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Cartesian equation of plane using two lines?

  1. Apr 28, 2012 #1
    Can someone help with this multi-part question. i did the first three but it doesnt seem right!
    (a) Find the equation of the line l through P(1, 1, 2) and Q(1, 0, 4) in vector, parametric and Cartesian forms.
    (b) Find the vector form of the line k through R(0, 3, 1) which is perpendicular to l.
    (c) Find the Cartesian form of the plane containing the lines l and k.
    (d) Find the vector form of the line through T(0, 0, 3) which is normal to this plane.
    so i guess i did the first bit alright : cartesian : z=6+4x , y=6
    for the second one i got this line : (-5,4,2)+t(28,-37,7) , i timed it by 17 to get rid of the denominator
    third part : the lines seem to have no intersection so i am stuck!

    appreciate it if anyone can help out
  2. jcsd
  3. Apr 29, 2012 #2


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    Science Advisor

    To which is this an answer? Clearly not (a) because x is always equal to 1 but y is NOT always equal to 6 and z is not always equal to 6+ 4(1)= 10.

    It's impossible to tell what you are saying because you talk about ""first", "second", and "third" when the problems are labeled "a", "b", "c", and "d" and "a" has three parts!
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