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Math Help - move a point along a plane in 3d space

  1. #1
    ave
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    move a point along a plane in 3d space

    I have a triangle in 3d space, and know all the angles and the coordinates of the 3 corners.
    I also have a point on one of the edges
    from that point I want to draw a perpendicular line, that follows the plane of the triangle- how would I do this mathematically
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  2. #2
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    Quote Originally Posted by ave View Post
    I have a triangle in 3d space, and know all the angles and the coordinates of the 3 corners. I also have a point on one of the edges from that point I want to draw a perpendicular line, that follows the plane of the triangle- how would I do this mathematically
    I not at all sure that I understand what it is you require.

    Is this it? Say that A,~B,~\&~C are the three vertices of the triangle and point <br />
D \in \overline {AB} .
    Now you want a line in the plane of the triangle through D which is perpendicular to <br />
\overline {AB} .
    If that is the case, the solution a straightforward.

    Otherwise, you need to try again to explain the question.
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  3. #3
    ave
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    Quote Originally Posted by Plato View Post
    I not at all sure that I understand what it is you require.

    Is this it? Say that A,~B,~\&~C are the three vertices of the triangle and point <br />
D \in \overline {AB} .
    Now you want a line in the plane of the triangle through D which is perpendicular to <br />
\overline {AB} .
    If that is the case, the solution a straightforward.

    Otherwise, you need to try again to explain the question.
    HI Plato- unfortunately my maths has a few holes when it comes to terminology and symbols, I can understand it quite easily, but have never studied it at uni, this unfortunately is my downfall, so I will try explain what I am looking for in picture form

    the red dot will be a point that lies on the edge of a triangle- I need to draw a perpendicular line- (in green) from that point.

    if you look in the side view (z/y axis), it must lie on the same plane as the triangle- hope that makes sense??


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    Then the equation of the line is \ell (t) = D + t\left[ {\overrightarrow {AB}  \times \left( {\overrightarrow {AB}  \times \overrightarrow {AC} } \right)} \right].
    Where the points are as I said above.
    But if you are weak in mathematics that may not help you at all.
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