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Math Help - Vector part of calculus course question

  1. #1
    s3a
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    Vector part of calculus course question

    Could someone please explain to me why the formula that is being used is being used in part c particularly (as in explaining the formula)?

    To be clear, I am talking about the (r - r_0) * [(r_1 - r_0) x (r_2 - r_0)] = 0 formula. If you're unsure of what I am asking or need more information or something, just ask.

    Sorry if this is not the right place to post but I don't know where else to post it since it's in my calculus course.

    Any input would be appreciated!
    Thanks in advance!
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  2. #2
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    Re: Vector part of calculus course question

    I think I can.

    First, the way I would do it. You have three points P_0, P_1 and P_2 all lying on the plane. If you create two vectors connecting two of the three points, say \vec{P_0P_1} and \vec{P_0P_2}. If you cross these, this will give the normal to the plane

    \vec{n} = \vec{P_0P_1} \times \vec{P_0P_2}.

    Suppose that (x,y,z) is a point lying on the plane then the vector throught this point and P_0 is

    <x-1,y-2,z-3> (every line on the plane will contain this vector)

    Now you want every line on the plane perpendicular to the normal so

    \vec{n} \cdot <x-1,y-2,z-3> = 0.

    What I think the author is doing.

    If you associate vectors for each point P_0, P_1 and P_2, from the origin to the point and denote these by {\bf r}_0, {\bf r}_1 and {\bf r}_2. Now

    {\bf r}_0- {\bf r}_1 = \vec{P_0P_1} and {\bf r}_0- {\bf r}_2 = \vec{P_0P_2}

    so

    \left({\bf r}_0- {\bf r}_1\right) \times  \left({\bf r}_0- {\bf r}_2\right) =  \vec{P_0P_1} \times \vec{P_0P_2} = \vec{n}.

    Now associate a vector from origin to (x,y,z) as denote this as {\bf r}. Thus, the vector from (x,y,z) to P_0 is

    {\bf r} - {\bf r}_0 = <x-1,y-2,z-3>.

    Now dot these two

    \left( {\bf r} - {\bf r}_0\right) \cdot \left({\bf r}_0- {\bf r}_1\right) \times  \left({\bf r}_0- {\bf r}_2) = 0

    so you see the answers are the same.
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  3. #3
    s3a
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    Re: Vector part of calculus course question

    Thanks!
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