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Vector projections

  1. Feb 13, 2008 #1

    tony873004

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    Let [tex]\overrightarrow v [/tex] and [tex]\overrightarrow w [/tex] be vectors in R3. Prove that [tex]\overrightarrow w - {\rm{proj}}_{\overrightarrow v } \overrightarrow w [/tex] is perpendicular to [tex]\overrightarrow v [/tex] .

    Here's my attempt:
    [tex]\begin{array}{l}
    \left( {\overrightarrow w - {\rm{proj}}_{\overrightarrow v } \overrightarrow w } \right) \cdot \overrightarrow v \mathop = \limits^? 0 \\
    \\
    {\rm{proj}}_{\overrightarrow v } \overrightarrow w = \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }}\overrightarrow v \\
    \\
    \overrightarrow v \cdot \overrightarrow w = v_1 w_1 + v_2 w_2 + v_3 w_3 \\
    \\
    \left| {\overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + v_3^2 } \\
    \left| {\overrightarrow v } \right|^2 = v_1^2 + v_2^2 + v_3^2 \\
    \\
    \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }} = \frac{{v_1 w_1 + v_2 w_2 + v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }} = \frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }} \\
    \\
    \frac{{\overrightarrow v \cdot \overrightarrow w }}{{\left| {\overrightarrow v } \right|^2 }}\overrightarrow v = \left( {\frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }}} \right)\left\langle {v_1 ,\,v_2 ,\,v_3 } \right\rangle \\
    \\
    \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{v_1 w_1 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_2 w_2 }}{{v_1^2 + v_2^2 + v_3^2 }} + \frac{{v_3 w_3 }}{{v_1^2 + v_2^2 + v_3^2 }}} \right)\left\langle {v_1 ,\,v_2 ,\,v_3 } \right\rangle \\
    \end{array}[/tex]

    Things are starting to get real ugly. Am I missing an easier way?
     
  2. jcsd
  3. Feb 13, 2008 #2

    Dick

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    Yes, you are. You want to show v.(w-(v.w)*v/(v.v))=0. Just multiply the outer dot product through.
     
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