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Thread: Find an orthogonal matrix

  1. #1
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    Find an orthogonal matrix

    Find an orthogonal matrix whose first row is $\displaystyle ( \frac {1}{3} , \frac {2}{3} , \frac {2}{3} ) $

    Solution so far:

    I know that the rows of an orthogonal make up of an orthogonal basis.

    But I can't remember what method I need to use to find it.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Find an orthogonal matrix whose first row is $\displaystyle ( \frac {1}{3} , \frac {2}{3} , \frac {2}{3} ) $

    Solution so far:

    I know that the rows of an orthogonal make up of an orthogonal basis.

    But I can't remember what method I need to use to find it.
    You need to solve:

    $\displaystyle a + 2b + 2c = 0$ .... (1)

    $\displaystyle a^2 + b^2 + c^2 = 1$ .... (2)

    Since there are an infinite number of solutions and you only need a matrix, let c = 0.

    Then

    $\displaystyle a + 2b = 0$ .... (1')

    $\displaystyle a^2 + b^2 = 1$ .... (2')

    A solution to (1') and (2') is $\displaystyle a = -\frac{2}{\sqrt{5}}$ and $\displaystyle b = \frac{1}{\sqrt{5}}$.

    So the elements of the second row in the matrix are $\displaystyle -\frac{2}{\sqrt{5}}$, $\displaystyle \frac{1}{\sqrt{5}}$, 0.

    To get the elements of the third row, you could construct a unit vector normal to (1, 2, 2) and (-2, 1, 0) and use its components .....

    For checking purposes, I get $\displaystyle -\frac{2}{3\sqrt{5}}$, $\displaystyle -\frac{4}{3\sqrt{5}}$, $\displaystyle -\frac{5}{3\sqrt{5}}$.
    Last edited by mr fantastic; Apr 14th 2008 at 09:43 PM. Reason: Thought I better check
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