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Math Help - Vectors on a plane

  1. #1
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    Question Vectors on a plane

    Hi

    I really need some assistance with the following problem:

    "Find all unit vectors in the plane determined by u = (3,0,1) and v = (1,-1,1) that are perpendicular to the vector w = (1,2,0)".

    The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors u and v?

    What would u and v look like if they were serparate planes?

    If someone know a website that explains the basics of vector planes I would appreciate the adress.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Drdumbom View Post
    Hi

    I really need some assistance with the following problem:

    "Find all unit vectors in the plane determined by u = (3,0,1) and v = (1,-1,1) that are perpendicular to the vector w = (1,2,0)".

    The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors u and v?

    What would u and v look like if they were serparate planes?

    If someone know a website that explains the basics of vector planes I would appreciate the adress.

    Thanks in advance
    This link will help you get started.
    Pauls Online Notes : Calculus II - Equations of Planes
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  3. #3
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    Hello Drdumbom
    Quote Originally Posted by Drdumbom View Post
    Hi

    I really need some assistance with the following problem:

    "Find all unit vectors in the plane determined by u = (3,0,1) and v = (1,-1,1) that are perpendicular to the vector w = (1,2,0)".

    The problem for me is that I donīt really know how to approach planes in general. For example, how do I calculate what plane is determined by the two vectors u and v?

    What would u and v look like if they were serparate planes?

    If someone know a website that explains the basics of vector planes I would appreciate the adress.

    Thanks in advance
    It may help if you think first of the unit vectors \textbf{i} and \textbf{j} that are used to define the x-y plane. Any vector in this plane can be written p\textbf{i} +q\textbf{j}, representing as it does, displacements p and q parallel to Ox and Oy respectively. ( p and q are, of course, any scalar numbers.)

    In the same way, any vector of the form p\textbf{u}+q\textbf{v} will lie in the plane defined by \textbf{u} and \textbf{v}. We may think of the grid of unit squares defined by \textbf{i} and \textbf{j} as having been replaced by a grid of parallelograms defined by \textbf{u} and \textbf{v}.

    So if \textbf{u} = \begin{pmatrix}3\\0\\1\end{pmatrix} and \textbf{v} = \begin{pmatrix}1\\-1\\1\end{pmatrix}, any vector lying in the plane determined by \textbf{u} and \textbf{v} will be of the form p\begin{pmatrix}3\\0\\1\end{pmatrix}+q\begin{pmatr  ix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}3p+q\\-q\\p+q\end{pmatrix}.

    Now two vectors are perpendicular if their dot (scalar) product is zero. So this vector is perpendicular to \textbf{w}=\begin{pmatrix}1\\2\\0\end{pmatrix}, if:
    \begin{pmatrix}3p+q\\-q\\p+q\end{pmatrix}\cdot\begin{pmatrix}1\\2\\0\end  {pmatrix}=0

    \Rightarrow (3p+q).(1) +(-q).(2)+(p+q).(0) = 0

     \Rightarrow 3p-q=0

     \Rightarrow q = 3p
    So the vector will be of the form \begin{pmatrix}3p+3p\\-3p\\p+3p\end{pmatrix}=\begin{pmatrix}6p\\-3p\\4p\end{pmatrix}=p\begin{pmatrix}6\\-3\\4\end{pmatrix}

    All that remains is to find unit vectors in this form. So that will be where
    p^2(6^2+(-3)^2+4^2) = 1

    \Rightarrow 61p^2=1

    \Rightarrow p = \pm\frac{1}{\sqrt{61}}
    So I reckon there's our answer: the unit vectors are \pm\frac{1}{\sqrt{61}}\begin{pmatrix}6\\-3\\4\end{pmatrix}

    Grandad
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