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    I worked out the cross product of B x C which gave {4, -14, -17}, and the cross product of A x C which gave {4, 6, 13}

    Unsure how to proceed.
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    You will need to find expressions for the three planes in the standard vector form r.n = p, where r is a general point on the plane, n is a vector normal to the plane and p is a scalar constant (different for each plane).

    For P1 you can reason that n is perpendicular to AC and BC since those two vectors are parallel to the plane. You can use any of the three known points in the plane to fix p.

    P2 and P3 are a bit easier as you are effectively given the two normal vectors.

    Once you have vector expressions for the three planes you will need to find the line intersection of two of them and then find where that line cuts the third plane.
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    (Original post by old_engineer)
    You will need to find expressions for the three planes in the standard vector form r.n = p, where r is a general point on the plane, n is a vector normal to the plane and p is a scalar constant (different for each plane).

    For P1 you can reason that n is perpendicular to AC and BC since those two vectors are parallel to the plane. You can use any of the three known points in the plane to fix p.

    P2 and P3 are a bit easier as you are effectively given the two normal vectors.

    Once you have vector expressions for the three planes you will need to find the line intersection of two of them and then find where that line cuts the third plane.
    For P1 I got -20y+10z+40=0
    P2: 4x-14y+17z+40=0
    P3: 4x+6y+13z-40=0

    Don't really know how to solve this
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    (Original post by old_engineer)
    ..
    I think there may be a slightly better way here:

    We can write a general point r on P_1 in the form {\bf r} = {\bf a} + \lambda ({\bf b-a}) + \mu ({\bf c-a})

    We can more or less write down conditions (of the form r.x = K for suitable vector x and constant K) for the other two planes. Sub in the general point for P_1 and you should end up with 2 simultaneous equations for \lambda, \mu. Solve that, and finally calculate r.
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    (Original post by DarkEnergy)
    For P1 I got -20y+10z+40=0
    P2: 4x-14y+17z+40=0
    P3: 4x+6y+13z-40=0

    Don't really know how to solve this
    P1 looks OK, albeit you could divide through by 10. But it looks as though you've made P2 perpendicular to OB and OC rather than perpendicular to BC. Similar comment for P3.
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    (Original post by DFranklin)
    I think there may be a slightly better way here:

    We can write a general point r on P_1 in the form {\bf r} = {\bf a} + \lambda ({\bf b-a}) + \mu ({\bf c-a})

    We can more or less write down conditions (of the form r.x = K for suitable vector x and constant K) for the other two planes. Sub in the general point for P_1 and you should end up with 2 simultaneous equations for \lambda, \mu. Solve that, and finally calculate r.
    Good call.
 
 
 
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