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Thread: vectors

  1. #1
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    vectors

    1. Prove that $\displaystyle \alpha^{2} + \beta^{2} + \gamma^{2} = 1 $ (direction cosines).

    So $\displaystyle \alpha = \frac{A_x}{|\bold{A}|} $ , $\displaystyle \beta = \frac{A_y}{|\bold{A}|} $, $\displaystyle \gamma = \frac{A_z}{|\bold{A}|} $ which implies $\displaystyle \alpha^{2} + \beta^{2} + \gamma^{2} = 1 $.

    2. Show that if $\displaystyle |\bold{A} - \bold{B}| = |\bold{A}+ \bold{B}| $ then $\displaystyle \bold{A} $ is perpendicular to $\displaystyle \bold{B} $.

    So $\displaystyle \sqrt{(A_x-B_x)^{2} + (A_y-B_y)^{2} + (A_z - B_z)^{2}} $ $\displaystyle = \sqrt{(A_x+B_x)^{2} + (A_y+B_y)^{2} + (A_z + B_z)^{2}} $.

    Then $\displaystyle (A_x - B_x)^{2} = (A_x+B_x)^{2} $ or $\displaystyle A_{x}^{2} - 2A_{x}B_{x} + B_{x}^{2} = A_{x}^{2} + 2A_{x}B_{x} + B_{x}^{2} $ so that $\displaystyle A_{x}B_{x} = A_{y}B_{y} = A_{z}B_{z} = 0 $ which implies that $\displaystyle \bold{A} \cdot \bold{B} = 0 $.

    Are these correct?
    Last edited by shilz222; Nov 26th 2007 at 11:47 AM.
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  2. #2
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    In # 2, Square both sides.
    $\displaystyle \begin{array}{rcl}
    \left\| {A + B} \right\|^2 & = & \left( {A + B} \right) \cdot \left( {A + B} \right) \\ & = & A \cdot A + 2A \cdot B + B \cdot B \\ \end{array}$

    $\displaystyle \begin{array}{rcl}
    \left\| {A - B} \right\|^2 & = & \left( {A - B} \right) \cdot \left( {A - B} \right) \\
    & = & A \cdot A - 2A \cdot B + B \cdot B \\ \end{array}$

    This implies that $\displaystyle A \cdot B = 0$
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