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Math Help - Simple method, finding position vector

  1. #1
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    Simple method, finding position vector

    Line L is 3i-5j+5k + t(2i+j-k)
    Point A has position vector i+2j+2k
    Point B is on the line L at 5i-4j+4k

    The line through A and B is perpendicular to L.
    Points C and D are on line L and 4*sqrt(5) units from A.

    Find the position vectors for C,D and the area of triangles ABC, ABD

    I got the solutions to be 9i-2j+2k and i-6j+6k and I got the area to be 4*sqrt(21)

    My problem is that I am not really happy with my method and I intuitively feel there is some nice simple way to do this. I also think that the question is looking for the use of the cross product in working out the area. I did not use it.

    Andrew
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    Re: Simple method, finding position vector

    What method did you actually use? Why are you unhappy with it?
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    Re: Simple method, finding position vector

    I was hoping you wouldn't ask, in case it really is clunky. I figured out the magnitude of AB to be 2*sqrt(14) which allowed me to get the third side of the right triangle as 2*srt(6). That is how I got the area without cross product. Then I had the direction vector for BC and the magnitude, so I was able to work out the position of C relative to B.

    Andrew
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