Here is a list of properties of the cross product.
The cross product satisfies the following rules, where #\vec{v}#, #\vec{w}#, #\vec{x}# are vectors in space, and #\lambda# is a scalar.
- #\vec{v}\times \vec{v} = \vec{0}#.
- The cross product of #\vec{v}# and #\vec{w}# is perpendicular to #\vec{v}# and #\vec{w}#, that is, the dot product of these two vectors is equal to #0#: \[(\vec{v}\times \vec{w})\cdot \vec{v}= 0\quad\text{en}\quad(\vec{v}\times \vec{w})\cdot \vec{w}= 0\]
- Antisymmetry: #\vec{v}\times \vec{w}=-(\vec{w}\times \vec{v})#.
- Mixed associativity with respect to scalar multiplication: \[\lambda \cdot(\vec{v}\times \vec{w}) = (\lambda\cdot \vec{v})\times \vec{w} = \vec{v}\times (\lambda \cdot \vec{w})\]
- Additivity of each argument \[\vec{x} \times (\vec{v}+ \vec{w}) = \vec{x} \times \vec{v} + \vec{x} \times \vec{w} \quad \hbox{en}\quad(\vec{v}+ \vec{w}) \times \vec{x} = \vec{v}\times \vec{x} + \vec{w} \times \vec{x}\]
- The cross product is exactly the same as #\vec{0}# if and only if #\vec{v}# and #\vec{w}# are on a line with #\vec{0}#.
- Orientation: if #\vec{v}# and #\vec{w}# are not on a line with #\vec{0}#, the threesome #\vec{v}#, #\vec{w}#, #\vec{v}\times\vec{w}# is right-hand oriented.
Let #\vec{u}# and #\varphi# be as in the definition of #\vec{v}\times\vec{w}#.
1. #\vec{v}\times \vec{v} = \vec{0}# is part of the definition of the cross product.
2. #\vec{v}\times \vec{w}# is a scalar multiple of #\vec{u}#. This vector is perpendicular to #\vec{v}# and #\vec{w}#, thus also #\vec{v}\times \vec{w}#.
3. Upon switching #\vec{v}# and #\vec{w}#, the angle of #\varphi# remains the same, as do the lengths of #\vec{v}# and #\vec{w}#. But the symbol changes because #\vec{u}# merges with the opposite vector #-\vec{u}# because of the right-hand rule.
4. If #\lambda=0# then all parts are equal to #0# and we are finished with the proof. Assume, therefore, that #\lambda# is not equal to #0#. #\varphi'# means the angle between #\vec{v}# and #\lambda\cdot\vec{w}#, and #\vec{u}'# means the unit vector perpendicular to #\vec{v}# and #\lambda\cdot\vec{w}#, which points in the direction so that #\vec{v}#, #\lambda\cdot\vec{w}#, #\vec{u}'# is right-handed. lf #\lambda\gt0#, we have #\varphi'=\varphi#, #\vec{u}'=\vec{u}#, and #|\lambda|=\lambda#, so \[\begin{array}{rcl}\vec{v}\times\left(\lambda\cdot\vec{w}\right) &=& \parallel\vec{v}\parallel\cdot\parallel\lambda\cdot\vec{w}\parallel\cdot\sin(\varphi')\cdot\vec{u}'\\ &=&|\lambda|\cdot\parallel\vec{v}\parallel\cdot\parallel\vec{w}\parallel\cdot\sin(\varphi)\cdot\vec{u}\\ &=&\lambda\cdot\parallel\vec{v}\parallel\cdot\parallel\vec{w}\parallel\cdot\sin(\varphi)\cdot\vec{u}\\ &=&\lambda\cdot\left(\vec{v}\times\vec{w}\right)\end{array}\] If #\lambda\lt0# the following applies #\varphi'=180^\circ-\varphi#, #\vec{u}'=-\vec{u}# en #|\lambda|=-\lambda#, so \[\begin{array}{rcl}\vec{v}\times\left(\lambda\cdot\vec{w}\right) &=& \parallel\vec{v}\parallel\cdot\parallel\lambda\cdot\vec{w}\parallel\cdot\sin\left(\varphi'\right)\cdot\vec{u}'\\ &=&|\lambda|\cdot\parallel\vec{v}\parallel\cdot\parallel\vec{w}\parallel\cdot\sin\left(180^\circ-\varphi\right)\cdot(-\vec{u})\\ &=&\lambda\cdot\parallel\vec{v}\parallel\cdot\parallel\vec{w}\parallel\cdot\sin(\varphi)\cdot\vec{u}\\ &=&\lambda\cdot\left(\vec{v}\times\vec{w}\right)\end{array}\] In any case, we conclude that #\vec{v}\times\left(\lambda\cdot\vec{w}\right) = \lambda\cdot\left(\vec{v}\times\vec{w}\right)# applies. The other equality can be proven in the same way.
