Vector Geometry – A Level Maths Year 1

1 Parallel Vectors and Collinear Points

Vectors turn geometry into algebra: to prove two lines are parallel, or three points lie on a line, or a point divides a line in a given ratio, you show that one vector is a multiple of another. This page completes Year 1 DfE content area J · Vectors, and applies it to forces and velocities.

Parallel

$\mathbf{a}$ is parallel to $\mathbf{b}$ $\iff$ $\mathbf{a} = \lambda\mathbf{b}$ for some scalar $\lambda \neq 0$

Collinear

$A$, $B$, $C$ are collinear $\iff$ $\overrightarrow{AB} = \lambda\overrightarrow{BC}$ (parallel) and they share the point $B$

Worked Example 1 — Proving three points are collinear

Show that $A(1, 2)$, $B(4, 8)$ and $C(6, 12)$ are collinear, and find the ratio $AB : BC$.

①$\overrightarrow{AB} = 3\mathbf{i} + 6\mathbf{j}$, $\overrightarrow{BC} = 2\mathbf{i} + 4\mathbf{j}$

②$\overrightarrow{AB} = \tfrac32\overrightarrow{BC}$, so $AB$ is parallel to $BC$.

③They share the point $B$, so $A$, $B$, $C$ are collinear.

④$AB : BC = \tfrac32 : 1 = 3 : 2$.

Writing only "$\overrightarrow{AB}$ is a multiple of $\overrightarrow{BC}$" is not enough for the final mark. The phrase "and they have the point $B$ in common" completes the proof.

2 Dividing a Line in a Given Ratio

$P$ on $AB$ with $AP : PB = m : n$

$\overrightarrow{OP} = \mathbf{a} + \dfrac{m}{m+n}(\mathbf{b} - \mathbf{a}) = \dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$

Don't memorise — route-find. Get to $P$ by going to $A$ and then the right fraction of the way along $\overrightarrow{AB}$. For $AP : PB = 2 : 1$, $P$ is $\tfrac23$ of the way from $A$ to $B$.

Worked Example 2 — A point in a ratio

$A(1, -1)$ and $B(7, 8)$. Find the coordinates of $P$ on $AB$ such that $AP : PB = 2 : 1$.

①$\overrightarrow{AB} = 6\mathbf{i} + 9\mathbf{j}$

②$\overrightarrow{OP} = \mathbf{a} + \tfrac23\overrightarrow{AB} = (\mathbf{i} - \mathbf{j}) + (4\mathbf{i} + 6\mathbf{j}) = 5\mathbf{i} + 5\mathbf{j}$

③$P(5, 5)$

3 Proving Geometric Results with Vectors

Choose two non-parallel "base" vectors (usually given, e.g. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$).
Write every other vector you need as a route through known vectors, simplified to $p\mathbf{a} + q\mathbf{b}$.
Compare: a multiple means parallel; equal expressions mean the same point.

Worked Example 3 — The midpoint theorem

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ and $N$ are the midpoints of $OA$ and $OB$. Prove that $MN$ is parallel to $AB$ and half its length.

①$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$

②$\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\tfrac12\mathbf{a} + \tfrac12\mathbf{b} = \tfrac12(\mathbf{b} - \mathbf{a})$

③$\overrightarrow{MN} = \tfrac12\overrightarrow{AB}$: so $MN \parallel AB$ and $MN = \tfrac12 AB$. ■

Worked Example 4 — Where two lines cross

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $P$ is the midpoint of $AB$, and $Q$ is on $OA$ with $OQ : QA = 2 : 1$. The lines $OP$ and $BQ$ meet at $X$. Find $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, and the ratio $OX : XP$.

①$\overrightarrow{OP} = \mathbf{a} + \tfrac12(\mathbf{b} - \mathbf{a}) = \tfrac12(\mathbf{a} + \mathbf{b})$. $X$ is on $OP$, so $\overrightarrow{OX} = \lambda \cdot \tfrac12(\mathbf{a} + \mathbf{b}) = \tfrac{\lambda}{2}\mathbf{a} + \tfrac{\lambda}{2}\mathbf{b}$.

