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Thread: dot product and vectors.

  1. #1
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    Exclamation dot product and vectors.

    Let w be a vector in R3. Show that ||w, u x v || = ||w||(||u||)||v||sin(theta)(sin(psi)) where psi is the angle between the vector w and the plane determined by the vectors u and v.
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  2. #2
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    Quote Originally Posted by carpark View Post
    Let w be a vector in R3. Show that ||w, u x v || = ||w||(||u||)||v||sin(theta)(sin(psi)) where psi is the angle between the vector w and the plane determined by the vectors u and v.
    $\displaystyle
    {\bf u} \wedge {\bf v} = \left\| {\bf u} \right\|\;{\kern 1pt} \left\| {\bf v} \right\|\;\sin (\theta )\;{\bf n} $

    $\displaystyle
    {\bf w}.({\bf u} \wedge {\bf v}) = \left\| {\bf w} \right\|\; \left[ {\left\| {\bf u} \right\|\;{\kern 1pt} \left\| {\bf v} \right\|\;\sin (\theta )} \right]\;\cos (\pi /2 - \varphi ) = \left\| {\bf w} \right\|\;\left\| {\bf u} \right\|\;{\kern 1pt} \left\| {\bf v} \right\|\;\sin (\theta )\;\sin (\varphi )
    $


    Where $\displaystyle \theta$ is the angle between $\displaystyle \bf{u}$ and $\displaystyle \bf{v}$, and $\displaystyle \bf{n}$ is the normal to the plane defined by $\displaystyle \bf{u}$ and $\displaystyle \bf{v}$, and $\displaystyle \pi/2-\varphi$ is the angle between the normal and $\displaystyle \bf{w}$, and so $\displaystyle \varphi$ is the angle between the plane and $\displaystyle \bf{w}$.

    RonL
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