Changes

==Theorem==

==[[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  

In [[Coordinate plane|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>

<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>

Let's set out to prove these three.

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

The area of a a parallelogram with sides <math>x,y \ </math> and interior [[acute angle|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>

<math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)</math>

===First Identity===

===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:

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>

<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>

[[Image:Cross product plane.jpg|right|200px|thumb|We imagine <math>\vec{p}</math> sticking out of the screen towards the reader.  You may need to enlarge this picture.]]

[[Image:Cross product plane.jpg|right|200px|thumb|We imagine <math>\vec{p}</math> sticking out of the screen towards the reader.  You may need to enlarge this picture.]]

So, let us assume that <math>\vec{q}_*, \vec{r}_*</math> both have non-zero magnitude and are not collinear.  If we imagine <math>\vec{p}</math> sticking out of our computer monitors, pointing towards the reader, then the vectors in questions all look something like the picture on the right.

So, let us assume that <math>\vec{q}_*, \vec{r}_*</math> both have non-zero magnitude and are not [[collinear]].  If we imagine <math>\vec{p}</math> sticking out of our computer monitors, pointing towards the reader, then the vectors in questions all look something like the picture on the right.

If <math>\vec{s}</math> is any vector in the plane of your screen (ie, <math>p_1x+p_2y+p_3z=0</math>), then <math>\vec{p}\times\vec{s}</math> is ''also'' in the plane of your screen.  It will make a <math>90^\circ=\pi/2</math> angle with <math>\vec{s}</math> and have magnitude <math>\left\|{  \vec{p}  }\right\| \left\|{  \vec{s}  }\right\|</math>, since <math>\sin(90^\circ)=1</math>.  This gives rise to the picture at right.  Letting <math>\vec{s}</math> be each of <math>\vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*</math>, we see that the vectors <math>\vec{p}\times\vec{q}_*, \vec{p}\times\vec{r}_*, \vec{p}\times(\vec{q}_*+\vec{r}_*)</math> have the same configuration as <math>\vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*</math>, just rotated <math>90^\circ</math> and multiplied by <math>\left\|{  \vec{p}  \ }\right\|</math>. Convince yourself that this picture demonstrates <math>\vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ </math>.

If <math>\vec{s}</math> is any vector in the plane of your screen (ie, <math>p_1x+p_2y+p_3z=0</math>), then <math>\vec{p}\times\vec{s}</math> is ''also'' in the plane of your screen.  It will make a <math>90^\circ=\pi/2</math> angle with <math>\vec{s}</math> and have magnitude <math>\left\|{  \vec{p}  }\right\| \left\|{  \vec{s}  }\right\|</math>, since <math>\sin(90^\circ)=1</math>.  This gives rise to the picture at right.  Letting <math>\vec{s}</math> be each of <math>\vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*</math>, we see that the vectors <math>\vec{p}\times\vec{q}_*, \vec{p}\times\vec{r}_*, \vec{p}\times(\vec{q}_*+\vec{r}_*)</math> have the same configuration as <math>\vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*</math>, just rotated <math>90^\circ</math> and multiplied by <math>\left\|{  \vec{p}  \ }\right\|</math>. Convince yourself that this picture demonstrates <math>\vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ </math>.