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Basis for a Plane that is given.

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the plane 3x1-x2+2x3 = 0 in R3. Find a basis for this plane. Hint: It's not hard to find vectors in this plane.

    2. Relevant equations
    Plane: 3x1-x2+2x3 = 0 in R3.

    3. The attempt at a solution
    Let,
    A = [itex]\left[3,\right.\left.-1,\right.\left.2\right][/itex] [itex]\rightarrow[/itex] [itex]\left[\frac{3}{3},\right.\left.\frac{-1}{3},\right.\left.\frac{2}{3}\right][/itex] (Row Reduced)

    [itex]\Rightarrow[/itex] x1 = [itex]\frac{1}{3}[/itex]x2-[itex]\frac{2}{3}[/itex]x3, x2 is free, x3 is free.

    [itex]\Rightarrow[/itex] [itex]\left\{x_2[\frac{1}{3},1,0]+x_3[\frac{-2}{3},0,1] | x_2, x_3 \in R\right\}[/itex]

    [itex]\Rightarrow[/itex] Basis of Plane = [itex]\left\{[\frac{1}{3},1,0],[\frac{-2}{3},0,1]\right\}[/itex]
     
  2. jcsd
  3. Jun 1, 2014 #2

    jedishrfu

    Staff: Mentor

    you didn't take advantage of the hint. A o B = |A| |B| cos(ABangle) and in your case you've got <3,-1,2> o <x1,x2,x3> = 0
     
  4. Jun 1, 2014 #3
    Don't I just get the plane then, because
    <3,-1,2> DOT <x1,x2,x3> = 3x1-x2+2x3
    which then is equal to 0.
     
  5. Jun 1, 2014 #4

    jedishrfu

    Staff: Mentor

    it means that the vectors you are looking for are perpendicular to <3,-1,2>

    It also means that <0,0,0> is in the plane

    so what if you set x1=0 and solve for x2 and x2 then you'd have one vector and similarly set x2=0 and solve for x1 and x3 to get the other vector.
     
  6. Jun 1, 2014 #5
    I set x1 = 0, so then -x2 + 2x3 = 0, so x2 = 2, x3 = -1 ... <0, 2, -1> = u
    I set x2 = 0, so then 3x1 + 2x3 = 0, so x1 = 2, x3 = -3 ... <2, 0, -3> = v

    <0, 2, -1> DOT <3, -1, 2> = 0
    <2, 0, -3> DOT <3, -1, 2> = 0

    So then my basis for the plane is <0, 2, -1>, <2, 0, -3>.

    If this is the correct basis, then I'm wondering how come the first basis was wrong? If in my first attempt I solved the problem every other way that I do to get the basis's for other matrices.
     
  7. Jun 1, 2014 #6

    jedishrfu

    Staff: Mentor

    I never said they were wrong. I just showed you how I would attack the problem using the hint that was given. In your case any two non parallel vectors in the plane can form a basis from which other vectors can be generated.

    I didn't quite understand your method and saw that you didn't consider the hint. My apologies for any confusion.

    Anyway, one last check you can do is to dot the basis vectors with each other to make sure they aren't parallel or anti-parallel (parallel but pointing in opposite directions).
     
  8. Jun 1, 2014 #7
    Thank you for the feed back! I dotted my original basis's with the vector that forms the plane and got 0 for both the vectors in the basis.
     
  9. Jun 1, 2014 #8

    jedishrfu

    Staff: Mentor

    Also dot them against each other as a check against them being parallel.
     
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