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Let A, B, and C be vectors in R3...prove r lies in a plane

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    ##\vec { \dot { r } } =(t+1)\vec { A } +(1-sint)\vec { B } \quad \vec { r(0) } =\vec { C } ##
    a. Find an equation of the tangent line to the curve at ##\vec { r(0) } =\vec { C } ##.
    b. Use a definite integral to find ##\vec { r(t) } ##
    c. If ##A## and ##B## are non parallel, prove ##\vec { r(t) } ## lines in a plane.

    2. Relevant equations


    3. The attempt at a solution
    a. ##\vec { T } =\vec { C } +t(\vec { A } +\vec { B } )##
    b. ##\vec { r(t) } =\frac { 1 }{ 2 } \vec { A } { t }^{ 2 }+\vec { A } t+\vec { B } t+\vec { B } cost-\vec { B } +\vec { C } ##
    c. How can I use an equation of a plane to prove this part?
     
  2. jcsd
  3. Oct 19, 2016 #2

    andrewkirk

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    A vector equation of a plane through ##\vec C## is
    $$\{\vec C+s\vec A +t\vec B\ :\ s,t\in\mathbb R\}$$
    where ##\vec A,\vec B## are two non-parallel vectors.

    The equation you have derived in (b) can be expressed in that format, hence all the points in the curve lie in the plane defined by that formula.

    If you have been given a different form for the vector equation of a plane, a few more steps may be necessary.
     
  4. Oct 19, 2016 #3
    Oh I haven't seen that equation before. The one we were taught is ##\vec { N } \cdot (\vec { P } -\vec { { P }_{ 0 } } )=0##, where ##\vec { N } ## is normal to ##(\vec { P } -\vec { { P }_{ 0 } } )##. I can't figure out where to get ##\vec { N } ## from

    Where can I find more on that equation? I've tried google already
     
  5. Oct 19, 2016 #4

    andrewkirk

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    Actually, on reflection, what I gave is a formula for a plane rather than an equation. The question doesn't say whether to use a formula or an equation, but since you've been given the equation you quote in your last post, it's best to use that.

    In your equation, the arbitrary point on the plane is ##\vec P##, the specified point on the plane is ##\vec P_0## and the normal to the plane is ##\vec N##.

    So the challenge is to find values for ##\vec P_0## and ##\vec N## such that all points ##\vec r(t)## on the curve satisfy the equation when we substitute them for ##\vec P##.

    Finding a satisfactory ##P_0## is easy. Given the equation you derived in (b), what is a vector (point) that must be on the plane containing the curve?

    Then, to find ##\vec N##, how can we write a formula for a vector that is perpendicular to the vector obtained by subtracting ##\vec P_0## from the right hand side of the formula in (b)?
     
  6. Oct 20, 2016 #5
    not sure if I have the right idea but if the curve does lie in a plane, then then any derivative lies in the same plane as any position. So, their cross product will give the appropriate normal vector. If the curve doesn't lie in a plane, then the dot product will not equal zero when we plug in ##r(t)##.
    ##\vec { \dot { r(0) } } =\vec { A } +\vec { B } \quad \vec { r(0) } =\vec { C } \\ \vec { N } =\vec { r(0) } \times \vec { \dot { r(0) } } =\left( \vec { A } +\vec { B } \right) \times \vec { C } \\ \vec { N } \cdot \left( \vec { r(t) } -\vec { r(0) } \right) =0\\ \left[ \left( \vec { A } +\vec { B } \right) \times \vec { C } \right] \cdot \left( \vec { r(t) } -\vec { C } \right) =0##
    when I do the computation above, I end up with ##\left( cost-\frac { 1 }{ 2 } { t }^{ 2 }-1 \right) \left( \vec { A } \times \vec { C } \cdot \vec { B } \right) =0##?
     
  7. Oct 20, 2016 #6

    andrewkirk

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    You have an acceptable answer for ##\vec P_0##, but not for ##\vec N##.

    To work out ##\vec N##, you might find it easier to think about the plane that is parallel to the plane of the curve but passes through the origin. If you can find an expression for the normal to that plane, using only ##\vec A,\vec B,\vec C## (and not necessarily all of them) then you can use the same normal for the plane of the curve, since parallel planes have the same normal.
     
  8. Oct 20, 2016 #7
    Ah, I think I have to sort of "guess" to get the normal vector I tried ##\vec { A } \times \vec { B } ## and it worked! So I have to somehow "use" the given information and hope it works.
     
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