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Math Help - Find a unit vector that is orthogonal to the plane

  1. #1
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    Question Find a unit vector that is orthogonal to the plane

    I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
    and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

    Also, can the desired vector be generated by linear transformation?

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  2. #2
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    Quote Originally Posted by nemixus View Post
    I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
    and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

    Also, can the desired vector be generated by linear transformation?

    If you take the cross product of the two vectors, you will get a vector that is orthogonal to both of the vectors, and as such, is also orthogonal to the plane they lie in.


    So A \times B = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 2 & 2 \\ 4 & 6 & 5\end{matrix}\right|

     = \mathbf{i}\left|\begin{matrix}2&2\\6&5\end{matrix}  \right| - \mathbf{j}\left|\begin{matrix}2&2\\4&5\end{matrix}  \right| + \mathbf{k}\left|\begin{matrix}2&2\\4&6\end{matrix}  \right|

     = \mathbf{i}(2\cdot 5 - 2\cdot 6) - \mathbf{j}(2\cdot 5 - 2\cdot 4) + \mathbf{k}(2\cdot 6 - 2\cdot 4)

     = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{j}.


    Now, if you were to take the dot product of this new vector with either A or B, you should get 0.


    You need a UNIT vector in this direction. So divide it by its length.

    |A\times B| = \sqrt{(-2)^2 + (-2)^2 + 4^2}

     = \sqrt{4 + 4 + 16}

     = \sqrt{24}

     = 2\sqrt{6}.


    So the unit vector is

    \frac{A\times B}{|A \times B|} = \frac{-2\mathbf{i} - 2\mathbf{j} + 4\mathbf{j}}{2\sqrt{6}}

     = -\frac{1}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} + \frac{2}{\sqrt{6}}\mathbf{k}

     = -\frac{\sqrt{6}}{6}\mathbf{i} - \frac{\sqrt{6}}{6}\mathbf{j} + \frac{\sqrt{6}}{3}\mathbf{k}.
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  3. #3
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    Quote Originally Posted by nemixus View Post
    I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
    and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

    Also, can the desired vector be generated by linear transformation?

    Let \vec n = (n_1, n_2, n_3) denote the orthogonal vector to the plane.

    Then according to the hint:

    2n_1+2n_2+2n_3=0
    4n_1+6n_2+5n_3=0

    Multiply the 1st equation by 2 and subtract it from the 2nd one:

    2n_2+n_3=0~\implies~n_3=-2n_2

    Plug in this term into one of the original equations and you'll get:

    n_1=n_2

    Set n_2 = t and you'll have:

    \vec n=\left\{\begin{array}{l}n_1=t\\n_2=t\\n_3=-2t\end{array}\right.

    Therefore \vec n = (1,1,-2)

    Use \overrightarrow{n^0} = \dfrac{\vec n}{|\vec n|} to calculate the components of the unit vector.
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  4. #4
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    Find a unit vector that is orthogonal to the plane by Linear Transformation

    Quote Originally Posted by nemixus View Post
    I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
    and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?


    Also, can the desired vector be generated by linear transformation?

    Can the desired vector be generated by linear transformation?
    Follow Math Help Forum on Facebook and Google+

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