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Linear Algebra Question

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    How to convert an implicit representation to a parametric

    2. Relevant equations
    Implicit: ax + by + cz = d.
    Parametric: (x,y,z) = (xp, yp, zp) + s(xv,yv,zv,) + t(xw,yw,wz)

    3. The attempt at a solution
    For parametric to implicit it would look like this:
    (x,y,z) = (0, 7, 6)T + s(-8, 7, -2)T + t(4, 1, -2)T
    1. Calculate the crossproduct of (-8,7,-2) x (4,1,-2) = (12,24,36) = (1,2,3) = (a,b,c)
    2. Calculate d using the first point: (1,2,3) (0,7,6) = d = 32
    3. Result: 1x + 2y + 3z = 32

    But how to go the other way around? I have looked all over google and youtube, but couldn't find it anywhere! :(.
    Example could be: x + 2y + 3z = 32

    I'm sorry that I couldn't try it before posting this. But I have no idea how to start :(.

    Please help me.. my exam is on tuesday and I've been learning very hard.
    Last edited: Oct 7, 2011
  2. jcsd
  3. Oct 7, 2011 #2


    Staff: Mentor

    Starting with your equation x + 2y + 3z = 32,
    1. Use this equation to find three points in the plane - any three, as long as they aren't all on the same line. Here are two points in this plane: P(0, 16, 0) and Q(32, 0, 0).
    2. Form the vectors PQ and PR. Neither vector should be a scalar multiple of the other.
    You now have enough information to write parametric equations for the plane.

    <x, y, z> = OP + r * PQ + s * PR
  4. Oct 8, 2011 #3
    Hi, thanks very much! It's much more clear now! Three more questions and my attempt:

    1. What is the key to create three points on different lines?
    Attempt: I suppose make sure that the combination should not be a scalar multiply of the other points?. So point R could be (3, 6, 5[itex]\frac{2}{3}[/itex]) right?

    2. By forming vectors; do you mean substract the points, like P - Q, P - R?
    P-Q = (32, 0, 0) - (0, 16, 0) = (32, -16, 0)
    P-R = (3, 6, 5[itex]\frac{2}{3}[/itex]) - (0, 16, 0) - = (3, -10, 5[itex]\frac{2}{3}[/itex])
    OP = (3, 6, 5[itex]\frac{2}{3}[/itex])

    So a solution would be: <x, y, z> = (3,6,5[itex]\frac{2}{3}[/itex]) + r(32, -16, 0) + s(3, -10, 5[itex]\frac{2}{3}[/itex]) is that right?

    3. How can you be sure that the points you choose won't result in vectors that are linearly dependent? Attempt: find three points that aren't on the same line. You can be sure that the vectors formed by substracting those points are not linearly dependent.
    Last edited: Oct 8, 2011
  5. Oct 8, 2011 #4


    Staff: Mentor

    Any point is fine whose coordinates satisfy the equation x + 2y + 3z = 32, and the one you show does so. However, you should write the last coordinate as 17/3 rather than as a mixed number - it makes calculations easier.
    Sort of. You're not really subtracting points, though, so you wouldn't be understood if you wrote "P - Q".

    Here's how you get the vector PQ. Here O is the origin.

    OP + PQ = OQ
    This means that adding the vector OP to the vector PQ results in the vector OQ

    Subtracting the vector OP from both sides gives
    PQ = OQ - OP

    For example, working in R2, if P = (3, 2) and Q = (2, 5), PQ = <2 - 3, 5 - 2> = <-1, 3>. I'm using angle brackets to distinguish vectors from points.

    The same ideas can be extended to vectors in higher dimensions.
    Again, don't write coordinates as mixed numbers. It's not wrong to do this, but you set yourself up for making mistakes. In mathematics at the level of algebra and above, mixed numbers are rarely used.

    I'll let you check for yourself. Pick values for r and s (r = 1 and s = 0 are pretty convenient). What point do you get? Does it satisfy the equation x + 3y + 3z = 32?

    Pick another set of values for r and s (this time, try r = 0 and s = 1). What point do you get this time? Does it satisfy the plane equation?

    If both points satisfy your equation, you can be almost certain you have the right vector equation. To be absolutely sure, pick one more set of values for r and s, and check that point. As long as all three points aren't on the same straight line, your work is good, because three points that are not collinear completely determine a plane.
  6. Oct 8, 2011 #5
    Thanks so much!!

    For the record: I made some miscalculations. I've fixed them in my first post :).
    Last edited: Oct 8, 2011
  7. Oct 8, 2011 #6
    Would you also like to give an answer to my last question?

    3. How can you be sure that the points you choose won't result in vectors that are linearly dependent?
    Attempt: Because a plane is non-collinear; if the points on the plane aren't on the same line, the vectors formed by those points are not linearly dependent.
    Last edited: Oct 8, 2011
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