②$X$ is also on $BQ$: $\overrightarrow{OX} = \mathbf{b} + \mu\left(\tfrac23\mathbf{a} - \mathbf{b}\right) = \tfrac{2\mu}{3}\mathbf{a} + (1 - \mu)\mathbf{b}$.

③$\mathbf{a}$ and $\mathbf{b}$ are not parallel, so compare coefficients: $\tfrac{\lambda}{2} = \tfrac{2\mu}{3}$ and $\tfrac{\lambda}{2} = 1 - \mu$.

④$\tfrac{2\mu}{3} = 1 - \mu \;\Rightarrow\; \tfrac{5\mu}{3} = 1 \;\Rightarrow\; \mu = \tfrac35$, and $\lambda = 2\left(1 - \tfrac35\right) = \tfrac45$.

⑤$\overrightarrow{OX} = \tfrac25(\mathbf{a} + \mathbf{b})$, and since $\overrightarrow{OX} = \tfrac45\overrightarrow{OP}$, $OX : XP = 4 : 1$.

Why you may compare coefficients: if $p\mathbf{a} + q\mathbf{b} = r\mathbf{a} + s\mathbf{b}$ with $\mathbf{a}$, $\mathbf{b}$ non-zero and not parallel, then $p = r$ and $q = s$. Say this in your answer.

Worked Example 5 — Diagonals of a parallelogram

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Prove that its diagonals bisect each other.

①$\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, so the midpoint of $OB$ has position vector $\tfrac12(\mathbf{a} + \mathbf{c})$.

②Midpoint of $AC$: $\mathbf{a} + \tfrac12(\mathbf{c} - \mathbf{a}) = \tfrac12(\mathbf{a} + \mathbf{c})$.

③The same point, so each diagonal passes through the midpoint of the other. ■

4 Vectors in Context: Forces and Velocities

Forces and velocities are vectors, so they add as vectors. The resultant of several forces is their vector sum. A particle is in equilibrium when the resultant force is zero.

Worked Example 6 — Resultant and equilibrium

Forces $\mathbf{F}_1 = (3\mathbf{i} + 2\mathbf{j})$ N and $\mathbf{F}_2 = (-5\mathbf{i} + 4\mathbf{j})$ N act on a particle. (a) Find the magnitude of the resultant and the angle it makes with $\mathbf{i}$. (b) Find the third force needed for equilibrium.

①$\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 = -2\mathbf{i} + 6\mathbf{j}$ N

②$|\mathbf{R}| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \approx 6.32$ N

③$\mathbf{R}$ points up and left: angle with $-\mathbf{i}$ is $\tan^{-1} 3 = 71.6°$, so the angle with $\mathbf{i}$ is $180° - 71.6° = 108.4°$.

④(b) For equilibrium $\mathbf{F}_3 = -\mathbf{R} = (2\mathbf{i} - 6\mathbf{j})$ N.

Worked Example 7 — A boat and a current

A boat steers due north at $5$ m s$^{-1}$ through the water, while a current flows due east at $1.2$ m s$^{-1}$. With $\mathbf{i}$ east and $\mathbf{j}$ north, find the resultant velocity, the speed, and the bearing on which the boat actually travels.

①$\mathbf{v} = 1.2\mathbf{i} + 5\mathbf{j}$ m s$^{-1}$

②Speed $= \sqrt{1.44 + 25} = \sqrt{26.44} \approx 5.14$ m s$^{-1}$

③Angle east of north: $\tan^{-1}\dfrac{1.2}{5} = 13.5°$, so bearing $013.5°$.

Velocity versus speed. Velocity is the vector $1.2\mathbf{i} + 5\mathbf{j}$; speed is its magnitude, $5.14$ m s$^{-1}$. Give the one the question asks for.

5 Quick Reference

Parallel

$\mathbf{a} = \lambda\mathbf{b}$

Collinear

Parallel and a common point.

