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Math Help - Finding equation of plane?

  1. #16
    A Plied Mathematician
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    I would agree that you've found two points on the line. Now what should you do?
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  2. #17
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    find the equation by using the two points:

    i get:

    6i+6j-k+t(7i+4j-k)
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  3. #18
    A Plied Mathematician
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    What is that the equation of? The line? You already have that. What does the equation of a plane look like?
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  4. #19
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    ok, let me start again i have the three points (6,6,-1) (13,10,-2) (1,3,2)

    n=OB-OA x OC-OA
    n= 7i+4j-k x -5i-3j+3k

    find the cross product

    i(12+3)-j(21-5)+k(-21+20)
    15i-16j-k

    sub points (1,3,2)

    i get 15(1)-16(3)-(2)
    15-48+2=-35

    15x-16y-z=-35

    this is incorrect, however shouldn't it be still correct even if the points are different
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  5. #20
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    Name the three points: 6,6,-1),~Q13,10,-2),~\&~R1,2,3)" alt="P6,6,-1),~Q13,10,-2),~\&~R1,2,3)" />.

    Now find the normal: N = \overrightarrow {PQ}  \times \overrightarrow {PR} .

    Then write the equation of the plane determined by the three points:
    N \cdot  < x - 1,y - 2,z - 3 >  = 0

    BTW: [tex]\times [/tex] gives \times in LaTeX.
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