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**3dhelp****Member**- Registered: 2009-09-09
- Posts: 1

I have a point with coordinates x,y,z on a plane that is defined by an angle from north (33 degrees clockwise) and a dip of 65 degrees down from the horizontal.

I want to move this point a distance of 50 in a direction normal to the plane and calculate the new coordinates of the point.

Can anyone tell me how to do this please

Thanks

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

Hi!

I am sure there will be an easier way to solve it, but I will try to say how would I do it.

I am going to assume you have an original point in the XY plane, heading north, which I will consider in the negative part of the X axis. Then, the original coordinates would be (-x, 0, 0), being the distance to the origin of coordinates equal to x.

I also consider that rotation from north and clockwise is keep the point in the XY plane, so rotate in Z an angle q1 = -33. And then doing the same for a "dip"angle, I assume it is rotate with respect to X, an angle q2.

First, you have to calculate the new coordinates of the point, for example from http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/3drota.htm

1. Rotation in Z:

2. Likewise, rotation in X:

Yo can actually find the 2 matrices, multiply one with the other and obtain the total displacement.

Now have point (x'', y'', z''). Then, we have to move it in the direction normal to certain plane, a distance of 50. Ok, the equation characterizing a plane using a vector normal to it, (a, b, c) that passes through point (x0, y0, z0) is:

And the line passing through (x'', y'', z'') normal to the plane can be written using an independent variable t:

In total, you have to calculate what new point (x''', y''', z''') is at distance 50 from (x'', y'', z''):

So the point must satisfy the equation of the line and of the distance.

And that's what I can say with the info you provided

Jose

Make everything as simple as possible, but not simpler. -- Albert Einstein

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