Results 1 to 11 of 11

Math Help - Finding points of rectangular box in R3

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    156

    Finding points of rectangular box in R3

    The faces of a rectangular box are parallel to the coordinate planes. Two of the vertices are at the points (1,2,3) and (7,5,4).

    *Draw the box and label the vertices

    Since it said the faces of the box are parallel to the coordinate planes, I tried finding 2 more vertices parallel to the z-axis. I said the point (1,2,3) on the other side of the z-axis = (1,-2,3) (red) since all y points are negative there and used the same reasoning for the point (7,5,4) (blue). However if I try to join the points (1,2,3), (1, -2, 3), (7,5,4) and (7,-5,4) together, I dont get a rectangular face on the z-axis (yellow). The orange and green line shows how I drew the points (7,-5,4) and (1,-2,3) respectively.

    What have I done wrong?
    Attached Thumbnails Attached Thumbnails Finding points of rectangular box in R3-r3.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Co-ordinates of four points on the lower surface of the recanguler box are

    (1,2,3), (1,5,3), (7,5,3) and (7,2,3)

    Similarly can you find the co-ordinates of the four points on the upper surface of the rectangular box?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    156
    Quote Originally Posted by sa-ri-ga-ma View Post
    Co-ordinates of four points on the lower surface of the recanguler box are

    (1,2,3), (1,5,3), (7,5,3) and (7,2,3)

    Similarly can you find the co-ordinates of the four points on the upper surface of the rectangular box?
    I am not sure how to find the correct points.What method did you use to find those 4 points?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    In lower surface of the rectangular box, z co-ordinate of all the four points must remain constant and that is equal to z co-ordinate of the given point. Similarly in the case of the upper surface.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2010
    Posts
    156
    Quote Originally Posted by sa-ri-ga-ma View Post
    In lower surface of the rectangular box, z co-ordinate of all the four points must remain constant and that is equal to z co-ordinate of the given point. Similarly in the case of the upper surface.
    Oh ok I understand that. I made a rough sketch of the rectangular box and got the 4 points for the one face. Like you said for the z-coordinates on the other side of the top surface = 4 (blue x), but how would you determine its x and y coordinate (also for the bottom surface with constant z = 3 (yellow x)?)
    Attached Thumbnails Attached Thumbnails Finding points of rectangular box in R3-box.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by SyNtHeSiS View Post
    The faces of a rectangular box are parallel to the coordinate planes. Two of the vertices are at the points (1,2,3) and (7,5,4).

    *Draw the box and label the vertices
    Might help to think this way: there are 8 vertices. For each point, the x-coordinate is either 1 or 7; the y coordinate is either 2 or 5; the z-coordinate is either 3 or 4. There are 8 possibilities (2^3). These are the 8 vertices we're after.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Apr 2010
    Posts
    156
    Quote Originally Posted by undefined View Post
    Might help to think this way: there are 8 vertices. For each point, the x-coordinate is either 1 or 7; the y coordinate is either 2 or 5; the z-coordinate is either 3 or 4. There are 8 possibilities (2^3). These are the 8 vertices we're after.
    I understand what you saying, but when I try this by drawing a diagram to find the other vertices, I dont know what to do . I could only find the one face (shaded in grey) in the attachment above.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by SyNtHeSiS View Post
    I understand what you saying, but when I try this by drawing a diagram to find the other vertices, I dont know what to do . I could only find the one face (shaded in grey) in the attachment above.
    If you understand what I'm saying, then you know the coordinates of all 8 vertices. Are you saying you don't know how to plot them?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,748
    Thanks
    648
    Hello, SyNtHeSiS!

    The faces of a rectangular box are parallel to the coordinate planes.
    Two of the vertices are at the points (1,2,3) and (7,5,4).

    Draw the box and label the vertices.

    Don't try to make a sketch now . . . wait until the end.


    The two points are diametrically opposite vertices.

    . . Since \,x goes from 1 to 6, the length of the box is 5.
    . . Since \,y goes from 2 to 5, the width of the box is 3.
    . . Since \,z goes from 3 to 4, the height of the box is 1.


    Consider the "bottom" of the box.
    All four vertices of this rectangle have \,z= 3.

    Plot the points on a two-dimensional graph.


    Code:
          |
          |(1,5,3)        6        (7,5,3)
          |   * → → → → → → → → → → → *
          |   ↑                       ↑
          |   ↑                       ↑
          |  3↑                       ↑3
          |   ↑                       ↑
          |   ↑                       ↑
          |   * → → → → → → → → → → → *
          |(1,2,3)        6        (7,2,3)
          |
          |
      - - + - - - - - - - - - - - - - - - - - -
          |


    We start at (1,2,3).

    We know the length is 6, so we move right 6 units . . . to (7,2,3).

    We know the width is 3, so we move up 3 units . . . to (1,5,3).

    And we do both: move right 6 units, up 3 units . . . to (7,5,3).


    And we have the coordinates of the bottom of the box:
    . . (1,2,3),\;(7,2,3),\;(1,5,3),\;(7,5,3)


    Since the height of the box is 1, add 1 to the \,z-coordinates.

    And we have the corordinates of the top of the box:
    . . (1,2,4),\;(7,2,4),\;(1,5,4),\;(7,5,4)



    Now I can draw the box . . . and then label the coordinates.


    Code:
                 (1,2,4)     (1,5,4)
                    * - - - - - * 
                   /           /|
                  /           / |
                 /           /  *(1,5,3)
                /           /  /
               /           /  /
              /           /  /
             /           /  /
     (7,2,4)* - - - - - *(7,5,4)
            |           | /
            |           |/
            * - - - - - *
         (7,2,3)     (7,5,3)

    And the unseen vertex (in the back) is (1,2,3)

    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Apr 2010
    Posts
    156

    Question

    Thanks a lot, that post helped a lot. I tried to draw the box, but I am having a problem projecting the surfaces through the z-axis. Referring to the attachment I drew the bottom surface in R3 with the coordinates: (1,5,0), (7,5,0), (1,2,0) and (7,2,0), then I tried projecting these points up by 3 to get (1,5,3), (7,5,3), (1,2,3) and (7,2,3), but they all lie on the line z = 3 (in blue) and look like a line as opposed to a plane
    Attached Thumbnails Attached Thumbnails Finding points of rectangular box in R3-box-plane.jpg  
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by SyNtHeSiS View Post
    Thanks a lot, that post helped a lot. I tried to draw the box, but I am having a problem projecting the surfaces through the z-axis. Referring to the attachment I drew the bottom surface in R3 with the coordinates: (1,5,0), (7,5,0), (1,2,0) and (7,2,0), then I tried projecting these points up by 3 to get (1,5,3), (7,5,3), (1,2,3) and (7,2,3), but they all lie on the line z = 3 (in blue) and look like a line as opposed to a plane
    z=3 is a plane, not a line.

    To plot a point (a,b,c) in R3, start at the origin, go a units in the x-direction, b units in the y-direction, and c units in the z-direction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Interpolating z(x,y) data point from 4 data points (rectangular)?
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: June 20th 2011, 06:04 PM
  2. Replies: 16
    Last Post: June 10th 2011, 06:49 AM
  3. Replies: 1
    Last Post: April 3rd 2010, 08:50 AM
  4. Finding two points...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 25th 2009, 04:45 AM
  5. Finding the right points...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2008, 03:23 AM

Search Tags


/mathhelpforum @mathhelpforum