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Vector geometry

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  Cambridge O Level Mathematics (Syllabus D) 4024

Vector geometry

The big idea: In a vector diagram you never need a length or an angle. Every route between two points is a chain of known vectors, and a proof is just two routes shown to point the same way.

Summary — What This Chapter Covers
  • Represent vectors by directed line segments: an arrow from start to finish.
  • A position vector starts at the origin O: ⟶OA = a. Then ⟶AB = b − a.
  • Express any vector in terms of two given vectors such as a and b by following a route.
  • Use vectors to prove lines parallel, prove three points collinear, and solve problems with ratios and similar triangles.
1. Directed Line Segments
Syllabus 7.4 · Paper 1 & 2

An arrow carries the vector

A vector can be drawn as a directed line segment: a line with an arrow showing which way it goes. Any segment with the same length and direction represents the same vector. In a parallelogram OABC with ⟶OA = a and ⟶OC = c, the opposite sides are equal vectors:

VectorRouteIn terms of a and c
⟶CBparallel to and the same as ⟶OAa
⟶ABparallel to and the same as ⟶OCc
⟶OB⟶OA + ⟶ABa + c
⟶AC⟶AO + ⟶OC−a + c = c − a
Going against an arrow

If your route travels backwards along a vector, subtract it. ⟶AO = −a because ⟶OA = a.

2. Position Vectors
Syllabus 7.4 · Paper 1 & 2

Vectors that start at the origin

The position vector of a point A is ⟶OA, the vector from the origin to A. If A is (2, 5), its position vector is 25. With position vectors a and b:

Key results ⟶AB = b − a    (end minus start)
midpoint M of AB: ⟶OM = ½(a + b)
P on AB with AP : PB = m : n: ⟶OP = a + m⁄m+n(b − a)
Worked Example

A is (2, 5) and B is (8, −1).
(a) Find ⟶AB.
(b) Find the position vector of the midpoint M of AB.
(c) P lies on AB with AP : PB = 1 : 2. Find the coordinates of P.

3. Expressing Vectors in Terms of Two Vectors
Syllabus 7.4 · Paper 1 & 2

Follow a route you know

Most questions give two vectors, often a and b, along two sides of a shape. To find any other vector, write a route from its start to its end using only sides you know, then simplify. A ratio on a line tells you the fraction of that line: AN : NB = 3 : 1 means AN is ¾ of AB.

Worked Example

OAB is a triangle with ⟶OA = a and ⟶OB = b. M is the midpoint of OB. N lies on AB with AN : NB = 3 : 1.
Find, in terms of a and b, simplifying your answers: (a) ⟶AB   (b) ⟶ON   (c) ⟶MN

4. Proving Lines Parallel and Points Collinear
Syllabus 7.4 · Paper 1 & 2

Scalar multiples

Two vectors are parallel when one is a number times the other. Three points are collinear (on one straight line) when two of the vectors joining them are parallel and share a point. Always finish with a sentence that says what you have shown.

Proof facts ⟶PQ = k⟶RS ⇒ PQ is parallel to RS and PQ = k × RS
⟶CE = k⟶CD ⇒ C, D and E are collinear (parallel, and C is common)
Worked Example

OAB is a triangle with ⟶OA = 6a and ⟶OB = 6b. C lies on OA with OC : CA = 2 : 1. D is the midpoint of AB. E is the point such that B is the midpoint of OE.
(a) Find ⟶CD and ⟶CE in terms of a and b.
(b) Show that C, D and E lie on a straight line, and find CD : DE.

Practice Question 1

⟶OA a ⟶OB b

5. Ratio and Similarity Problems
Syllabus 7.4 · Paper 1 & 2
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  • Two routes, compare the coefficients
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