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Homework Help: Cartesian coordinates in 3D problem.

  1. Aug 18, 2004 #1
    I've no idea what to do with this, the examples didn't have anything of this style:

    The point A has coordinates (3,0,0), the point B has coordinates, (0,3,0), the point C has coordinates (0,0,7). Find, to 0.1 degrees, the sizes of the angle between the planes OAB and ABC, where O is the origin.

    Could someone give me an idea of what to do? I've only just introduced myself to cartesian coordinates/vectors in 3 dimensions.
  2. jcsd
  3. Aug 18, 2004 #2


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    The plane OAB is simply the xy-plane.
    By symmetry you can see (draw a picture) that the angle between the planes is
    the same as the angle between the vector [1,1,0] and the vector [-1/2,-1/2,3].
  4. Aug 18, 2004 #3
    The thing is, i have no idea how to calculate angles in 3d. I always thought of angles as 2 dimensional.
  5. Aug 18, 2004 #4
    I suggust drawing the points first then sketching the plane OAB and ABC. Then you can treat this as a simple Pythagorean then trig question. As Galileo has stated, the drawing will be in symmetry.

    a hint would be finding the length of AB and the line from origin to the midpt of AB
  6. Aug 18, 2004 #5
    I can't draw 3d shapes but i've sketched it as best as i can. Where did you get the coordinates 1,1,0 and -.5,-.5,3 from? What mathematical process? I hate crappy 3d vectors especially with this crappy book that throws you a question without any method for doing it.
    Last edited: Aug 18, 2004
  7. Aug 18, 2004 #6
    I know the length of AB is root3. The line AB has equation 3y+3x-9=0. Neither do i understand how you can have a three dimensional angle without making a whole lot of extra degrees or using two angles and reinventing trigonometry.
    Last edited: Aug 18, 2004
  8. Aug 18, 2004 #7


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    A is on the x-axis and B is on the y-axis. The plane OAB is just the xy-plane.
    The plane OAB crosses OAB in the line AB. The "angle" between the planes is the angle between a line in OAB perpendicular to AB and a line in ABC perpendicular to AB, that is between the line through (0,0,7) perpendicular to AB and the line through (0,0,0) perpendicular to AB.
  9. Aug 20, 2004 #8


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    Two nonzero, nonparallel vectors (in whatever dimension) determine a plane.
    The angle between those vectors is an angle on that plane.

    Given nonzero vectors [itex]\vec A[/itex] and [itex]\vec B[/itex], the angle [itex]\theta[/itex] between them can be determined by using two expressions for the dot-product [itex]\vec A \cdot \vec B[/itex].

    [tex]\vec A \cdot \vec B = A_x B_x + A_y B_y +A_z B_z= |\vec A| |\vec B| \cos\theta [/tex]
  10. Aug 21, 2004 #9
    Sometimes the angle between two planes is determined by first computing the angle between the normal vectors to a plane. The normal to a plane [tex]\vec n[/tex] is simply a vector perpendicular to the plane. The vector equation of a plane (fyi) is written concisely as

    [tex]\vec n \cdot \vec r = 0[/tex]

    Here [tex]\vec r[/tex] is a vector perpendicular to n, which means r lies in the plane.
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