# Vectors Coordinates of a cube.

• August 13th 2010, 08:53 PM
olski1
Vectors Coordinates of a cube.
Hello,

I have been asked to find the three dimensional coordinates of a cube which has been sunk into the ground. The ground can be considered the x-y plane. and i have been given the coordinates of two of the corners of the cube on the top face (looking down on the cube) point A = (13.0492,30.9498,9.01115) and C = (1.97687,4.68868,17.3632). it is not clear where the origin is.
Points B and D are the same height above the ground.

I started of by concluding that the z coordinates of B and D where halfway between the z co-ordinates of A and C. ( 13.181175).

I then proceeded to find the resultant vectors of AB and BC. I then proceeded with knowledge that the length of BC and AB are equal, thus using the vector length formula i was able to solve for one of the coordintes of Bx = (1102.183741 - 52.52224By)/22.14466 and i subbed this value into a dot product of AB and BC and using quadratic equation i obtained values for B coordinates. i repeated this to find the D coordinates.

B= (-6.197,23.5882,13.1872)
D= (21.1958, 12.0503, 13.1872)

However the next task of this problem is to find the co-ordinates of the A', B' ,C' and D'. which are the points at where the cube intersects the ground. A is conected to A' and similary for the other points. This is where i have become completely stuck!!!

any help would be gratefully appreciated!!
• August 13th 2010, 08:55 PM
olski1
Attachment 18553

here is a picture of the cube
• August 13th 2010, 09:19 PM
Vlasev
This is not very simple but I think it'll work. Since you have 4 points A,B,C and D, you can use any 3 to find the normal vector to the plane ABCD. Say you use vectors AB and AD. Let N = ADxAB. This normal vector faces towards the ground by the right hand rule of cross product. This will be your direction vector. Note that N is in the same direction and parallel to AA', BB', CC' and DD'. Take point A. Then you can get the equation of the line passing through A and A' using N. i.e. L = A + tN where t is a free parameter. This is the vector equation of the line that passes thru AA'. Then you need to set the z coordinate of L to find where it intersects the xy plane. Sub t back into L to get the coordinates of A'. I hope this helped.
• August 14th 2010, 12:55 AM
olski1
Thank for your help. it was extremely usefull. :)

althought now i am up to the last part of the problem. i need to find the volume of the cube, above and below the ground. I have calculated the area A',B',C',D' to be 460.42 m^2. And i have all the coordinates of the corners and intersections with the x-y plane.
also, the lengths of the sides AA',BB',CC',DD' and AB. however as the heights above ground are different heights except B and D. i am not sure how to go about it.
• August 14th 2010, 01:48 AM
Vlasev
An inefficient, though correct way to do this problem is to make a coordinate change so that say vertex A is at the origin and the axes run along the edges of the cube. Then we find the equation of the plane of the ground with respect to this new coordinate system and then integrate to find the volume. It would be like this where z is the equation and L is the length of the side of the cube.

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This will find the volume of the piece that is above ground. Of course the piece that is below ground will have volume $L^3 - V$
• August 14th 2010, 05:42 AM
olski1
thanks once again.

Is there an efficent, not so correct way to do it?
• August 14th 2010, 01:49 PM
Vlasev
Most likely but I cannot think of one at the moment. I will think about it.