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Planes and parametric equations

  1. Nov 1, 2014 #1
    • OP warned about not using the homework template
    Hi everyone! I'm having some issues with this problem for linear algebra. I understand parametric equations fairly but I'm confused about the unit vector notation

    1) Consider the plane r(s,t)=2i + (t-s) j + (1+3s-5t) k find the z component of the point (2,-1, z0)

    For what values of s and t is this the case?

    I don't really know how to start the problem because it isn't in vector or parametric form like I'm used to.
     
  2. jcsd
  3. Nov 1, 2014 #2

    Mark44

    Staff: Mentor

    It's easy enough to get from the vector form to the parametric form of this plane.
    Here x = 2, y = t - s, and z = 1 + 3s - 5t, and you're given a point (2, -1, z0).
     
  4. Nov 1, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't know what you are "used to" but it certainly is in "vector form" and, as Mark44 says, it is easy to convert to parametric form:
    x= 2, y= t- s, z= 1+ 3s- 5t. In order to have [tex](x, y, z)= (2, -1, z_0)[/tex] you must have 2= 2, t- s= -1, and [tex]1+ 3s- 5t= z_0[/tex].

    Perhaps it is the fact that there is not a single "unique" answer that is bothering you?

    There are an infinite number of points, in fact an entire line, with x= 2, y= -1. From t- s= -1, we can get t= s+ 1 and so write [tex]z_0= 1+ 3s- 5(s+ 1)= 1+ 3s- 5s- 5= -4- 2s[/tex]. The set of such points consists of the line x= 2, y= -1, z= -4- 2s, for any s.
     
  5. Nov 1, 2014 #4
    Thank you for the replies! I left a little something out of the problem, it said find the z component so that it lies on the plane. Wouldn't that make it just one specific point?
     
  6. Nov 1, 2014 #5

    Mark44

    Staff: Mentor

    Work the problem through and see.
     
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