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Math Help - Vector proof

  1. #1
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    Vector proof

    Hey guys,

    I need help proving that if |a+b| = |a-b|, then a and b are perpendicular vectors.

    I also need to prove that prove that the angle formed by joining any point on a circle to the end points of a diameter is 90 degrees.

    Thanks!
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  2. #2
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    \left\| {A - B} \right\|^2  = \left( {A - B} \right) \cdot \left( {A - B} \right) = A \cdot A - 2A \cdot B + B \cdot B
    And \left\| {A + B} \right\|^2  = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B
    From the given we know that \left\| {A + B} \right\|^2  = \left\| {A - B} \right\|^2  \Rightarrow \quad 4A \cdot B = 0 \Rightarrow \quad A \cdot B = 0
    QED
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  3. #3
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    Quote Originally Posted by Plato View Post
    \left\| {A - B} \right\|^2  = \left( {A - B} \right) \cdot \left( {A - B} \right) = A \cdot A - 2A \cdot B + B \cdot B
    And \left\| {A + B} \right\|^2  = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B
    From the given we know that \left\| {A + B} \right\|^2  = \left\| {A - B} \right\|^2  \Rightarrow \quad 4A \cdot B = 0 \Rightarrow \quad A \cdot B = 0
    QED
    Thanks, but I'm still not following...

    First off I still don't see how they're equal, (one had +2A, the other had -2A) then where did the 4A dot B = 0 come from??
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  4. #4
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    Hello, TheJacksonater!

    Prove: .if |\vec a+\vec b| \:= \:|\vec a-\vec b|, then: . \vec a \perp \vec b
    Let: . \begin{array}{ccc}\vec a &=& \langle x_1,y_1\rangle \\ \vec b &=& \langle x_2,y_2\rangle \end{array}

    Then: . \vec a + \vec b \:=\:\langle x_1+x_2,\:y_1+y_2\rangle \quad\Rightarrow\quad |\vec a + \vec b| \:=\:\sqrt{(x_1+x_2)^2 + (y_1+y_2)^2}

    . And: . \vec a - \vec b \:=\:\langle x_1-x_2,\:y_1-y_2\rangle \quad\Rightarrow\quad |\vec a - \vec b| \:=\:\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}


    Since |a + b| \:=\:|a - b|, we have: . (x_1+x_2)^2 + (y_1+y_2)^2 \:=\:(x_1-x_2)^2 + (y_1-y_2)^2

    . . x_1^2 + 2x_1x_2 + x_2^2 + y_1^2 + 2y_1y_2 + y_2^2 \;=\;x_1^2 - 2x_1x_2 + x_2^2 + y_1^2 - 2y_1y_2 + y_2^2

    . . 4x_1x_2 + 4y_1y_2 \:=\:0 \quad\Rightarrow\quad x_1x_2 + y_1y_2\:=\:0 \quad\Rightarrow\quad \langle x_1,y_1\rangle \cdot \langle x_2,y_2\rangle \:=\:0

    Therefore: . \vec a\cdot\vec b \:=\:0 \quad\Rightarrow\quad \vec a \perp \vec b




    Prove that the angle formed by joining any point
    on a circle to the endpoints of a diameter is 90°.
    Let the circle be: . \begin{array}{ccc}x &=& r\cos\theta \\ y &=& r\sin\theta \end{array}

    Let A(r,0) \text{ and }B(\text{-}r,0) be the endpoints of the diameter.

    Let C(r\cos\theta, r\sin\theta) be any point on the circle.

    Draw segments AC and BC.
    Code:
                    |
                  * * *
              *     |     *   C
            *       |       o(r·cosθ,r·sinθ)
           *        |     /  *
                    |   /
          *         | / θ     *
      - B o - - - - * - - - - o A - -
    (-r,0)*         |         *(r,0)
                    |
           *        |        *
            *       |       *
              *     |     *
                  * * * 
                    |

    \overrightarrow{AC} \;=\;\bigg\langle r\cos\theta - r,\:r\sin\theta\bigg\rangle \;=\;\bigg\langle r(\cos\theta - 1),\:r\sin\theta\bigg\rangle

    \overrightarrow{BC} \;=\;\bigg\langle r\cos\theta + r,\:r\sin\theta\bigg\rangle \;=\;\bigg\langle r(\cos\theta + 1),\:r\sin\theta\bigg\rangle

    Then: . \overrightarrow{AC}\cdot\overrightarrow{BC} \;=\;[r(\cos\theta - 1)][r(\cos\theta+1)] + (r\sin\theta)(r\sin\theta)

    . . . . . . . . . . . = \;r^2(\cos^2\!\theta-1) + r^2\sin^2\!\theta

    . . . . . . . . . . . =\;-r^2(1-\cos^2\!\theta) + r^2\sin^2\!\theta

    . . . . . . . . . . . = \;-r^2\sin^2\!\theta + r^2\sin^2\!\theta

    . . . . . . . . . . . = \;0


    Therefore: . \overrightarrow{AC} \perp \overrightarrow{BC} \quad\Rightarrow\quad \angle C \:=\:90^o

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