Unit 28 · Vectors (Combined Maths II)
Describe points by position vectors from an origin, find $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, distances, midpoints and the centroid, and use them to describe parallelograms.
Fix a point $O$, the origin. The position vector of a point $A$ is $\overrightarrow{OA}$, written $\mathbf{a}$ (the lower-case letter matching the point). If $A$ has coordinates $(x, y)$ then $\mathbf{a} = x\mathbf{i} + y\mathbf{j}$, so points and position vectors carry the same information.
To go from $A$ to $B$, go back to $O$ and then out to $B$: $$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}.$$ Remember it as head minus tail. The distance between the points is $AB = |\mathbf{b} - \mathbf{a}|$.
If $M$ is the midpoint of $AB$ then $\overrightarrow{AM} = \tfrac{1}{2}\overrightarrow{AB}$, so $$\mathbf{m} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{\mathbf{a} + \mathbf{b}}{2}.$$ The centroid $G$ of triangle $ABC$ (where the medians meet) has $\mathbf{g} = \dfrac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}$; the next lesson proves it. Notice that these are the usual coordinate formulas written in vector form.
Example: $A(1, 2)$ and $B(4, 6)$. Then $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$, $\mathbf{b} = 4\mathbf{i} + 6\mathbf{j}$, $\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j}$ and $AB = \sqrt{9 + 16} = 5$. The midpoint has position vector $\tfrac{1}{2}(5\mathbf{i} + 8\mathbf{j}) = \tfrac{5}{2}\mathbf{i} + 4\mathbf{j}$, that is the point $\left(\tfrac{5}{2}, 4\right)$.
$ABCD$ (lettered in order round the figure) is a parallelogram exactly when opposite sides are equal as vectors: $\overrightarrow{AB} = \overrightarrow{DC}$. In position vectors, $\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d}$, so $$\mathbf{d} = \mathbf{a} + \mathbf{c} - \mathbf{b}.$$ Also $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$, that is the midpoints of the diagonals $AC$ and $BD$ coincide: the diagonals bisect each other.
Example: $A(1, 1)$, $B(4, 2)$, $C(2, 5)$ are three vertices of a parallelogram $ABCD$. Find $D$. $\mathbf{d} = \mathbf{a} + \mathbf{c} - \mathbf{b} = (\mathbf{i} + \mathbf{j}) + (2\mathbf{i} + 5\mathbf{j}) - (4\mathbf{i} + 2\mathbf{j}) = -\mathbf{i} + 4\mathbf{j}$, so $D(-1, 4)$. Check: $\overrightarrow{AB} = 3\mathbf{i} + \mathbf{j}$ and $\overrightarrow{DC} = (2 + 1)\mathbf{i} + (5 - 4)\mathbf{j} = 3\mathbf{i} + \mathbf{j}$.
Example: The origin is $O$, $\overrightarrow{OP} = 5\mathbf{i} - 2\mathbf{j}$ and $\overrightarrow{PQ} = -2\mathbf{i} + 6\mathbf{j}$. Then $\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = 3\mathbf{i} + 4\mathbf{j}$, so $|\overrightarrow{OQ}| = 5$, and the midpoint of $PQ$ has position vector $\tfrac{1}{2}(8\mathbf{i} + 2\mathbf{j}) = 4\mathbf{i} + \mathbf{j}$.
Exam tip: The commonest slip is $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$. Say "head minus tail" every time, and test it on numbers: from $A(1, 1)$ to $B(4, 2)$ the move is $+3$ in $x$, so $\overrightarrow{AB}$ must have $\mathbf{i}$-component $+3$. For "find the fourth vertex" questions, the diagonal-midpoint idea ($\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$) avoids drawing the figure, but check the letter order stated in the question.