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Thread: Find an equation of a plane normal to a given vector

  1. #1
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    Find an equation of a plane normal to a given vector

    I have to find an equation for a plane normal to$\displaystyle r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 > $
    when $\displaystyle t=1$

    Is the equation in the form:

    $\displaystyle <a,b,c> . <x,y,z> - r(1) = 0?$

    would i take the derivative of x, y, and z?

    Would this require the product rule for x, and y individually? not z of course!


    here is my attempt at just the product rule of x alone:

    $\displaystyle e^t d/dx(sin(\frac{\pi{}}2{t}) + sin(\frac{\pi{}}2{t}) . d/dx (e^t) $

    would it then be?

    $\displaystyle e^t cos(\frac{\pi{}}2{t}) + e^t sin(\frac{\pi{}}2{t}) $

    I guess my 2 questions are. is my generic form correct of the equation for a plane normal to the vector correct above?
    and when i take the derivative do i use the product rule and did i do it correctly?

    I am not sure how to get <a,b,c> and the point <x,y,z> either.

    I am completely lost

    Thank you
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  2. #2
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    Re: Find an equation of a plane normal to a given vector

    Quote Originally Posted by icelated View Post
    I have to find an equation for a plane normal to$\displaystyle r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 > $
    when $\displaystyle t=1$
    Is the equation in the form:
    $\displaystyle <a,b,c> . <x,y,z> - r(1) = 0?$

    You have a great deal of that wrong.
    The equation in the form:
    $\displaystyle \nabla r(1) \cdot \left( {\left\langle {x,y,z} \right\rangle - r(1)} \right) = 0$
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  3. #3
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    Re: Find an equation of a plane normal to a given vector

    We haven't learned del yet.
    Last edited by icelated; May 21st 2013 at 11:04 AM.
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  4. #4
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    Re: Find an equation of a plane normal to a given vector

    Quote Originally Posted by icelated View Post
    We haven't learned del yet.
    Sorry, but I read your post too quickly.
    You need the tangent (velocity) vector to the curve: $\displaystyle r'(t)$.
    That is the normal of the plane. In this case it is $\displaystyle \left\langle {{x_t}(1),{y_t}(1),{z_t}(1)} \right\rangle $
    Last edited by Plato; May 21st 2013 at 12:11 PM.
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  5. #5
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    Re: Find an equation of a plane normal to a given vector

    Quote Originally Posted by icelated View Post
    I have to find an equation for a plane normal to$\displaystyle r(t)=< e^t sin(\frac{\pi{}}2{t}), e^t cos(\frac{\pi{}}2{t}), t^2 > $
    when $\displaystyle t=1$

    Is the equation in the form:

    $\displaystyle <a,b,c> . <x,y,z> - r(1) = 0?$

    would i take the derivative of x, y, and z?

    Would this require the product rule for x, and y individually? not z of course!


    here is my attempt at just the product rule of x alone:

    $\displaystyle e^t d/dx(sin(\frac{\pi{}}2{t}) + sin(\frac{\pi{}}2{t}) . d/dx (e^t) $

    would it then be?

    $\displaystyle e^t cos(\frac{\pi{}}2{t}) + e^t sin(\frac{\pi{}}2{t}) $
    Not quite. The derivative of $\displaystyle sin(\frac{\pi}{2}t)$ is $\displaystyle \frac{\pi}{2}cos(\frac{\pi}{2}t)$. You forgot the $\displaystyle \frac{\pi}{2}$ multiplying.

    I guess my 2 questions are. is my generic form correct of the equation for a plane normal to the vector correct above?
    and when i take the derivative do i use the product rule and did i do it correctly?

    I am not sure how to get <a,b,c> and the point <x,y,z> either.

    I am completely lost

    Thank you
    Follow Math Help Forum on Facebook and Google+

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