Vectors in problem solvingAQA A-Level Maths: Revision notes

Most vector problems follow the same steps. Sketch the figure and mark every known vector. To find any vector, choose a route along known vectors from its start to its end, adding each one and reversing the sign of any you travel against: for example $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\mathbf b-\mathbf a$. Simplify the result and write it in terms of the given vectors only. Use ratios: if $M$ is on $AB$ with $AM:MB=m:n$, then $\overrightarrow{AM}=\frac{m}{m+n}\overrightarrow{AB}$.

Points are collinear if they lie on one straight line. To prove it, show that two vectors between the points are scalar multiples, for example $\overrightarrow{PR}=\lambda\overrightarrow{PQ}$, and that they share a point. For $\mathbf p=\mathbf i+2\mathbf j$, $\mathbf q=3\mathbf i+8\mathbf j$, $\mathbf r=k\mathbf i+14\mathbf j$: $\overrightarrow{PQ}=2\mathbf i+6\mathbf j$ and $\overrightarrow{PR}=(k-1)\mathbf i+12\mathbf j$. Setting $\overrightarrow{PR}=\lambda\overrightarrow{PQ}$ gives $12=6\lambda$, $\lambda=2$, so $k-1=4$ and $k=5$. Lines are parallel when their direction vectors are scalar multiples.

If $\mathbf a$ and $\mathbf b$ are not parallel (and non-zero), then $p\mathbf a+q\mathbf b=r\mathbf a+s\mathbf b$ implies $p=r$ and $q=s$. This lets you find where two lines meet. Write the position of the intersection $X$ in two ways, one for each line, each with its own unknown scalar, then equate coefficients of $\mathbf a$ and $\mathbf b$ and solve the simultaneous equations.

In triangle $OAB$, $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$, $AM:MB=1:3$ and $C$ is the midpoint of $OA$. Find where $OM$ meets $CB$. $\overrightarrow{OM}=\mathbf a+\frac14(\mathbf b-\mathbf a)=\frac34\mathbf a+\frac14\mathbf b$ and $\overrightarrow{CB}=\mathbf b-\frac12\mathbf a$. Along $OM$: $\overrightarrow{OX}=\lambda\left(\frac34\mathbf a+\frac14\mathbf b\right)$. Along $CB$: $\overrightarrow{OX}=\frac12(1-\mu)\mathbf a+\mu\mathbf b$. Equate: $\frac{3\lambda}{4}=\frac{1-\mu}{2}$, $\frac\lambda4=\mu$. So $\lambda=4\mu$, $6\mu=1-\mu$, $\mu=\frac17$, $\lambda=\frac47$ and $\overrightarrow{OX}=\frac37\mathbf a+\frac17\mathbf b$.

A force is a vector, written $(x\mathbf i+y\mathbf j)$ N. The resultant of several forces is their vector sum. A particle is in equilibrium when the resultant is the zero vector, so an unknown force is the negative of the sum of the others. By Newton's second law, $\mathbf F=m\mathbf a$, so acceleration is $\mathbf a=\frac{\mathbf F}{m}$ in the direction of the resultant. If $\mathbf F_1=3\mathbf i+2\mathbf j$ and $\mathbf F_2=-5\mathbf i+4\mathbf j$ act on a $2$ kg particle, the resultant is $-2\mathbf i+6\mathbf j$ N and the acceleration is $(-\mathbf i+3\mathbf j)$ m s⁻². The magnitude of a force or acceleration comes from $\sqrt{x^2+y^2}$.

Position vectors work in problems about ships, aircraft and drones. With $\mathbf i$ east and $\mathbf j$ north, a ship starting at $\mathbf r_0$ with constant velocity $\mathbf v$ has position $\mathbf r=\mathbf r_0+t\mathbf v$ after $t$ hours. For $\mathbf r_0=8\mathbf i+6\mathbf j$ and $\mathbf v=-2\mathbf i+\mathbf j$, $\mathbf r=(8-2t)\mathbf i+(6+t)\mathbf j$. The ship is due north of $O$ when the $\mathbf i$-component is zero: $t=4$, at $10\mathbf j$. Read the question for a condition (due north, collinear with, equidistant from), turn it into an equation for the components, solve it, and answer in context with units.