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Math Help - To find the parametric equation of a line in Euclidian space knowing the following.

  1. #1
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    To find the parametric equation of a line in Euclidian space knowing the following.

    If known that a line "m" passes through the point P (2,-1,3) in 3d vector space

    and knowing it is parallel to the plane pi (x-y+z=0)


    and cuts the line l { (1-2g)i + (3+2g)j +(2-g)k } (where "g" is just a prameter, i don't have "lambda" on this keyboard )


    how would you determine the equation of the line "m"?

    Help very appreciated thanks!!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Find the plane \pi_1 parallel to \pi and contains P . Find the intersection point Q between \pi_1 and the line l . The solution is the line passing trough P and Q .


    Fernando Revilla
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  3. #3
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    Here's how I would do it. Any line through (2, -1, 3) can be written in the form x= At+ 2, y= Bt- 1, z= Ct+ 3 where <A, B, C> is a vector in the direction of the line. Since the line is to be parallel to the plane x- y+ z= 0, <A, B, C> must be perpendicular to the normal to that plane, <1, -1, 1>. That is, <A, B, C>.<1, -1, 1>= A- B+ C= 0. Since that is true for any multiple of the vector, we can, for example, take A= 1 and have 1- B+ C= 0 so that B= C+ 1. That is, we can write the equation of the line as x= t+ 2, y= (C+ 1)t- 1, and z= Ct+ 3 and we only need to find C. The last condition is that the line intersect the line given by x= 1- 2g, y= 3+ 2g, z= 2- g. Putting those into the equation of the line gives 1- 2g= t+ 2, 3+ 2g= (C+ 1)t- 1, and 2- g= Ct+ 3. From the first, t= -1- 2g. Putting that into the second equation gives two equations for C and g. Of course, C gives you the equation of the line you want.
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