Vector Geometry

Find where two lines in a shape cross, and the ratios they make, by writing one point two ways and matching the a and b terms.

In the exam for Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, IGCSE Additional Maths 0606 · 63 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What Vector Geometry Asks

The big questions draw two lines across a shape. They cross at a point P. You find the vector to P, or the ratio P makes on a line.

Steps

1
P is on the line OM: $O P = λ O M$
2
P is also on AN: $A P = μ A N$
3
Two answers for one point: match the a terms and the b terms.

Key idea

The Greek letters $λ$ (lambda) and $μ$ (mu) are unknown fractions: how far along each line P sits.

Two Routes to the Same Point

The 5 or 6-mark question, the most common by far. Here $O A = 3 a$ , $O B = 3 b$ , M is the midpoint of AB, N is on OB with $O N : N B = 1 : 2$ , and P is where OM meets AN. First write $O P$ two ways.

Worked example

1
$A B = - 3 a + 3 b$ — Walk from A back to O (that is −3a), then out to B (that is 3b).
2
$A M = \frac{1}{2} (- 3 a + 3 b)$ — M is the midpoint of AB, so AM is half of AB.
3
$O M = 3 a + \frac{1}{2} (- 3 a + 3 b)$ — Walk from O to A, then from A to M.
4
$O M = \frac{3}{2} a + \frac{3}{2} b$ — Collect the a terms: 3a − (3/2)a = (3/2)a. The b term is (3/2)b.
5
$O P = λ O M$ — P is on OM, so OP is some fraction λ of OM.
6
$O P = \frac{3}{2} λ a + \frac{3}{2} λ b$ — Route 1: multiply λ into each term.
7
$O N = \frac{1}{3} \times 3 b = b$ — ON : NB = 1 : 2 makes ON one part out of three of OB.
8
$A N = - 3 a + b$ — Walk from A back to O, then out to N.
9
$O P = 3 a + μ (- 3 a + b)$ — Route 2: walk to A, then a fraction μ of the way along AN. It is a different fraction, so it gets a different letter.
10
$O P = (3 - 3 μ) a + μ b$ — Multiply μ into the bracket, then collect the a terms: 3a − 3μa.

Common mistake

Using the same letter for both routes. P is a different fraction of the way along each line, so it needs two letters.

Match the a and b Terms

Both routes end at the same point, so the amounts of $a$ must match and the amounts of $b$ must match. That gives two equations.

Worked example

1
$b : \frac{3}{2} λ = μ$ — Both routes end at P, so the amount of b in each must match.
2
$a : \frac{3}{2} λ = 3 - 3 μ$ — The amount of a must match too.
3
$μ = 3 - 3 μ$ — The left sides of both equations are the same, so the right sides are equal. This leaves only one letter.
4
$4 μ = 3$ — Add 3μ to both sides.
5
$μ = \frac{3}{4}$ — Divide both sides by 4.
6
$\frac{3}{2} λ = \frac{3}{4}$ — Put μ = 3/4 into the b equation.
7
$λ = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ — Multiply both sides by 2/3 to get λ alone.
8
$O P = \frac{3}{4} a + \frac{3}{4} b$ — Put μ = 3/4 into route 2: 3 − 3(3/4) = 3/4 for the a term, and μ for the b term.
9
$O P : P M = λ : (1 - λ) = \frac{1}{2} : \frac{1}{2} = 1 : 1$ — P is λ of the way along OM, so OP is λ and PM is the rest, 1 − λ.

Key idea

If the question asks for a ratio, read it off the letter: $λ = \frac{1}{2}$ means P is half of the way from O to M.

Part (a): Write a Side

The 1-mark part (a) nearly always asks for one side, like $A B$ . Use it again in part (b).

Worked example

1
$O A = 5 a, O B = 3 b$ — These are the vectors from O. The side AB does not start at O.
2
$A B = A O + O B$ — Walk from A back to O, then out to B.
3
$A O = - 5 a$ — Going back along OA is the opposite of OA, so it gets a minus.
4
$A B = - 5 a + 3 b$ — Put the two parts together.

Common mistake

Writing 5a − 3b. That is BA, the side walked the wrong way. The routes are in Expressing Vectors in Terms of a and b.

Finding a Ratio From a Multiple

The 3-mark question gives $A P$ and asks for $A P : P M$ . Find $A M$ , then see what multiple of it $A P$ is.