5. In order to derive the rule #\vec{x}\times(\vec{v}+\vec{w}) = (\vec{x}\times \vec{v}) + (\vec{x}\times \vec{w})#, we use the parallelepiped rule and the additivity of the dot product for an arbitrary vector #\vec{y}#, the following applies \[ \begin{array}{rcl}\left(\vec{x}\times\left(\vec{v}+\vec{w}\right)\right)\cdot\vec{y} &=&-\left(\vec{x}\times\vec{y}\right) \cdot \left(\vec{v}+ \vec{w})\right)\\ &&\phantom{xxx}\color{blue}{\text{parallelepiped }}\\&=&-(\vec{x}\times \vec{y})\cdot\vec{v}-(\vec{x}\times \vec{y})\cdot\vec{w}\\ &&\phantom{xxx}\color{blue}{\text{additivy dot product}}\\&=&(\vec{x}\times \vec{v})\cdot\vec{y}+(\vec{x}\times \vec{w})\cdot\vec{y}\\ && \phantom{xxx}\color{blue}{\text{parallelepiped}}\\&=&\left((\vec{x}\times \vec{v})+(\vec{x}\times \vec{w})\right)\cdot\vec{y}\\&&\phantom{xxx}\color{blue}{\text{additivity dot product}}\end{array}\] If we subtract the right-hand side from the left-hand side and apply additivity and mixed associativity with respect to the scalar #-1#, we find for each vector #\vec{y}# that: \[\left(\left(\vec{x}\times\left(\vec{v}+\vec{w}\right)\right)-\left((\vec{x}\times \vec{v})+(\vec{x}\times \vec{w})\right)\right)\cdot\vec{y}=0\] The vector #\left(\vec{x}\times\left(\vec{v}+\vec{w}\right)\right)-\left((\vec{x}\times \vec{v})+(\vec{x}\times \vec{w})\right)# is thus perpendicular to each vector of the space. The only vector which meets these requirements is the zero vector, so \[ \vec{x}\times\left(\vec{v}+\vec{w}\right)-\left((\vec{x}\times \vec{v})+(\vec{x}\times \vec{w})\right)=0\] The statement we are trying to prove follows by bringing the second term to the right.
6. By definition, the cross product of #\vec{v}# and #\vec{w}# is equal to #\vec{0}#, if #\vec{v}# and #\vec{w}# are on the same line as #\vec{0}#. Suppose #\vec{v}# and #\vec{w}# are not on the same line as #\vec{0}#. Then #\vec{u}# would be a vector of a length #1#, and the cross product would be a scalar multiple of this, which is the product of the three factors #\parallel\vec{v}\parallel#, #\parallel\vec{w}\parallel#, #\sin(\varphi)#, wherein #\varphi# is located strictly between #0^\circ# and #180^\circ#. All three factors are not equal to #0#, so the scalar multiple of #\vec{u}# is not equal to #\vec{0}#. This proves statement 6.
7. This is a consequence of the fact that #\vec{v}#, #\vec{w}#, #\vec{u}# is right-hand oriented and #\vec{v}\times\vec{w}# is a scalar multiple of #\vec{u}# with a positive scalar.
