Vector product and scalar triple productEdexcel A-Level Further Maths: Revision notes

The vector product of $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$ is $\mathbf{a}\times\mathbf{b}=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}.$ The result is a vector. Its magnitude is $|\mathbf{a}||\mathbf{b}|\sin\theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, and its direction is perpendicular to both, given by the right-hand rule. Key properties: $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}$, $\mathbf{a}\times\mathbf{a}=\mathbf{0}$, and $\mathbf{i}\times\mathbf{j}=\mathbf{k}$. For $\mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $\mathbf{b}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$: $\mathbf{a}\times\mathbf{b}=5\mathbf{i}+5\mathbf{j}-5\mathbf{k}$.

Because $\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, it gives a vector perpendicular to two given directions, such as a normal to a plane containing two lines. Any non-zero multiple also works. To check, take the scalar product with each original vector: it must be $0$. For $\mathbf{a}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$ and $\mathbf{b}=\mathbf{j}+3\mathbf{k}$: $\mathbf{a}\times\mathbf{b}=5\mathbf{i}-3\mathbf{j}+\mathbf{k}$. Check: $\mathbf{a}\cdot(5\mathbf{i}-3\mathbf{j}+\mathbf{k})=5-6+1=0$ and $\mathbf{b}\cdot(5\mathbf{i}-3\mathbf{j}+\mathbf{k})=-3+3=0$. For a unit vector, divide by the magnitude.

$|\mathbf{a}\times\mathbf{b}|$ is the area of the parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$, because $|\mathbf{a}||\mathbf{b}|\sin\theta$ is base times perpendicular height. The area of a triangle is half of this: $\text{Area}=\tfrac12|\mathbf{a}\times\mathbf{b}|.$ For a triangle $ABC$, use two sides from the same vertex, e.g. $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Example: $A(1,0,2)$, $B(2,3,0)$, $C(4,1,1)$ give $\overrightarrow{AB}\times\overrightarrow{AC}=-\mathbf{i}-5\mathbf{j}-8\mathbf{k}$, with magnitude $\sqrt{90}$, so the triangle has area $\frac{3\sqrt{10}}{2}$.

The scalar triple product of three vectors is $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}),$ calculated by finding $\mathbf{b}\times\mathbf{c}$ first and then taking the scalar product with $\mathbf{a}$. The result is a number. It can be written as a determinant of the components of $\mathbf{a},\mathbf{b},\mathbf{c}$. Cycling the vectors 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 its sign, so $\mathbf{a}\cdot(\mathbf{c}\times\mathbf{b})=-\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$. Example: $\mathbf{a}=\mathbf{i}+2\mathbf{k}$, $\mathbf{b}=3\mathbf{j}+\mathbf{k}$, $\mathbf{c}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ give $\mathbf{b}\times\mathbf{c}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}$, so $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=2-6=-4$.

A parallelepiped with edges $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ from one vertex has volume $V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.$ A tetrahedron with the same three edges has one sixth of this: $V=\tfrac16|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.$ Always take the modulus, as the triple product can be negative. For the vectors in the previous section, the parallelepiped has volume $4$ and the tetrahedron $\frac46=\frac23$. If the vertices are given as points, form the three edge vectors from one vertex first, e.g. $\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{AD}$.

Volume $=$ base area $\times$ perpendicular height for a parallelepiped, and Volume $=\frac13\times$ base area $\times$ height for a tetrahedron. Combining these with the vector product and the triple product gives perpendicular distances. For the parallelepiped above, take $\mathbf{b}$ and $\mathbf{c}$ as the base: the base area is $|\mathbf{b}\times\mathbf{c}|=\sqrt{14}$ and the volume is $4$, so the height from the end of $\mathbf{a}$ to the base is $\frac{4}{\sqrt{14}}$.