Vectors in three dimensionsEdexcel A-Level Maths: Revision notes

In three dimensions a point has coordinates $(x,y,z)$ and a vector has three components. The unit vectors along the three perpendicular axes are $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$, so $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}=\begin{pmatrix}x \\ y \\ z\end{pmatrix}.$ The position vector of the point $(x,y,z)$ is $\overrightarrow{OP}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$. The $x$ and $y$ axes usually lie in a horizontal plane, with $z$ vertically upwards. Two vectors are equal only if all three components are equal. Vectors are added and multiplied by scalars component by component, exactly as in two dimensions: $(2\mathbf{i}-\mathbf{j}+3\mathbf{k})+2(\mathbf{i}+4\mathbf{j}-\mathbf{k})=4\mathbf{i}+7\mathbf{j}+\mathbf{k}$.

As in two dimensions, $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ (end minus start). For $A(2,-1,3)$ and $B(5,3,15)$: $\overrightarrow{AB}=3\mathbf{i}+4\mathbf{j}+12\mathbf{k}$. The position vector of the midpoint of $AB$ is $\frac12(\mathbf{a}+\mathbf{b})$. To find a point such as $C$ with $\overrightarrow{AC}=2\overrightarrow{AB}$, write $\overrightarrow{OC}=\mathbf{a}+2\overrightarrow{AB}=8\mathbf{i}+7\mathbf{j}+27\mathbf{k}$.

The magnitude of $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ is $\sqrt{x^2+y^2+z^2}$, from applying Pythagoras twice (once across the base, once up to the point). The distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2.$ Example: $AB=\sqrt{3^2+4^2+12^2}=13$. The diagonal of a $12\times4\times3$ room is $\sqrt{144+16+9}=13$ m, because $12^2+4^2+3^2=169$. Never drop a component: $\sqrt{3^2+4^2}=5$ is the length across the floor only.

The same ideas used in two dimensions work in three. Triangle sides are found from $\mathbf{b}-\mathbf{a}$, $\mathbf{c}-\mathbf{a}$ and $\mathbf{c}-\mathbf{b}$; you can show a triangle is isosceles by comparing lengths, or right-angled using the converse of Pythagoras, $AB^2+AC^2=BC^2$. Example: $A(1,-1,2)$, $B(3,1,3)$, $C(-1,0,4)$ give $AB=AC=3$ and $BC^2=18=9+9$, so angle $A=90^\circ$ and area $=\frac12\times3\times3=\frac92$. Points along a line: the point halfway from $O$ to $G(12,4,3)$ is $\frac12\overrightarrow{OG}=6\mathbf{i}+2\mathbf{j}+\frac32\mathbf{k}$.

A drone flies from $P(4,-2,7)$ to $Q(-2,4,4)$, in metres with $\mathbf{k}$ upwards. $\overrightarrow{PQ}=-6\mathbf{i}+6\mathbf{j}-3\mathbf{k}$. Distance $PQ=\sqrt{36+36+9}=9$ m. The midpoint of the flight is $\frac12(\mathbf{p}+\mathbf{q})=\mathbf{i}+\mathbf{j}+\frac{11}{2}\mathbf{k}$, at height $5.5$ m. Interpretation: the $-3\mathbf{k}$ shows the drone descends $3$ m during the flight. Always read the components in context to say what they mean.