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Math Help - Ratio help

  1. #1
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    Ratio help

    Hi, I need a hint to start this problem:

    If  (a^2 +b^2+c^2)(x^2+y^2+z^2) = (ax+by+cz)^2

    prove that  \frac{x}{a} = \frac{y}{b} = \frac{z}{c}

    Thanks.
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  2. #2
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    Quote Originally Posted by MagicS06 View Post
    Hi, I need a hint to start this problem:

    If  (a^2 +b^2+c^2)(x^2+y^2+z^2) = (ax+by+cz)^2

    prove that  \frac{x}{a} = \frac{y}{b} = \frac{z}{c}

    Thanks.
    That's not a ratio... but since you asked for a hint...

    Expand all the brackets and collect like terms, you should get...

    a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 = 2abxy + 2acxz + 2bcyz

    b^2x^2 - 2abxy + a^2y^2 + a^2z^2 -2acxz + c^2x^2 + b^2z^2 - 2bcyz + c^2y^2 = 0

    Recall that (p-q)^2 = p^2 - 2pq + q^2? Using this we get...

    (bx - ay)^2 + (az - cx)^2 + (bz - cy)^2 = 0

    Now, assuming a, b, c, x, y and z are all real... we have 3 square terms adding to become 0. Since a square term is never non-negative... all the terms must be 0.

    So we have (bx - ay)^2 = 0 which gives bx = ay or \frac{x}{a} = \frac{y}{b}.

    Similarly, (az - cx)^2 = 0 gives \frac{z}{c}=\frac{x}{a},

    and (bz - cy)^2 = 0 gives \frac{z}{c} = \frac{y}{b}.

    Thus

    \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.
    Last edited by Prove It; September 26th 2008 at 04:53 PM.
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