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Homework Help: Consider the lines & find an equation of a plane [Cal III]

  1. Feb 7, 2015 #1
    Consider the Lines Below:

    = <2, 3, 0> + t<3, -3, 1>

    r = <5, 0, 1> + s<-3, 3, 0>

    A.) Find the point at which the given lines intersect.
    B.) Find an equation of the plane that contains these lines.

    2. Relevant equations

    The vector equation of a line is given by r = r0 - tv

    The equation of a plane is given by N . (r - r0)

    3. The attempt at a solution

    I worked everything to the best of my ability but I ended up getting an equation of a line as opposed to the equation of a plane... Did I do something wrong? Here is a picture of my work:

  2. jcsd
  3. Feb 7, 2015 #2


    Staff: Mentor

    I don't see anything wrong with your work. The two lines intersect at (5, 0, 1) (I checked), and the equation of the plane is indeed x + y = 5. In three dimensions this is a plane.
  4. Feb 7, 2015 #3
    Thank you so much. I think I figured out the error in my way of thinking. For some reason I was thinking of my answer as looking like a cube in R3. But since x + y = 5 is a line in R2, I could just scale the Z value infinitely in either direction in R3, as long as I'm restricted to that line. Thus, I would have a plane. Interesting. Thanks so much for your response!!!!
  5. Feb 7, 2015 #4


    User Avatar
    Science Advisor

    You say " I ended up getting an equation of a line" but that is not true. A line cannot be written as a single equation in 3 dimensions. What you have is a plane in which the coefficient of z happens to be 0. That simply means that the plane is parallel to the z-axis. You can think of it as the line defined by x+ y= 5 in the xy-plane extended parallel to the z-axis.
  6. Feb 8, 2015 #5
    Yes, you are absolutely right. Having 0 as the coefficient of Z is what scared me. My professor hadn't covered what that meant very well, so I was confused. Thank you very much for your input.
    Last edited: Feb 8, 2015
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