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Homework Help: Pythagoras and Vector coordinates of a cube.

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Basically we are given the coordinates for A=(13.0492, 30.9498, 9.01115) and C=(1.97687, 4.68868, 17.3632) which are two top corners of a cube that is sunk in the ground partway (specifically it's based on the cube near the Melbourne museum). But we don't know where the origin is of the coordinate system. We do know that the x-y plane sits level with the ground and that points B and D are at equal heights above the ground. All values are in meters.

    We are told that A', B', C' and D' are the points level with the ground below their respective points of A, B, C and D. We are then asked to find the following:
    A = (13.0492, 30.9498, 9.01115)
    B = (?, ?, ?)
    C = (1.97687, 4.68868, 17.3632)
    D = (?, ?, ?)
    A' = (?, ?, ?)
    B' = (?, ?, ?)
    C' = (?, ?, ?)
    D' = (?, ?, ?)
    |AA'| =?m
    |BB'| =?m
    |CC'| =?m
    |DD'| =?m
    sidelength =?m
    area(A'B'C'D') =?m2
    volume above =?m3
    volume below =?m3

    2. Relevant equations

    Pythagoras's Theorem, vector methods and vector calculus.

    3. The attempt at a solution

    I attempted to find the distance between A and C since this value can be used to find the side lengths. So I went |A-C| which came out as 29.6985m. Then I used a^2+b^2=C^2 to find the values of the side lengths seeing as the distance between A and C is the hypotenuse and we know a and b will be equal. So [tex]\sqrt{(29.6985)^2/2}[/tex]=21.0000m.

    Finally I was able to find the heights of points B and D above the ground since I know it pivots on these points. So using the gradient from C to A and taking the height at the midpoint between C and A I was able to go:
    17.3632-9.01115=8.35205m (the distance it lowers over the distance between C and A)
    Then 8.35205/29.6985=0.28123m (how far it lowers per meter from C to A)
    Then (29.6985/2)X0.28123=4.17604m it lowers half way from C to A or raises From A to C. So the height of points B and D is 17.3632-4.1760=13.1872m.

    So that's as far as I can go. I'm not sure how to go about finding the x and y coordinates for points B and D. I was thinking there might be a way to solve for the dot product to make it equal to 0 since it meets at right angles but I'm not sure how to do that. Do we just sub in any values for x and y? Because I think there would be thousands of possibilities in that case.

    I also don't know how to calculate the coordinates for A', B', C' and D'. At first I thought you could assume that z will equal 0 and the x, y part of their coordinates would be the same as their upper counterparts but then I realized that the straight edge of the cube makes an angle with the ground so these points are not directly under A, B, C and D respectively. Still I believe if you formed a right angled triangle by making a straight line from say A to the ground you can then calculate the length of the remaining sides if you had one angle but I don't know how to find any of the angles.

    Please correct me if I've made any mistakes any help is welcome. If anything doesn't make sense let me know and I'll try to explain it as I've cut it down from 2pgs so you wont have to read so much.
     
  2. jcsd
  3. Aug 23, 2010 #2

    LCKurtz

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    The numbers you have so far appear to be correct. Call R = < x, y, z> a variable position vector. You have the length of a side of the cube, call it s and the z coordinate of the other two vertices, call it h for now. Your points B and D must lie on the sphere

    |R - A|2=s2 and also on the sphere

    |R - C|2=s2.

    Also you know the z coordinates for B and D are both h. Substitute that into the above two equations and solve for the two unknowns x and y. This should give you two (x,y) solutions which, together with z = h, gives you the coordinates for B and D. Let me know what you get.
     
  4. Aug 23, 2010 #3
    Ok, so far this is what I have. I hope I'm on the right track. I've narrowed it down to two equations:
    [PLAIN]http://img189.imageshack.us/img189/3694/equat.gif [Broken]

    I'm hoping from there I can factorize both sides of the equation so I'll have four x values and four y values. I then presume I'll be able to eliminate the ones that don't meet at right angles with A and C by using the dot product to see if they equal 0. Does this seem right to you? Thanks for all your help too. I really appreciate it.
     
    Last edited by a moderator: May 4, 2017
  5. Aug 23, 2010 #4

    LCKurtz

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    Hopefully you are using some mathematics software to help you with crunching all those decimal numbers, eh?
     
