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Thread: Vector how to find the equation..

  1. #1
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    Vector how to find the equation..

    here is the question that I have and I have confuse

    find the equation of the straight line which is perpendicular to the plane
    2x+3y+4z=5
    and which goes through the point (1,1,7)
    is the point (5,7,15) on this line?



    r=(1,1,7)+t(2,3,4)
    that maybe is something!?

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  2. #2
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    yes that's correct
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  3. #3
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    so the solution is that: r=(1,1,7)+t(2,3,4)?!
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  4. #4
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    yes because the normal vector of a plane given by the equation ax+by+cz+d=0 is [a,b,c]
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  5. #5
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    ok thx!
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  6. #6
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    You still have to answer the question "is the point (5,7,15) on this line?"
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  7. #7
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    for that i have to say x=1+2t y=1+3t and z=7+4t and then solve with respect to t right!?
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  8. #8
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    Quote Originally Posted by NaNa View Post
    for that i have to say x=1+2t y=1+3t and z=7+4t and then solve with respect to t right!?
    Yes. x= 1+ 2t= 5, y= 1+ 3t= 7, z= 7+ 4t= 15. Obviously you can solve the first equation, 1+ 2t= 5, then check to see if that value works for the other two equations.
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  9. #9
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    Yeah i have done it!!!thx!!!
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