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Calc3.2/Cross Product Proof (view source)
Revision as of 04:51, February 28, 2010
, 04:51, February 28, 2010preparing for expansion
==Theorem==
In Cartesian coordinates, the cross product, defined as having length equal to the area of the parallelogram bordered on two sides by <math>\vec{u} \ </math> and <math>\vec{w} \ </math>, and direction perpendicular to either vector, with a right handed orientation, is given by
<math>\vec{u}\times\vec{w}= \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} \times\begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix} =\begin{pmatrix}u_2 w_3 - u_3 w_2\\u_3w_1-u_1w_3\\u_1w_2-u_2w_1\end{pmatrix}</math>
==Proof==
To prove this theorem, we will need several identities about the cross product:
<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>
<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>
<math>\vec{p}\times(\vec{q}+\vec{r})=(\vec{p}\times\vec{q})+(\vec{p}\times\vec{r})</math>
Let's set out to prove these three.
The area of a a parallelogram with sides <math>x,y \ </math> and interior acute<ref>The acute descriptor is unnecessary, since the obtuse angle will then be <math>\pi-\theta \ </math>, and hence the sine will be equal.</ref> angle <math>\theta \ </math> is <math>xy\sin(\theta) \ </math>. Therefore, by definition, we have
<math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)</math>
where <math>\theta \ </math> is the angle between the two vectors.
===First Identity===
This definitional equation for the magnitude of the cross product immediately gives us an identity - for scalar (ie, non-vector) value ''c'', we have:
<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>
===Second Identity===
Since our definitional equation for the magnitude of the cross product also has the property that <math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{w} \times \vec{u}}\right\|</math>, we can combine this fact with the "right-handed orientation" part of the definition to get another identity:
<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>
===Third Identity===
The final identity we will require takes a little bit more finesse. Suppose we have three vectors, <math>\vec{p},\vec{q},\vec{r} \ </math>. Suppose <math>\vec{p}\neq\vec{0} \ </math>.
==Notes==
{{reflist}}
In Cartesian coordinates, the cross product, defined as having length equal to the area of the parallelogram bordered on two sides by <math>\vec{u} \ </math> and <math>\vec{w} \ </math>, and direction perpendicular to either vector, with a right handed orientation, is given by
<math>\vec{u}\times\vec{w}= \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} \times\begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix} =\begin{pmatrix}u_2 w_3 - u_3 w_2\\u_3w_1-u_1w_3\\u_1w_2-u_2w_1\end{pmatrix}</math>
==Proof==
To prove this theorem, we will need several identities about the cross product:
<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>
<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>
<math>\vec{p}\times(\vec{q}+\vec{r})=(\vec{p}\times\vec{q})+(\vec{p}\times\vec{r})</math>
Let's set out to prove these three.
The area of a a parallelogram with sides <math>x,y \ </math> and interior acute<ref>The acute descriptor is unnecessary, since the obtuse angle will then be <math>\pi-\theta \ </math>, and hence the sine will be equal.</ref> angle <math>\theta \ </math> is <math>xy\sin(\theta) \ </math>. Therefore, by definition, we have
<math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)</math>
where <math>\theta \ </math> is the angle between the two vectors.
===First Identity===
This definitional equation for the magnitude of the cross product immediately gives us an identity - for scalar (ie, non-vector) value ''c'', we have:
<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>
===Second Identity===
Since our definitional equation for the magnitude of the cross product also has the property that <math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{w} \times \vec{u}}\right\|</math>, we can combine this fact with the "right-handed orientation" part of the definition to get another identity:
<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>
===Third Identity===
The final identity we will require takes a little bit more finesse. Suppose we have three vectors, <math>\vec{p},\vec{q},\vec{r} \ </math>. Suppose <math>\vec{p}\neq\vec{0} \ </math>.
==Notes==
{{reflist}}