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Math Help - Vector Proof

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    Vector Proof

    Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.
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    Quote Originally Posted by tdat1979 View Post
    Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.
    x.y = (|v|u + |u|v) . (|v|u - |u|v)

    = |v|^2|u|^2+|u||v|v.u-|v||u|u.v-|u|^2|v|^2

    = 0

    Hence, x and y are perpendicular.
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  3. #3
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    Quote Originally Posted by tdat1979 View Post
    Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.
    I hope you're familiar with the matrix notation of vectors...

    Let u = [a, b] and v = [c, d].

    Then

    |u| = \sqrt{a^2 + b^2} and |v| = \sqrt{c^2 + d^2}.


    Therefore

    x = |v|u + |u|v

     = \sqrt{c^2 + d^2}[a, b] + \sqrt{a^2 + b^2}[c, d]

     = [\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d].


    y = |v|u - |u|v

     = \sqrt{c^2 + d^2}[a, b] - \sqrt{a^2 + b^2}[c, d]

     = [\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d]



    To show x and y are perpendicular, we need to show that their dot product is 0.


    So x \cdot y = [\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d]
    \cdot[\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d]

     = (\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c)(\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c)
     + (\sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d)(\sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d)

     = (c^2 + d^2)a^2 - (a^2 + b^2)c^2 + (c^2 + d^2)b^2 - (a^2 + b^2)d^2

    a^2c^2 + a^2d^2 - a^2c^2 - b^2c^2 + b^2c^2 + b^2d^2 - a^2d^2 - b^2d^2

     = 0


    So x and y are perpendicular (PHEW!)
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