Ratio $m : n$

$\mathbf{a} + \dfrac{m}{m+n}(\mathbf{b} - \mathbf{a})$

Midpoint

$\tfrac12(\mathbf{a} + \mathbf{b})$

Routes

$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}$

Intersection

Two expressions with $\lambda$, $\mu$; compare coefficients.

Comparing

Allowed only for non-parallel $\mathbf{a}$, $\mathbf{b}$.

Resultant

Vector sum of the forces.

Equilibrium

Resultant $= \mathbf{0}$.

Speed

Magnitude of velocity.

6 Practice Questions

Question 1

Show that $6\mathbf{i} - 9\mathbf{j}$ is parallel to $4\mathbf{i} - 6\mathbf{j}$.

▶ Show solution

$6\mathbf{i} - 9\mathbf{j} = \tfrac32(4\mathbf{i} - 6\mathbf{j})$, a scalar multiple, so they are parallel.

Question 2

Find $k$ such that $k\mathbf{i} + 6\mathbf{j}$ is parallel to $2\mathbf{i} - 3\mathbf{j}$.

▶ Show solution

$k\mathbf{i} + 6\mathbf{j} = \lambda(2\mathbf{i} - 3\mathbf{j})$. From $\mathbf{j}$: $\lambda = -2$.

$k = 2\lambda = -4$.

Question 3

Show that $A(-2, 1)$, $B(1, 3)$ and $C(7, 7)$ are collinear.

▶ Show solution

$\overrightarrow{AB} = 3\mathbf{i} + 2\mathbf{j}$, $\overrightarrow{BC} = 6\mathbf{i} + 4\mathbf{j} = 2\overrightarrow{AB}$.

So $AB \parallel BC$, and they share the point $B$: $A$, $B$, $C$ are collinear.

Question 4

$A(2, 3)$ and $B(10, -5)$. $P$ lies on $AB$ with $AP : PB = 3 : 1$. Find the coordinates of $P$.

▶ Show solution

$\overrightarrow{AB} = 8\mathbf{i} - 8\mathbf{j}$; $P$ is $\tfrac34$ of the way along.

$\overrightarrow{OP} = (2\mathbf{i} + 3\mathbf{j}) + (6\mathbf{i} - 6\mathbf{j}) = 8\mathbf{i} - 3\mathbf{j}$, so $P(8, -3)$.

Question 5

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AB$. Find $\overrightarrow{OM}$ and $\overrightarrow{CM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

▶ Show solution

In a parallelogram $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$.

$\overrightarrow{OM} = \mathbf{a} + \tfrac12\mathbf{c}$

$\overrightarrow{CM} = \overrightarrow{CO} + \overrightarrow{OM} = -\mathbf{c} + \mathbf{a} + \tfrac12\mathbf{c} = \mathbf{a} - \tfrac12\mathbf{c}$

Question 6

Forces $(2\mathbf{i} + 5\mathbf{j})$ N, $(4\mathbf{i} - 3\mathbf{j})$ N and $(p\mathbf{i} + q\mathbf{j})$ N act on a particle in equilibrium. Find $p$ and $q$.

▶ Show solution

Sum $= (6 + p)\mathbf{i} + (2 + q)\mathbf{j} = \mathbf{0}$.

$p = -6$, $q = -2$.

Question 7

Forces $(5\mathbf{i} + 2\mathbf{j})$ N and $(7\mathbf{i} + 3\mathbf{j})$ N act on a particle. Find the magnitude of the resultant and the angle it makes with $\mathbf{i}$.

▶ Show solution

$\mathbf{R} = 12\mathbf{i} + 5\mathbf{j}$ N, so $|\mathbf{R}| = \sqrt{144 + 25} = 13$ N.

Angle $= \tan^{-1}\tfrac{5}{12} = 22.6°$ above $\mathbf{i}$.

Question 8

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $D$ is the point on the diagonal $OB$ with $OD : DB = 1 : 2$, and $N$ is the midpoint of $OC$. Prove that $A$, $D$ and $N$ are collinear, and find the ratio $AD : DN$.

▶ Show solution

$\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, so $\overrightarrow{OD} = \tfrac13(\mathbf{a} + \mathbf{c})$; and $\overrightarrow{ON} = \tfrac12\mathbf{c}$.

