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Is there a mathmatician in the house?

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  • Yenzarill
    replied
    I flew a plane in 3D space once.

    Normal? what? ohh... me? hell no.

    Leave a comment:


  • MortalMonkey
    replied
    w00t!
    Thanks Thebos, it's working perfectly!
    Methinks me might have to script you a funny mutator with this...

    Leave a comment:


  • MortalMonkey
    replied
    Thanks everyone.

    If I know the distance from A to Q, it's really easy:

    Q = A + Normal( B - A ) * distance

    So the whole formula in UScript would be:

    Q = A + Normal( B - A ) * ((( B - A ) dot ( P - A )) / VSize( A - B ));

    Would that be correct?

    Leave a comment:


  • ElJoelio
    replied
    Hmm.. maybe I misunderstood the question - I thought he was looking for the normal of the plane created by the 3 points.

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  • Thebos
    replied
    Not sure how the cross-product references above are to be applied (unless I misunderstood the intial question). Apply the definitions from my first comment and add plane(ABP) as the plane defined by points A,B and P. { Vec(AB) X vec(AP) } results in a vector perpendicular to plane(ABP) that intersects plane(ABP) at point A. Not sure how that helps find point Q on line(AB) such that line(PQ) is perpendicular to line(AB). If it does, please elaborate as I am curious. Thanks.

    Leave a comment:


  • Thebos
    replied
    Been a while, but I believe it goes like this:

    Consider points A,B,P,Q.

    Points A and B form vector AB (ie vec(AB)).

    Point P is a known point in space. It also forms (with point A) vec(AP).

    Point Q exist on line AB and line PQ is perpendicular to to line AB

    1) Find the dot product [(vec(AP))*(vec(AB))]. Result is a scalar.

    2) Find the distance of point Q from point A by dividing the magnitude of vec(AB) by the dot product just obtained. (ie { (vec(AP))*(vec(AB)) } / { |vec(AB)| } ).

    Hope that helps.

    <Edited because I made a mistake>

    Leave a comment:


  • ElJoelio
    replied
    I sez: http://www.netcomuk.co.uk/~jenolive/vect8.html

    Dalai sez: http://physics.syr.edu/courses/java-suite/crosspro.html

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  • LaughingRat
    replied
    Adjust your coordinate axes so that P and AB are co-planar, and solve as for 2D?

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  • Lurox
    replied
    Ask Ordinator, he made a nice tutorial on UEd to real scale.

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  • MortalMonkey
    started a topic Is there a mathmatician in the house?

    Is there a mathmatician in the house?

    I just can't figure it out:
    How on earth does one calculate the base point of a normal (N) from a point (P) to a line (AB) in 3d space?
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