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Math Help - Find perpendicular line given parametric eqn and point

  1. #1
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    Find perpendicular line given parametric eqn and point

    Hi all

    I need to find a line perpendicular to a known line.

    I have the parametric equation for the known line and the point on the known line through which the perpendicular line should pass, how do I calculate what the vector of the perpendicular line should be?

    I know its probably quite simple but I have learn t all I know about about this on YouTube.

    Thanks
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  2. #2
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    Re: Find perpendicular line given parametric eqn and point

    Quote Originally Posted by drcronik View Post
    Hi all
    I need to find a line perpendicular to a known line.
    I have the parametric equation for the known line and the point on the known line through which the perpendicular line should pass, how do I calculate what the vector of the perpendicular line should be?
    I know its probably quite simple but I have learn t all I know about about this on YouTube.
    If you have a given line then you know a point, P, on that line and the direction vector, D.
    Now pick any vector, N, that is perpendicular to D. That is N\cdot D=0
    The line you want is P+tN.
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  3. #3
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    Smile Re: Find perpendicular line given parametric eqn and point

    Hi Plato

    Thanks for the speed response

    Do I solve D x N = 0 for each vector ie X,Y,Z

    my line is in the form [X,Y,Z] x t[Vx,Vy,Vz]

    so for the perpendicular line

    Vxd x Vxn = 0 and Vyd x Vyn = 0 and Vzd x Vzn = 0.

    Did I understand this correctly?

    Thanks
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  4. #4
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    Re: Find perpendicular line given parametric eqn and point

    Quote Originally Posted by drcronik View Post
    Do I solve D x N = 0 for each vector ie X,Y,Z
    my line is in the form [X,Y,Z] x t[Vx,Vy,Vz]
    so for the perpendicular line
    Vxd x Vxn = 0 and Vyd x Vyn = 0 and Vzd x Vzn = 0.
    You are making far too much out of this.
    Say the line is <3,-2,5>+t<1,-3,4>
    The direction vector is <1,-3,4>.
    Pick any vector perpendicular to that vector.
    <-1,1,1> will work. <4,0,-1> will also work.
    Infinitely many will work. All that is necessary is <-1,1,1>\cdot<1,-3,4>=0.
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  5. #5
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    Re: Find perpendicular line given parametric eqn and point

    I understand, you mean the dot product o f the vectors must equal 0. I did a bit more research.

    Here is what I am trying to achieve. Find perpendicular line given parametric eqn and point-route.jpg

    Attached is a diagram of a possible route of a Cartesian robot.

    I need to link path AB and path BC with an arc, to ensure smooth transition into and out of the arc paths AB & BC need to be tangents to the arc. The arc should be between points D & G. Length DB = BG

    Using the cad software I have think I have found the solution but I do not know how to calculate it manually.

    This is what I have done.

    1. Draw 2 circles e & f with centers D & G and axis is path AB & BC respectively, the radius of the circles will be twice the distance of BD

    2. Find the intersecting points of circles e & f. There will be 2. Lets call these points H & I

    3. Find the midpoint of line HI. Lets call it J

    4. J will then be the center point of the arc starting at point D and ending at point G.

    I have tested this in 2d using the cad and replaced the circles with lines perpendicular to the path and it works perfectly so I am hoping this will work in 3d.

    Like I said I have the theory of what I need to do but am struggling with how to actually do the calculation, the cad program does all the calculations for me.

    I realize now that calculating the perpendicular line will not work as there are an infinite no. of these lines and I need the two to intersect.

    Can you help me? I have started this as a new thread aswell so that I get a different heading.

    Thanks
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