Vector productAQA A-Level Further Maths: Revision notes

The vector product of $\mathbf a$ and $\mathbf b$ is a vector, $\mathbf a\times\mathbf b=|\mathbf a||\mathbf b|\sin\theta\,\hat{\mathbf n},$ where $\theta$ is the angle between them and $\hat{\mathbf n}$ is the unit vector perpendicular to both, in the direction given by the right-hand rule. In components, $\begin{pmatrix}a_1 \\ a_2 \\ a_3\end{pmatrix}\times\begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix}=\begin{pmatrix}a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1\end{pmatrix}.$ Example: $(2,-1,3)\times(1,4,-2)=(2-12,\ 3+4,\ 8+1)=(-10,7,9)$. The unit vectors satisfy $\mathbf i\times\mathbf j=\mathbf k$, $\mathbf j\times\mathbf k=\mathbf i$, $\mathbf k\times\mathbf i=\mathbf j$, and reversing any pair gives the negative.

• Not commutative: $\mathbf b\times\mathbf a=-(\mathbf a\times\mathbf b)$.
• $\mathbf a\times\mathbf a=\mathbf 0$, and $\mathbf a\times\mathbf b=\mathbf 0$ exactly when $\mathbf a$ and $\mathbf b$ are parallel (or one is zero), because $\sin\theta=0$.
• $\mathbf a\times\mathbf b$ is perpendicular to both $\mathbf a$ and $\mathbf b$, so $(\mathbf a\times\mathbf b)\cdot\mathbf a=0$ and $(\mathbf a\times\mathbf b)\cdot\mathbf b=0$.
• Distributive: $\mathbf a\times(\mathbf b+\mathbf c)=\mathbf a\times\mathbf b+\mathbf a\times\mathbf c$.
• Scalars factor out: $(k\mathbf a)\times\mathbf b=k(\mathbf a\times\mathbf b)$.
• Not associative in general: $(\mathbf a\times\mathbf b)\times\mathbf c\ne\mathbf a\times(\mathbf b\times\mathbf c)$.

To find a vector perpendicular to two given directions, take their vector product.

The magnitude $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta$ is the area of the parallelogram with adjacent sides $\mathbf a$ and $\mathbf b$. So the area of a triangle with sides $\mathbf a$ and $\mathbf b$ from a common vertex is $\frac12|\mathbf a\times\mathbf b|$. For triangle $ABC$, $\text{Area}=\tfrac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|.$ Example: $A(1,0,2)$, $B(3,2,3)$, $C(2,-1,5)$ give $\overrightarrow{AB}\times\overrightarrow{AC}=(7,-5,-4)$, magnitude $\sqrt{90}$, so the area is $\frac{3\sqrt{10}}{2}$. Since area $=\frac12\times\text{base}\times\text{height}$, the perpendicular distance from $C$ to line $AB$ is $\dfrac{|\overrightarrow{AB}\times\overrightarrow{AC}|}{|AB|}=\sqrt{10}$.

A line through the point with position vector $\mathbf a$ in direction $\mathbf b$ can be written $(\mathbf r-\mathbf a)\times\mathbf b=\mathbf 0.$ This says that $\mathbf r-\mathbf a$ is parallel to $\mathbf b$, which is exactly the condition for $\mathbf r$ to lie on the line. It is equivalent to $\mathbf r=\mathbf a+\lambda\mathbf b$. To test whether a point lies on the line, substitute its position vector and check that the vector product is $\mathbf 0$. To write the equation from a vector equation, read off $\mathbf a$ and $\mathbf b$. A direction perpendicular to two given directions comes from their vector product, which is often how the direction of a new line is found.

$P(0,1,2)$, $Q(2,3,1)$, $R(1,0,4)$. Then $\overrightarrow{PQ}=(2,2,-1)$ and $\overrightarrow{PR}=(1,-1,2)$, so $\overrightarrow{PQ}\times\overrightarrow{PR}=(4-1,\ -1-4,\ -2-2)=(3,-5,-4),$ with magnitude $\sqrt{50}=5\sqrt2$. The triangle has area $\frac{5\sqrt2}{2}$ and the parallelogram $PQSR$ (with $S=Q+R-P=(3,2,3)$) has area $5\sqrt2$. The diagonals are $\overrightarrow{PS}=(3,1,1)$ and $\overrightarrow{QR}=(-1,-3,3)$, and $\frac12|\overrightarrow{PS}\times\overrightarrow{QR}|=\frac12\sqrt{200}=5\sqrt2$ as well.