# Thread: dot product and vectors.

1. ## 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.

2. Originally Posted by carpark
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.
$
{\bf u} \wedge {\bf v} = \left\| {\bf u} \right\|\;{\kern 1pt} \left\| {\bf v} \right\|\;\sin (\theta )\;{\bf n}$

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

RonL