Vectors in problem solvingEdexcel A-Level Maths: Revision notes

Most vector problems use three ideas: (1) $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$, (2) any journey can be split into a route through other points, such as $\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}$, and (3) equal vectors have equal components, while parallel vectors are scalar multiples. Method: sketch the situation, label the known vectors, write the required vector as a route from the origin, then simplify. State the final answer in terms of $\mathbf{i}$ and $\mathbf{j}$ (or the given letters) and, in context, with units.

If $P$ lies on $AB$ with $AP:PB=1:2$, then $AP$ is $\frac13$ of $AB$, so $\overrightarrow{OP}=\mathbf{a}+\frac13\overrightarrow{AB}$. For $\mathbf{a}=6\mathbf{i}+2\mathbf{j}$ and $\mathbf{b}=-3\mathbf{i}+11\mathbf{j}$: $\overrightarrow{AB}=-9\mathbf{i}+9\mathbf{j}$ and $\overrightarrow{OP}=3\mathbf{i}+5\mathbf{j}$. The midpoint is the special case $\frac12$. To show points are collinear, show two vectors between them are parallel and share a point. With $\overrightarrow{OR}=9\mathbf{i}+15\mathbf{j}=3\overrightarrow{OP}$, the points $O$, $P$, $R$ lie on a line. Write the conclusion in words: 'parallel and share the point $O$, so collinear'.

In a parallelogram $ABCD$, opposite sides are equal and parallel as vectors: $\overrightarrow{AD}=\overrightarrow{BC}$ and $\overrightarrow{AB}=\overrightarrow{DC}$. Given three position vectors, the fourth is found by a route: $\mathbf{d}=\mathbf{a}+\overrightarrow{BC}=\mathbf{a}+\mathbf{c}-\mathbf{b}$. Example: $\mathbf{a}=\mathbf{i}+2\mathbf{j}$, $\mathbf{b}=5\mathbf{i}+3\mathbf{j}$, $\mathbf{c}=8\mathbf{i}+7\mathbf{j}$ give $\mathbf{d}=4\mathbf{i}+6\mathbf{j}$. The diagonals bisect each other: the midpoint of $AC$, $\frac12(\mathbf{a}+\mathbf{c})=\frac92\mathbf{i}+\frac92\mathbf{j}$, is also the midpoint of $BD$. Remember that the order of letters in $ABCD$ matters: $AC$ and $BD$ are the diagonals. A parallelogram is a rectangle only if its diagonals are also equal in length.

A force has magnitude and direction, so it is a vector, and several forces acting on a particle are added to give the resultant: $\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2+\ldots$. A particle is in equilibrium when the resultant is zero. Example: $\mathbf{F}_1=5\mathbf{i}+2\mathbf{j}$ and $\mathbf{F}_2=-3\mathbf{i}+7\mathbf{j}$ give resultant $2\mathbf{i}+9\mathbf{j}$ N. For equilibrium a third force must be $\mathbf{F}_3=-2\mathbf{i}-9\mathbf{j}$ N. The size of the resultant is $\sqrt{85}$ N, and its direction makes angle $\tan^{-1}\frac92=77.5^\circ$ with $\mathbf{i}$. The use of forces in dynamics, such as $\mathbf{F}=m\mathbf{a}$, is part of the Mechanics content; here only the vector working is needed.

With constant velocity $\mathbf{v}$ from initial position $\mathbf{r}_0$, the position after $t$ hours is $\mathbf{r}=\mathbf{r}_0+t\mathbf{v}$. The speed is the magnitude $|\mathbf{v}|$ (a scalar). Example: a ship starts at $2\mathbf{i}-5\mathbf{j}$ km with $\mathbf{v}=3\mathbf{i}+4\mathbf{j}$ km h$^{-1}$. Speed $=5$ km h$^{-1}$ and $\mathbf{r}=(2+3t)\mathbf{i}+(4t-5)\mathbf{j}$. Due east of $O$ means the $\mathbf{j}$ component is zero: $t=1.25$, when $\mathbf{r}=5.75\mathbf{i}$, so the ship is $5.75$ km from $O$. In context, read $\mathbf{i}$ as east and $\mathbf{j}$ as north, 'due north' means a zero $\mathbf{i}$ component, and always state units.