Results 1 to 5 of 5

Math Help - Equation of a plane

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    37

    Equation of a plane

    Find the equation of the plane normal to r(t) = <e^t sin(pi/2 * t), e^t cos(pi/2 *t), t^2> when t = 1.

    Sorry not sure how to display pi.

    Where do I start?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kenshinofkin View Post
    Find the equation of the plane normal to r(t) = <e^t sin(pi/2 * t), e^t cos(pi/2 *t), t^2> when t = 1.

    Sorry not sure how to display pi.

    Where do I start?
    A vector tangent to r(t) is dr/dt. Evaluate at t = 1 and you have the normal vector to the plane.

    r(1) gives the position vector of a point in the plane.

    So you have a normal to the plane and a point in the plane.

    The equation of a plane is ax + by + cz = d where <a, b, c> is a normal to the plane and d is found by substituting a known point into ax + by + cz = d (once you've substituted the values of a, b and c of course).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    17
    I'm assuming that \textbf{r}(t)=\left<e^t\sin(\pi t/2),e^t\cos(\pi t/2),t^2\right> is meant to represent the x,y and z co-ordinates (respectively).
    If that is the case then, when t=1, we have \textbf{r}(t)=\left<e,0,1\right>.
    Recall the dot product - \textbf{a}.\textbf{b}=|\textbf{a}||\textbf{b}|\cos  (\theta), where \theta is the angle between the two vectors.
    Note that, if two vectors are orthogonal then \theta=\pi/2 and so \textbf{a}.\textbf{b}=|\textbf{a}||\textbf{b}|\cos  (\pi/2)=0 (as the cosine of \pi/2 is zero).
    Hence, you want to find another vector, call this \textbf{b}=\left<b_1,b_2,b_3\right> and find \textbf{r}(1).\textbf{b}=0.
    In other, words, you are looking for solutions to the equation eb_1+0*b_2+b_3=eb_1+b_3=0. This implies that b_1/b_3=-e and hence there are infinitely many solutions of this form


    Also, as a sidenote, if t is not constant, you get a really interesting kind of geometry based on spirals. It 'looks' a little bit like making your way from the top of a trumpet to the bottom (or vice versa, depending) only going around the inside of the horn (I think that's what it's called)


    Hope that helps

    Keith
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    37
    Quote Originally Posted by mr fantastic View Post
    A vector tangent to r(t) is dr/dt. Evaluate at t = 1 and you have the normal vector to the plane.

    r(1) gives the position vector of a point in the plane.

    So you have a normal to the plane and a point in the plane.

    The equation of a plane is ax + by + cz = d where <a, b, c> is a normal to the plane and d is found by substituting a known point into ax + by + cz = d (once you've substituted the values of a, b and c of course).
    So if I understand this right, I take the the derivative of the vector and evaluate it at t=1 then evaluate just the vector its self at t=1. <a,b,c> will be the derivative of the vector evaluated at t=1 and the point I get from the vector I substutute for (x,y,z). Is this correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kenshinofkin View Post
    So if I understand this right, I take the the derivative of the vector and evaluate it at t=1 then evaluate just the vector its self at t=1. <a,b,c> will be the derivative of the vector evaluated at t=1 and the point I get from the vector I substutute for (x,y,z). Is this correct?
    Yes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 1st 2011, 08:57 PM
  2. Replies: 3
    Last Post: December 5th 2010, 06:46 PM
  3. Replies: 3
    Last Post: October 12th 2010, 06:06 AM
  4. Replies: 4
    Last Post: May 26th 2010, 11:48 AM
  5. Replies: 2
    Last Post: May 23rd 2010, 11:46 AM

Search Tags


/mathhelpforum @mathhelpforum