The scalar productEdexcel International A Level Maths: Revision notes

The scalar product of two vectors is a number, not a vector. For $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$: $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3.$ It is also $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$, where $\theta$ is the angle between the two vectors when drawn tail to tail. Example: for $\mathbf{p}=2\mathbf{i}-3\mathbf{j}+\mathbf{k}$ and $\mathbf{q}=4\mathbf{i}+\mathbf{j}-5\mathbf{k}$, $\mathbf{p}\cdot\mathbf{q}=8-3-5=0$. In 2D, drop the third term. Note that $\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2$.

Rearranging $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$: $\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$ If $\mathbf{a}\cdot\mathbf{b}>0$ the angle is acute, and if $\mathbf{a}\cdot\mathbf{b}<0$ it is obtuse. Example: $\overrightarrow{OA}=2\mathbf{i}+2\mathbf{j}+\mathbf{k}$ and $\overrightarrow{OB}=3\mathbf{i}-6\mathbf{j}+2\mathbf{k}$ give $\mathbf{a}\cdot\mathbf{b}=-4$, $|\mathbf{a}|=3$, $|\mathbf{b}|=7$, so $\cos AOB=-\frac{4}{21}$ and $AOB=101.0^\circ$.

To find angle $ABC$ in a triangle, use the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$, because both must start at $B$. Using $\overrightarrow{AB}$ with $\overrightarrow{BC}$ gives the supplement, $180^\circ-\theta$. For points $A(1,-1,2)$, $B(3,1,3)$, $C(2,-3,0)$: $\overrightarrow{BA}=-2\mathbf{i}-2\mathbf{j}-\mathbf{k}$ and $\overrightarrow{BC}=-\mathbf{i}-4\mathbf{j}-3\mathbf{k}$, so $\cos ABC=\frac{13}{3\sqrt{26}}$ and $ABC=31.8^\circ$. The angle between two lines is found from their direction vectors; if the answer exceeds $90^\circ$, the acute angle between the lines is $180^\circ$ minus it.

If $\mathbf{a}$ and $\mathbf{b}$ are non-zero vectors and $\mathbf{a}\cdot\mathbf{b}=0$, then $\cos\theta=0$, so $\mathbf{a}$ and $\mathbf{b}$ are perpendicular. The converse also holds. To find an unknown constant, set the scalar product equal to zero. Example: $\mathbf{r}=\mathbf{i}+2\mathbf{j}+\lambda\mathbf{k}$ is perpendicular to $\mathbf{p}=2\mathbf{i}-3\mathbf{j}+\mathbf{k}$ when $2-6+\lambda=0$, so $\lambda=4$. This is the quickest test for a right angle in a triangle or rectangle: show the scalar product of the two sides at that vertex is $0$.

Combine the tools: find vectors between points, take scalar products, and use $|\mathbf{a}|$.

• A right angle at $A$ means $\overrightarrow{AB}\cdot\overrightarrow{AC}=0$, and then area $=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|$.
• For a rectangle $ABDC$, $\overrightarrow{BD}=\overrightarrow{AC}$, so $\overrightarrow{OD}=\overrightarrow{OB}+\overrightarrow{AC}$.
• The angle between diagonals uses their direction vectors; take the modulus of the cosine for the acute angle. Example: $A(1,2,3)$, $B(4,-1,5)$, $C(3,4,p)$ with a right angle at $A$ gives $2p-6=0$, so $p=3$ and the area is $2\sqrt{11}$.