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Hi,
Given two planes A and B, each of them are in the form of Parallelogram with known vertices.
Eg. For plane A there are four points(p1,p2,p3, and p4) as the four vertices of the parallelogram. And also there are for plane B.
I want to check whether a given point, eg.(6,10,3), is in between of these two planes or not. How can i do this? can you give me a step by step solution or example how to solve this.
Regards,
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I'm not sure what you mean by 'in between'. Are the planes restricted to being parallel to each other? If they're not, I can't think how a point could be between them.
Why did the vector cross the road?
It wanted to be normal.
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If you want to just do a really rough check, you can average all 8 points to
get the middle point between the two parallelograms. Then see if your
point of interest is within a certain radius from that middle.
The radius is half the distance between the centers of each parallelogram.
Those 2 centers are just the average of their 4 points.
And the distance formula is "distance" = squareRootOf(xDifference^2 + yDifference^2 + zDifference^2)
igloo myrtilles fourmis
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yes they are parallel.
Now i want to kno how to write a plane equation from the above vertices(vertices of plane A and B) in the form of plane equation Ax+By+Cz+D=0. After getting equation for two planes is there any way to check the given point is between of these two plane or not?
Regards,
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For more information the points are:
(10,10,10), (50,10,15),(55,20,10),(15,20,5)-----Plane A.
(20,10,0), (60,10,5),(65,20,0),(25,20,-5)-------Plane B.
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use these equations to get A,B,C, and D.
http://local.wasp.uwa.edu.au/~pbourke/geometry/planeeq/
Now the A, B, C will be the same for both planes.
Your D values will be the limits for your
determination of being between the planes.
When you get a point to check,
then calculate Ax + By + Cz and if it is in between
your 2 values of D, then you are between the planes!!!!
I hope I'm right, since I just learned this a bit.
igloo myrtilles fourmis
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Thank you very much
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