Worked example

1
$A B = 6 a, A C = 2 b$ — M is the midpoint of BC. First find AM.
2
$B C = - 6 a + 2 b$ — Walk from B back to A, then out to C.
3
$B M = \frac{1}{2} (- 6 a + 2 b) = - 3 a + b$ — M is the midpoint, so BM is half of BC.
4
$A M = 6 a + (- 3 a + b)$ — Walk from A to B, then from B to M.
5
$A M = 3 a + b$ — Collect the a terms: 6a − 3a = 3a.
6
$A P = 2 a + \frac{2}{3} b$ — This is the vector the question gives for AP.
7
$A P = \frac{2}{3} (3 a + b)$ — Take out 2/3: (2/3) × 3a = 2a and (2/3) × b = (2/3)b, so AP is two thirds of AM.
8
$P M = 1 - \frac{2}{3} = \frac{1}{3} of A M$ — The rest of AM is what is left over.
9
$A P : P M = \frac{2}{3} : \frac{1}{3} = 2 : 1$ — Two thirds to one third is the same as 2 parts to 1 part.

Common mistake

Writing 2 : 3. The fraction two thirds means AP is 2 parts of 3, so PM is the 1 part left.

Ratio From a Position Vector

Sometimes the question gives you the position vector of a point on a line and asks for the ratio it makes. Work out its piece of the line and compare.

Worked example

1
$O A = x, O B = y, O D = \frac{3}{7} x + \frac{4}{7} y$ — D is on the line AB. Find AD : DB.
2
$A D = A O + O D = - x + \frac{3}{7} x + \frac{4}{7} y$ — Walk from A back to O, then out to D.
3
$A D = - \frac{4}{7} x + \frac{4}{7} y$ — Collect the x terms: −1 + 3/7 = −4/7.
4
$A D = \frac{4}{7} (- x + y)$ — Take out 4/7 from both terms.
5
$A B = - x + y$ — Walk from A back to O, then out to B. It is the same bracket.
6
$A D = \frac{4}{7} A B$ — So AD is four sevenths of the whole line AB.
7
$D B = 1 - \frac{4}{7} = \frac{3}{7} of A B$ — DB is the rest of the line.
8
$A D : D B = 4 : 3$ — Four sevenths to three sevenths is 4 parts to 3 parts.

Key idea

Check: the two fractions in OD add to 1 (3/7 + 4/7), as they must for a point on AB. The bigger fraction is on y, so D is nearer B, and AD : DB = 4 : 3.

When a Line Is Extended

Sometimes the lines only meet when one is extended past the shape. You use the same two routes. The fraction just comes out bigger than 1.

$O A = 6 a$ , $O B = 2 b$ , $O P = 2 a$ and $O Q = b$ . The lines AB and PQ meet at R, beyond B. Find $O R$ and the ratio $A B : B R$ .

Worked example

1
$A B = - 6 a + 2 b$ — Walk from A back to O, then out to B.
2
$O R = 6 a + μ (- 6 a + 2 b)$ — Route 1: R is on AB. Walk to A, then a fraction μ along AB.
3
$O R = (6 - 6 μ) a + 2 μ b$ — Multiply μ into the bracket and collect.
4
$P Q = - 2 a + b$ — Walk from P back to O, then out to Q.
5
$O R = 2 a + λ (- 2 a + b)$ — Route 2: R is on PQ. Walk to P, then a fraction λ along PQ.
6
$O R = (2 - 2 λ) a + λ b$ — Multiply λ into the bracket and collect.
7
$b : 2 μ = λ$ — Both routes reach R, so the b terms match.
8
$a : 6 - 6 μ = 2 - 2 λ$ — The a terms match too.
9
$6 - 6 μ = 2 - 4 μ$ — Replace λ by 2μ, so 2λ becomes 4μ.
10
$4 = 2 μ$ — Add 6μ to both sides, then take 2 from both sides.
11
$μ = 2$ — Divide both sides by 2. It is bigger than 1, which is right: R is beyond B.
12
$O R = (6 - 6 (2)) a + 2 (2) b = - 6 a + 4 b$ — Put μ = 2 into route 1. Keep the brackets so the negative is not lost.
13
$A R = 2 A B$ — μ = 2 means R is twice as far from A as B is.
14
$A B : B R = 1 : 1$ — AB is one half of AR, so BR is the other half. B is the midpoint of AR.

Common mistake

Rejecting μ = 2 because a fraction should be less than 1. Past 1 only means the point is beyond the end of the line, and that is what “extended” means.

Which Method? Read the Question

Look at what the question gives and what it wants.

You see	You do
Write AB (1 mark)	one route: Expressing Vectors in Terms of a and b
Two lines meet at P	two routes, two unknown fractions, match a and b
Lines meet beyond the shape	same two routes: the fraction is bigger than 1
AP given, find AP : PM	write AP as a multiple of AM
Position vector of D given, find AD : DB	write AD as a multiple of AB
Distance between towns	Magnitude of a Vector
Prove parallel or a straight line	Vector Proofs

Key idea

Show your working clearly. The marks come from each route and each equation, not only the answer.

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