 1, 2005 #8

    HallsofIvy

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    How many perpendiculars to a plane do you think there are?! :biggrin: The z-axis is, of course, perpendicular to the xy-plane and every line perpendicular to the xy-plane must be parallel to it.
    The coordinate system in your picture, like standard coordinates systems is "right handed". The means basically that if you curl the fingers on your right hand from the positive x-axis toward the positive y-axis, your thumb will be pointing at the positive z-axis. If you are standing on a rope from (1, 0, 0) to (0, 1, 0) and (0,0,0) is a source of gravity then your head is pointing away from (0,0,0). (If you were at (1/2,1/2,0), your body would lie on the line y= x.) Assuming that you are facing (0,1,0), you left arm will extend parallel to the positive z-axis and your right arm parllel to the negative z-axis.
     
  10. Sep 1, 2005 #9
    Thanks for clearing that up guys, sorry i'm slow, some say i'm legally retarded. :tongue2:
     
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