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Math Help - Point of intersection

  1. #1
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    Point of intersection

    Finding the point of intersection of straight through the points A (3,-5,2) and B (11,-3,6) with the straight (r): (x,y,z) = (5,-3,2) + t (4,-2,4).


    Answer:
    R (7,-4,4)
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  2. #2
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    \vec{s} = \vec{s}_{0} + \lambda \vec{v}

    First find the direction vector of your line, i.e. \vec{v} = \vec{AB} = (8, 2, 4).

    Now, s_{0} can simply the vector from the origin to the point A (or B, doesn't matter as long as it's on the line). So, our line is: \vec{s} = (3, -5, 2) + \lambda(8,2,4)

    Now to see where lines s and r intersect, we set the them equal to each other, i.e. (3, -5, 2) + \lambda(8,2,4) =  (5,-3,2) + t (4,-2,4).
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  3. #3
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    ok, tanks. I managed to solve
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