Calc3.2.CrossProductProof

Theorem

In Cartesian coordinates, the cross product, defined as having length equal to the area of the parallelogram bordered on two sides by ${\vec {u}}\$ and ${\vec {w}}\$, and direction perpendicular to either vector, with a right handed orientation, is given by

${\vec {u}}\times {\vec {w}}={\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\times {\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}={\begin{pmatrix}u_{2}w_{3}-u_{3}w_{2}\\u_{3}w_{1}-u_{1}w_{3}\\u_{1}w_{2}-u_{2}w_{1}\end{pmatrix}}$

Proof

To prove this theorem, we will need several identities about the cross product:

$(c{\vec {u}})\times {\vec {w}}=c({\vec {u}}\times {\vec {w}})$

${\vec {u}}\times {\vec {w}}=-({\vec {w}}\times {\vec {u}})\$

${\vec {p}}\times ({\vec {q}}+{\vec {r}})=({\vec {p}}\times {\vec {q}})+({\vec {p}}\times {\vec {r}})$

Let's set out to prove these three.

The area of a a parallelogram with sides Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x,y \ } and interior acute angle Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta \ } is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xy\sin(\theta) \ } . Therefore, by definition, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta \ } 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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})}

Second Identity

Since our definitional equation for the magnitude of the cross product also has the property that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{w} \times \vec{u}}\right\|} , we can combine this fact with the "right-handed orientation" part of the definition to get another identity:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ }

Third Identity

The final identity we will require takes a little bit more finesse. Suppose we have three vectors, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p},\vec{q},\vec{r} \ } . Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p}\neq\vec{0} \ } . Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ } be the plane through the origin which is perpendicular to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p} \ } (ie, the plane Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_1x+p_2y+p_3z=0} ). If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q} \ } is any vector, let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}_*} be the projection of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}} into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ } . Clearly, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\|{ \vec{q}_* }\right\|=\left\|{ \vec{q} \ }\right\|\sin(\theta) \ } , and so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p}\times\vec{q}=\vec{p}\times\vec{q}_* \ } . The reader may also assure themselves if they wish that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\vec{q}+\vec{r})_* = \vec{q}_*+\vec{r}_* \ } .

Hence, to prove the third identity, it would suffice to prove instead Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ } . The advantage is that the vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}_*, \vec{r}_*, \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}_*+\vec{r}_* \ } are all either Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{0}} or perpendicular to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p} \ } .

If either Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}_*, \vec{r}_*=\vec{0}} , then the identity we seek is certainly true, and if they are collinear, then one is a multiple of the other and the identity we seek follows from application of our first two identities.

We imagine Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p}} sticking out of the screen towards the reader. You may need to enlarge this picture.

So, let us assume that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{q}_*, \vec{r}_*} both have non-zero magnitude and are not collinear. If we imagine Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{p}} sticking out of our computer monitors, pointing towards the reader, then the vectors in questions all look something like the picture on the right.