Results 1 to 4 of 4

Math Help - Show that curve is in a plane

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Show that curve is in a plane

    Show that the curve,

    \overline{r(t)} = 4cost\overline{i} + 4sint\overline{j} - sint\overline{k}

    is in a plane, and find the equation for the plane..

    If I substitute into the equation of a plane I get,

    a(4cost)+b(4sint)+c(-sint)+d=0,

    I see that it will hold if for example b=1 and c=4, and so I guess that I have showed that the curve is in a plane, and could use [0,1,4] as my normal vector.

    Is there a nice way of doing this so?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    There are two ways that I know of to show that a curve is in a plane. The basic idea is that you have to show either that all the tangent vectors to the curve are in a plane, or that all the secant vectors are in a plane. If you know calculus, probably the first option is easier; otherwise, do the second option.

    1. Show that all the tangent vectors are in a plane. Take the derivative of \vec{r}(t) - call it \vec{v}(t). Show that there exists a nonzero constant vector \hat{n} such that \vec{v}(t)\cdot\hat{n}=0 for all t.

    2. Show that all the secant vectors are in a plane. Produce all vectors of the form \vec{s}(u,v)=\vec{r}(u)-\vec{r}(v). Show that there exists a nonzero constant vector \hat{n} such that \vec{s}(u,v)\cdot\hat{n}=0 for all u,v.

    Unfortunately, your method is not going to work, because your method is trying to show that the radius vectors from the origin to points in the plane are in the plane. That will only work if the origin happens to be one of your radius vectors (i.e., the plane passes through the origin). If the set of radius vectors does not include the origin, then you're going to have to do what I am suggesting.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    \vec{v}(t) = -4sint\vec{i} + 4cost\vec{j} - cost\vec{k}

    I let \vec{n} = [a_1,a_2,a_3] and so,

    \vec{v}(t)\cdot \vec{n} = -a_14sint + a_24cost - a_3cost = 0

    This holds for \vec{n}=[0,1,4].

    Is this what you meant?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    That looks good to me!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that the plane...
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 2nd 2011, 03:20 AM
  2. Replies: 1
    Last Post: February 11th 2011, 09:38 PM
  3. Curve in the plane
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 24th 2010, 04:49 AM
  4. Length of a Plane Curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2009, 03:28 PM
  5. normal plane to the curve intersecting a plane
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 23rd 2009, 10:53 AM

Search Tags


/mathhelpforum @mathhelpforum