$\overrightarrow{AD} = \tfrac13\mathbf{a} + \tfrac13\mathbf{c} - \mathbf{a} = -\tfrac23\mathbf{a} + \tfrac13\mathbf{c}$

$\overrightarrow{AN} = \tfrac12\mathbf{c} - \mathbf{a}$

$\overrightarrow{AD} = \tfrac23\left(-\mathbf{a} + \tfrac12\mathbf{c}\right) = \tfrac23\overrightarrow{AN}$. The vectors are parallel and share the point $A$, so $A$, $D$, $N$ are collinear. ■

$D$ is $\tfrac23$ of the way from $A$ to $N$, so $AD : DN = 2 : 1$.

Question 9

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $C$ is the midpoint of $OA$ and $D$ is on $OB$ with $OD : DB = 1 : 2$. The lines $AD$ and $BC$ meet at $X$. Find $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, and the ratio $AX : XD$.

▶ Show solution

$\overrightarrow{OC} = \tfrac12\mathbf{a}$, $\overrightarrow{OD} = \tfrac13\mathbf{b}$.

On $AD$: $\overrightarrow{OX} = \mathbf{a} + \lambda\left(\tfrac13\mathbf{b} - \mathbf{a}\right) = (1 - \lambda)\mathbf{a} + \tfrac{\lambda}{3}\mathbf{b}$

On $BC$: $\overrightarrow{OX} = \mathbf{b} + \mu\left(\tfrac12\mathbf{a} - \mathbf{b}\right) = \tfrac{\mu}{2}\mathbf{a} + (1 - \mu)\mathbf{b}$

$\mathbf{a}$, $\mathbf{b}$ not parallel, so: $1 - \lambda = \tfrac{\mu}{2}$ and $\tfrac{\lambda}{3} = 1 - \mu$.

From the second, $\lambda = 3 - 3\mu$. Then $1 - 3 + 3\mu = \tfrac{\mu}{2}$, so $\tfrac{5\mu}{2} = 2$ and $\mu = \tfrac45$, $\lambda = \tfrac35$.

$\overrightarrow{OX} = \tfrac25\mathbf{a} + \tfrac15\mathbf{b}$.

$\overrightarrow{AX} = \tfrac35\overrightarrow{AD}$, so $AX : XD = 3 : 2$.

Question 10

A river is $60$ m wide with straight parallel banks and flows at $0.9$ m s$^{-1}$. A swimmer can swim at $1.5$ m s$^{-1}$ in still water. Take $\mathbf{i}$ in the direction of the current and $\mathbf{j}$ straight across the river.

(a) She aims straight across. Find her resultant velocity and speed. (b) How long does she take to cross, and how far downstream does she land? (c) Instead she wants to land directly opposite her start. Find the angle to the bank-perpendicular at which she must aim upstream, and the time this crossing takes. (d) State one modelling assumption.

▶ Show solution

(a) $\mathbf{v} = 0.9\mathbf{i} + 1.5\mathbf{j}$ m s$^{-1}$; speed $\sqrt{0.81 + 2.25} = \sqrt{3.06} \approx 1.75$ m s$^{-1}$.

(b) Only the $\mathbf{j}$ component carries her across: time $= 60 \div 1.5 = 40$ s.

Downstream drift $= 0.9 \times 40 = 36$ m.

(c) Her swimming velocity must be $-0.9\mathbf{i} + y\mathbf{j}$ so that the $\mathbf{i}$ components cancel, with $0.81 + y^2 = 2.25$, so $y = 1.2$.

Angle upstream from straight across: $\sin\alpha = \dfrac{0.9}{1.5} = 0.6$, so $\alpha = 36.9°$.

Resultant velocity $1.2\mathbf{j}$, so time $= 60 \div 1.2 = 50$ s.

(d) For example: the current has the same speed everywhere across the river (in reality it is slower near the banks), or she swims at a constant speed.

Vector Geometry · A Level Maths Year 1 · Created with MathJax