  6. Aug 24, 2010 #5
    Just my trusty scientific calculator and the good old quadratic formula. I've gone through several binder books and notepads with all my scribbling though. Anyway I've come up with these numbers so far and am just about to use the dot product to check they are perpendicular:
    [PLAIN]http://img265.imageshack.us/img265/5873/eqat2.gif [Broken]
    and then from there I got:
    [PLAIN]http://img37.imageshack.us/img37/3403/vect.gif [Broken]

    Just to be sure I can dot B with C or A and same with D and it should =0 for both instances right? Once more thanks for all your help.

    Oops, I just realized there should be a decimal point after the 0 in the y part of the first solution to B.
     
    Last edited by a moderator: May 4, 2017
  7. Aug 24, 2010 #6
    I'm not sure where I've gone wrong, but I've tried every which way to dot my two solutions for B and D and none of them equal 0 or anywhere near 0 for that matter. I've looked over my calculations about 5 times and can't see anywhere that I went wrong.
    [PLAIN]http://img163.imageshack.us/img163/784/working1.gif [Broken]
    [PLAIN]http://img409.imageshack.us/img409/831/working2.gif [Broken]
    [PLAIN]http://img84.imageshack.us/img84/2062/working3.gif [Broken]

    Above is my working if that helps.
     
    Last edited by a moderator: May 4, 2017
  8. Aug 24, 2010 #7

    LCKurtz

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    I haven't tried to follow and verify your steps with all those decimals. I crunched the numbers with Maple and I didn't get the same answers as you. Here's what I got. I didn't round off anything and I suppose they could be reversed depending on the orientation.

    B = [21.19581960,12.05024361,13.1871749999999999]
    D = [-6.169749602,23.58823639,13.1871749999999999]

    Just for amusement I plotted the cube this leads to:

    cube.jpg

    I admit to some curiosity whether this looks like the real cube near the Melbourne museum. :smile:
     
  9. Aug 24, 2010 #8
    Just a question, the two bars around the R-A is the modulus sign right? I've been trying all night to solve it in every which way I could but no matter what I can't get the same answers as you. I don't suppose you could show me step by step how to get there? I'm obviously doing something wrong.
     
  10. Aug 24, 2010 #9

    LCKurtz

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    The two equations I gave are:

    |R - A|2=s2
    |R - C|2=s2

    Here they are written out (after substituting the known value z = h):

    (x - 13.0492000000000008)2 + (y - 30.9497999999999998)2 + 17.43918480
    = 440.9998274

    (x - 1.97686999999999992)2 + (y - 4.68867999999999972)2 + 17.43918480
    = 440.9998274

    I didn't solve them myself (life is too short for that), Maple did giving:

    {x = 21.19581960, y = 12.05024361}, {x = -6.169749602, y = 23.58823639}

    which gives the points for B and D:

    B = [21.19581960,12.05024361,13.1871749999999999]
    D = [-6.169749602,23.58823639,13.1871749999999999]

    Perhaps you can use these to find your mistake. Of course, there is more to do. Good luck with it; I am going to be away from my computer for a few days.
     
  11. Aug 24, 2010 #10
    Also it says at the end of the question that the dot product can be used to determine weather the values are correct because they are supposed to meet at right angles, but I don't see how that will work. If I dot A with B it will definitely not equal 0.

    I'm really stuck on as how to find A', B', C' and D' as well. I've been trying to come up with a way to use Pythagoras to do it because if you make a right angle triangle with the ground and cube than the top point will be A (for this corner), a straight line down to the ground will have the same coordinates as A except z=0 and the length of this side will be z. That's as far as I can get. I know the angle with the ground will be 90 degrees but I can't seem to find the final point which will be A'.

    Edit: you posted while I was typing, thanks for those, I'm still not sure where I went wrong but I think it was in my rounding somewhere or perhaps I just used the wrong way of solving them. I believe I need to make y=o first than go through and solve for x, and like you said life is too short for that.

    oh and enjoy your holiday away from your computer after helping me I'm sure you deserve it.
     
    Last edited: Aug 24, 2010
  12. Feb 26, 2011 #11
    Maybe you should do your own work instead of getting random people to give you the answers. This is after all a university assignment, and this doesnt seem to be within the rules does it?

    May I urge people not to directly give him any results or answers but just hints as to how to get on the way to the solution. This is actually an important university assignment that is meant to be done by oneself.
     
  13. Feb 26, 2011 #12

    berkeman

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