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 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} \ } 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{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 $x,y\$ and interior acute angle $\theta \$ is $xy\sin(\theta )\$. Therefore, by definition, we have

$\left\|{{\vec {u}}\times {\vec {w}}}\right\|=\left\|{\vec {u}}\right\|\left\|{\vec {w}}\right\|\sin(\theta )$

where $\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:

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

Second Identity

Since our definitional equation for the magnitude of the cross product also has the property that $\left\|{{\vec {u}}\times {\vec {w}}}\right\|=\left\|{{\vec {w}}\times {\vec {u}}}\right\|$, we can combine this fact with the "right-handed orientation" part of the definition to get another identity:

${\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, ${\vec {p}},{\vec {q}},{\vec {r}}\$. Suppose ${\vec {p}}\neq {\vec {0}}\$. Let $P\$ be the plane through the origin which is perpendicular to ${\vec {p}}\$ (ie, the plane $p_{1}x+p_{2}y+p_{3}z=0$). If ${\vec {q}}\$ is any vector, let ${\vec {q}}_{*}$ be the projection of ${\vec {q}}$ into $P\$. Clearly, $\left\|{{\vec {q}}_{*}}\right\|=\left\|{{\vec {q}}\ }\right\|\sin(\theta )\$, and so ${\vec {p}}\times {\vec {q}}={\vec {p}}\times {\vec {q}}_{*}\$. The reader may also assure themselves if they wish that $({\vec {q}}+{\vec {r}})_{*}={\vec {q}}_{*}+{\vec {r}}_{*}\$.

Hence, to prove the third identity, it would suffice to prove instead ${\vec {p}}\times ({\vec {q}}_{*}+{\vec {r}}_{*})={\vec {p}}\times {\vec {q}}_{*}+{\vec {p}}\times {\vec {r}}_{*}\$. The advantage is that the vectors ${\vec {q}}_{*},{\vec {r}}_{*},\$ and ${\vec {q}}_{*}+{\vec {r}}_{*}\$ are all either ${\vec {0}}$ or perpendicular to ${\vec {p}}\$.

If either ${\vec {q}}_{*},{\vec {r}}_{*}={\vec {0}}$, 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 ${\vec {p}}$ sticking out of the screen towards the reader. You may need to enlarge this picture.

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

If ${\vec {s}}$ is any vector in the plane of your screen (ie, $p_{1}x+p_{2}y+p_{3}z=0$), then ${\vec {p}}\times {\vec {s}}$ is also in the plane of your screen. It will make a $90^{\circ }=\pi /2$ angle with ${\vec {s}}$ and have magnitude $\left\|{\vec {p}}\right\|\left\|{\vec {s}}\right\|$, since $\sin(90^{\circ })=1$. This gives rise to the picture at right. Letting ${\vec {s}}$ be each of ${\vec {q}}_{*},{\vec {r}}_{*},{\vec {q}}_{*}+{\vec {r}}_{*}$, we see 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{p}\times\vec{q}_*, \vec{p}\times\vec{r}_*, \vec{p}\times(\vec{q}_*+\vec{r}_*)} have the same configuration as 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}_*} , just rotated 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 90^\circ} and multiplied by 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{p} \ }\right\|} . Convince yourself that this picture demonstrates 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}_* \ } .