The vector productEdexcel International A Level Further Maths: Revision notes

The vector product of $\mathbf{a}$ and $\mathbf{b}$ is $\mathbf{a}\times\mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}},$ where $\theta$ is the angle between them and $\hat{\mathbf{n}}$ is a unit vector perpendicular to both, in the direction given by the right-hand rule. It is a vector, not a number. Properties: $\mathbf{b}\times\mathbf{a}=-\mathbf{a}\times\mathbf{b}$; $\mathbf{a}\times\mathbf{a}=\mathbf{0}$; $\mathbf{a}\times(\mathbf{b}+\mathbf{c})=\mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}$; and $\mathbf{a}\times\mathbf{b}=\mathbf{0}$ exactly when $\mathbf{a}$ and $\mathbf{b}$ are parallel (for non-zero vectors).

For $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$: $\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ a_1&a_2&a_3\\ b_1&b_2&b_3\end{vmatrix}=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}.$ Example: $\mathbf{a}=2\mathbf{i}+\mathbf{j}-\mathbf{k}$, $\mathbf{b}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}$ gives $(2+3)\mathbf{i}-(4+1)\mathbf{j}+(6-1)\mathbf{k}=5\mathbf{i}-5\mathbf{j}+5\mathbf{k}$. Check: $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}=0$ and $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{b}=0$. For the unit vectors, $\mathbf{i}\times\mathbf{j}=\mathbf{k}$, $\mathbf{j}\times\mathbf{k}=\mathbf{i}$, $\mathbf{k}\times\mathbf{i}=\mathbf{j}$.

$|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta$ is the area of the parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$. The triangle with those two sides has area $\frac12|\mathbf{a}\times\mathbf{b}|$. For points $P$, $Q$, $R$: area of triangle $PQR=\frac12\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|$. Example: $P(1,2,0)$, $Q(3,3,2)$, $R(2,5,1)$ gives $\overrightarrow{PQ}\times\overrightarrow{PR}=-5\mathbf{i}+5\mathbf{k}$ and area $\frac{5\sqrt2}{2}$. Because area $=\frac12\times$ base $\times$ height, the shortest distance from $C$ to the line $AB$ is $\frac{\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|}{|\overrightarrow{AB}|}$.

The scalar triple product is $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$, a number. In components, $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\begin{vmatrix}a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\end{vmatrix}.$ Cyclic order does not change it: $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})=\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})$. Swapping two vectors changes the sign. It equals $0$ exactly when $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are coplanar (they lie in one plane). Example: $\mathbf{a}=\mathbf{i}+\mathbf{j}+2\mathbf{k}$, $\mathbf{b}=3\mathbf{i}+\mathbf{j}$, $\mathbf{c}=2\mathbf{j}+\mathbf{k}$ gives $\mathbf{b}\times\mathbf{c}=\mathbf{i}-3\mathbf{j}+6\mathbf{k}$ and $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=10$.

$|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$ is the volume of the parallelepiped with edges $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$. The base $|\mathbf{b}\times\mathbf{c}|$ is a parallelogram area and $\mathbf{a}$ dotted with the unit normal gives the height. A tetrahedron with those three edges has volume $\frac16|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$, because the pyramid on a parallelogram base is $\frac13$ of the parallelepiped and the triangle base halves it again. The height of a tetrahedron above face $ABC$ comes from volume $=\frac13\times$ area $\times h$. For the points $A(1,0,2)$, $B(2,2,4)$, $C(0,0,4)$, $D(5,-1,4)$: $\overrightarrow{AB}\times\overrightarrow{AC}=4\mathbf{i}-4\mathbf{j}+2\mathbf{k}$, $\overrightarrow{AD}\cdot(\ldots)=24$, volume $=4$, area $ABC=3$, so $h=4$.