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Math Help - Proving this equation?

  1. #1
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    Proving this equation?

    l a x b l^2 + (a dot b)^2 = lal^2 lbl^2

    a and b are vectors

    well i know that a cross b = lallblsin theta and a dot b = lallblcos theta but i don't know how to proceed from here...

    how would i go about proving this using the definitions of a cross b and a dot b? why is this equation equivalent to the Pythagorean identity?
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  2. #2
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    \displaystyle | \vec A \times \vec B |^2 + \left ( \vec A \cdot \vec B \right ) ^2 = \left ( |\vec A||\vec B|\sin{\theta} \right ) ^2 + \left ( |\vec A||\vec B|\cos{\theta} \right ) ^2

    Factor \displaystyle |\vec A|^2|\vec B|^2.

    \displaystyle |\vec A|^2|\vec B|^2\sin^2{\theta} + |\vec A|^2|\vec B|^2\cos^2{\theta} = |\vec A|^2|\vec B|^2 \left(\sin^2{\theta} + \cos^2{\theta} \right)

    Use the trigonometric identity \displaystyle \sin^2{\theta} + \cos^2{\theta} = 1.

    \displaystyle |\vec A|^2|\vec B|^2 \left(\sin^2{\theta} + \cos^2{\theta} \right) = |\vec A|^2|\vec B|^2
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  3. #3
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    You're almost there...

    \displaystyle |\mathbf{a}\times\mathbf{b}| + (\mathbf{a}\cdot\mathbf{b})^2 = (|\mathbf{a}||\mathbf{b}|\sin{\theta})^2 + (|\mathbf{a}||\mathbf{b}|\cos{\theta})^2

    \displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2{\theta} + |\mathbf{a}|^2|\mathbf{b}|^2\cos^2{\theta}

    \displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2(\sin^2{\theta} + \cos^2{\theta})

    \displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2(1)

    \displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2.
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  4. #4
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    \mid a X b\mid=\mid a\mid\mid b\mid sin\theta...........1

    a.b=\mid a\mid \mid b \mid cos\theta................2

    add the square of both the equation and use cos^2\theta+sin^2\theta=1
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