Strange question with 3 dimensional grid

M

mathpunch!

New member
Joined
Sep 21, 2007
Messages
10



I do not have a strong higher math background so please bear with me here.

The values in brown represent the corners' distance off the ground in feet in three-dimensional space (grid), or the "Z" axis value, and you're looking down at the triangle from directly above.

Points A, B, and C are all on the same level in the "Z" axis and you don't know where. All you know is the distance in feet on the x and y axis only from each lettered point to each corner of the triangle. So in other words, you know how far each point would have to move--without being moved in the "Z" axis--to be directly above or below each of the corners.

What I need to find out is, at what point in the "Z" axis would points A, B, and C intersect the "face" of the three-dimensional triangle if they only moved up and down ("Z" axis) just using the previously mentioned distance values.

For example, A would equal about 4.5, B would be about 3-ish, and C would be 2.

Please tell me if I need to explain myself more clearly.

Any help you can give would be much appreciated!
 
Points A, B, and C are all on the same level in the "Z" axis and you don't know where. All you know is the distance in feet on the x and y axis only from each lettered point to each corner of the triangle.
Click to expand...

what (x,y) coordinates?
 
The pictured problem is incomplete because it would be very difficult to draw all of the (x,y) distance values from each lettered point to directly above or below each corner. That's fine because I'm not looking to solve this problem specifically, I'm just looking for some sort of function that uses the information given to find where exactly the lettered points would intersect the triangle face if they crossed it in the Z axis, and the diagram is just an incomplete aid I made.

Remember, you don't know the actual distance as a straight line from the lettered points directly to the corners. This is a little hard to explain but what you do know is the distance to the corners if you ignore the "Z" axis for now. That is, you know how far each point would have to move--without being moved in the "Z" axis!--to be directly above or below each of the corners. So if you drew the distance as a line, all lines would be on exactly the same Z axis plane as the lettered points.

I hope it is clearer now. I need to find a function here, not a solution. Thanks!

If it's impossible with the information given, please tell me what information I would have to know to know where a vertical line drawn through points A, B, and C would would intersect the 3d face that the triangle makes up
 
Ok, I need to revise my question a bit.

trisdm4.jpg



More diagrams!
--Two views of the same graph. Coordinates in blue represent the (x,y,z) coordinates of the corners of the triangle. Coordinates in red represent the (x,y,z) coordinates of points A and B. Also, the triangle is a right triangle.

btw -- I messed up one of the triangle's coordinate values on this diagram. The corner whose y value reads -3 should read -5

What I'm trying to figure out is how to find where a line drawn in the Z axis straight through one of the lettered points would intersect the triangle using just the information given (assuming the line would intersect the triangle) . So, even though it's labeled on my diagram, the lettered points' location in the Z axis doesn't matter as far as where its line (which aren't drawn in this diagram) would intersect the triangle.

Is there a function I can use to find out where exactly in the Z axis the imaginary straight up-and-down line through the lettered points would intersect the triangle?

If not, what other information would one need?
 
Beautiful diagrams - however, I am still a little unsure about what mean by the following statement:

.... where a line drawn in the Z axis straight through one of the lettered points ...

Specially the part where you say "..a line drawn through the z- axis"

There can be infinite number of lines through "A" (say) that intersects 'z' axis. That adjective "straight through" is not quite mathematical - can you describe it in some other way?

If you meant

the line joining A and B (which may or may not intersect z-axis)...

then of course the problem statement is complete and solvable.
 
What I mean by "a line in the Z axis" is a line through the lettered points and parallel to the Z axis so that both points respective lines would intersect the triangle (in this case). So there can be only one possible line that is parallel to the Z axis and intersects the lettered points.

I didn't say "straight through" the Z axis, I said straight up-and-down meaning that the lines drawn through the points are parallel to the line on my diagram labeled Z.

See now?
 
mathpunch! said:
What I mean by "a line in the Z axis" is a line through the lettered points and parallel to the Z axis so that both points respective lines would cross the triangle. So there can be only one possible line that is parallel to the Z axis and intersects the lettered points.

See now?
Click to expand...

Your points are (-4,1,6) and (2, -2.5,6)

A line joining those points will be normal (perpendicular) to the z-axis.

The joining line - to be parallel to the z-axis - must have same x co-ordinates and same y co-ordinates (and different z co-ordinates). For example line joining (2,4,6) and (2,4,8) will be parallel to z-axis.
 
I don't want a line joining these points. The two points have nothing to do with each other. Pretend point B doesn't exist. What I need to find is where a line that is drawn through point A (say), the points along said line would vary only in the Z axis, would intersect the triangle. So another point drawn anywhere along the line would have the same x and y value but a different Z value making that point directly above or below point A.

I need a function to find where this line would intersect the triangle.

Sorry for being ambiguous.
 
The equation of the plane that contains the points (-5,-5,5), (-5,-5,4) & (5,-5,2) is 3x+y+10z=30. Any line through point A that is parallel to the z-axis has this parametric form: \(\displaystyle \left\{ {\begin{array}{l} {x = - 4} \\ {y = 1} \\
{z = t + 6} \\ \end{array}} \right\) where t is any real number.

Now that line intersects the plane at \(\displaystyle \left\( { - 4,1,\frac{{41}}{{10}}} \right\)\).
 
The equation of your given plane is, (following standard method of finding equation of a plane passing through three given points):

3x + y + 10z = 30

Assume the line is through A.

Then assume that the co-ordinate of the point of intersection between the vertical line through A and the given plane is (-4, 1, z_1)

since this point lies on the plane:

3*(-4) + 1 + 10*(z_1) = 30

Solve for z_1. Iget

z_1 = 41/10 = 4.1
 
mathpunch! said:
Could you exaplain a bit more? Where does t come from?
Click to expand...
Sorry, but no. That is not what we are about here.
None of us has time to give you a course in vector geometry.
There are several very basic textbooks available.
It should not take more that two months to master the material.
 
Don't know - either one of us could have made a mistake.

Our processes are correct and similar.

You go through the numbers and find out the correct answer.

This is not a homework problem I assume - what is this related to?
 
No this is not school related, it is part of a personal programming project for a very simple pseudo 3d engine.
 
You must log in or register to reply here.
Top