The second property, and the length of #\vec{v}\times\vec{w}# do not quite capture the cross product of two vectors: based on these two parts, the cross product may point two other ways (perpendicular to the plane spanned by #\vec{v}# and #\vec{w}#). In the context of the space, we shall describe how to geometrically indicate which way the cross product points. In general, the following appears to apply: Choose an orthonormal basis #\vec{e}_1#, #\vec{e}_2#, #\vec{e}_3# in the space. If this basis is chosen in such a way that a corkscrew rotating from #\vec{e}_1# to #\vec{e}_2#, moves in the direction of #\vec{e}_3#, then a corkscrew rotating from #\vec{v}# to #\vec{w}# moves in the direction of #\vec{v}\times \vec{w}#. If the corkscrew that is rotated from #\vec{e}_1# to #\vec{e}_2# moves in the direction of #-\vec{e}_3#, then a corkscrew that rotates from #\vec{v}# to #\vec{w}# moves in the direction of #-\vec{v}\times \vec{w}#.
Combine the seventh property with the second, and the length is #\vec{v}\times\vec{w}# fully determined.
On the proof of property d) [obv old def]
As stated, the proof follows after writing it out using the definition. In part d) you should maneuver subtly in order to get the equality.
Here we explain the idea behind the calculation.
If you want to prove that
#\parallel\vec{v}\times \vec{w}\parallel = \parallel\vec{v}\parallel\cdot \parallel\vec{w}\parallel\cdot \sin( \varphi)#,
it is convenient to ignore the squares and show that
#\parallel\vec{v}\times \vec{w}\parallel^2# is equal to
\[\parallel\vec{v}\parallel^2\cdot \parallel\vec{w}\parallel^2\cdot \sin^2 (\varphi )\]
If you replace #\sin^2 (\varphi)# by #1-\cos^2 (\varphi)#, then you will get on track to the dot product: \[\parallel\vec{v}\parallel^2\cdot \parallel\vec{w}\parallel^2\cdot \sin^2 (\varphi) = \parallel\vec{v}\parallel^2\cdot \parallel\vec{w}\parallel^2\cdot\left (1- \cos^2( \varphi)\right ) =\parallel\vec{v}\parallel^2\cdot \parallel\vec{w}\parallel^2 - (\vec{v}\cdot \vec{w})^2 \] By writing it out, you are proving that #\parallel\vec{v}\times \vec{w}\parallel^2# equals #\parallel\vec{v}\parallel^2\cdot \parallel\vec{w}\parallel^2 - (\vec{v}\cdot \vec{w})^2 #, so #(v_2\cdot w_3-v_3\cdot w_2)^2 + (v_3\cdot w_1 - v_1\cdot w_3)^2 + (v_1\cdot w_2 - v_2\cdot w_1)^2# equals \[(v_1^2+v_2^2+v_3^2)\cdot (w_1^2 + w_2^2 + w_3^2) - (v_1\cdot w_1 + v_2\cdot w_2 + v_3\cdot w_3)^2\] We leave that to the reader.
The second feature is useful to determine a normal vector, if two directional vectors in a plane are given, for instance. This is particularly evident if we have found a formula for the cross product of coordinates.
It is a given that #\vec{x}#, #\vec{y}#, #\vec{z}#, #\vec{v}#, #\vec{w}# are vectors in the space with #\vec{x}\times\vec{v}=\vec{y}# and #\vec{x}\times\vec{w}=\vec{z}#. This means that \[\vec{x}\times\left(-6\cdot \vec{v}+9\cdot\vec{w}\right)\] is a linear combination of #\vec{y}# and #\vec{z}#. Which one?
#\vec{x}\times\left(-6 \cdot \vec{v}+9 \cdot\vec{w}\right)=-6 \cdot\vec{y}+9 \cdot\vec{z}#
After all,
\[\begin{array}{rcl}\vec{x}\times\left(-6 \cdot \vec{v}+9 \cdot\vec{w}\right)&=&\vec{x}\times\left(-6 \cdot \vec{v}\right)+\vec{x}\times\left(9 \cdot\vec{w}\right)\\ &&\phantom{xxx}\color{blue}{\text{additivity}}\\
&=&-6 \cdot\vec{x}\times \vec{v}+9 \cdot\vec{x}\times \vec{w}\\&&\phantom{xxx}\color{blue}{\text{mixed associativity}}\\&=&-6 \cdot\vec{y}+9 \cdot\vec{z}
\end{array}\]