Results 1 to 4 of 4

Thread: Equation of plane question.

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    24

    Equation of plane question.

    Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

    Question:
    Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

    Would the normal vector to the plane simply be the above vector? But then would I use (0,0,0) as a point in the plane?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by dmbocci View Post
    Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

    Question:
    Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

    Would the normal vector to the plane simply be the above vector? No.

    But then would I use (0,0,0) as a point in the plane? No.

    The line tangent to $\displaystyle \vec{\text{r}}(t)$ is in the direction of $\displaystyle \displaystyle {{d}\over{dt}}\vec{\text{r}}(t)$.

    Use the position specified by $\displaystyle \vec{\text{r}}(\pi/2)$ as a point in the plane.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    4
    Quote Originally Posted by dmbocci View Post
    Its been a little while since I've done these ones and I'm having a little trouble wrapping my head around this one.

    Question:
    Find the equation of the plane perpendicular to the line tangent to r(t)=(3sint)i - (2cost)j + (t)k at t=pi/2.

    Would the normal vector to the plane simply be the above vector? But then would I use (0,0,0) as a point in the plane?
    Dear dmbocci,

    First find the unit tangent vector to the curve.

    $\displaystyle \underline{T}=\dfrac{d\underline r}{ds}=\dfrac{d\underline r}{dt}\times\dfrac{dt}{ds}=\dfrac{\underline{r'}(t )}{\mid\underline{r'}(t)\mid}$

    Then find the unit normal using,

    $\displaystyle \dfrac{d\underline{T}}{ds}=\kappa\underline N$

    The vector equation of the plane perpendicular to the tangent line would then be,

    $\displaystyle \left(\underline{R}-\underline{r}\left(\frac{\pi}{2}\right)\right).\un derline N=0$ where $\displaystyle \underline R$ represent the position vector of any point in the plane.

    Hope this helps you to solve the problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    24
    Thanks guys, great help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 1st 2011, 07:57 PM
  2. Replies: 3
    Last Post: Dec 5th 2010, 05:46 PM
  3. Replies: 3
    Last Post: Oct 12th 2010, 05:06 AM
  4. Replies: 4
    Last Post: May 26th 2010, 10:48 AM
  5. Replies: 2
    Last Post: Mar 7th 2009, 02:20 PM

Search Tags


/mathhelpforum @mathhelpforum