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559
This chapter will show you …
● the properties of vectors
● how to add and subtract vectors
● how to use vectors to solve geometrical problems
Visual overview
What you should already know
● Vectors are used to describe translations
Quick check
Use column vectors to describe these translations.
a A to C
b B to D
c C to D
d D to E
C
B D
A E
1 Properties ofvectors
2 Vectors ingeometry
Properties
Addition and subtraction Solving geometrical problems
A vector is a quantity which has both magnitude and direction. It can be represented by a straight linewhich is drawn in the direction of the vector and whose length represents the magnitude of the vector.Usually, the line includes an arrowhead.
The translation or movement from A to B is represented by thevector a.
a is always printed in bold type, but is written as a.
a can also be written as AB⎯→
.
A quantity which is completely described by its magnitude, and has no direction associated with it, is called a scalar. The mass of a bus (10 tonnes) is an example of a scalar. Another example is a linearmeasure, such as 25.4 mm.
Multiplying a vector by a number (scalar) alters its magnitude (length) but not its direction. For example,the vector 2a is twice as long as the vector a, but in the same direction.
A negative vector, for example –b, has the same magnitude as the vector b, but is in the oppositedirection.
Addition and subtraction of vectors
Take two non-parallel vectors a and b, then a + b is defined to be the translation of a followed by thetranslation of b. This can easily be seen on a vector diagram.
ab b
a
a + b
b –b
a 2a
a
A
B
560
Properties of vectors25.1
Key wordsdirectionmagnitudevector
In this section you will learn how to:● add and subtract vectors
Similarly, a – b is defined to be the translation of a followed by the translation of –b.
Look at the parallelogram grid below. a and b are two independent vectors that form the basis of this grid.It is possible to define the position, with reference to O, of any point on this grid by a vector expressed interms of a and b. Such a vector is called a position vector.
For example, the position vector of K is⎯→OK or k = 3a + b, the position vector of E is
⎯→OE or e = 2b.
The vector ⎯→HT = 3a + b, the vector
⎯→PN = a – b, the vector
⎯→MK = 2a – 2b, and the vector
⎯→TP = –a – b.
Note ⎯→OK and
⎯→HT are called equal vectors because they have exactly the same length and are in the same
direction.⎯→MK and
⎯→PN are parallel vectors but
⎯→MK is twice the magnitude of
⎯→PN.
b
a A C F J
KGD
O
B
E H L P R
N
S TQMI
a a
a – b
–b–b
561
CHAPTER 25: VECTORS
EXAMPLE 1
a Using the grid above, write down the following vectors in terms of a and b.
i⎯→BH ii
⎯→HP iii
⎯→GT
iv→T I v
⎯→FH vi
⎯→BQ
b What is the relationship between the following vectors?
i⎯→BH and
⎯→GT ii
⎯→BQ and
⎯→GT iii
⎯→HP and
→T I
c Show that B, H and Q lie on the same straight line.
a i a + b ii 2a iii 2a + 2b iv – 4a v –2a + 2b vi 2a + 2b
b i⎯→BH and
⎯→GT are parallel and
⎯→GT is twice the length of
⎯→BH.
ii⎯→BQ and
⎯→GT are equal.
iii⎯→HP and
→TI are in opposite directions and
→T I is twice the length of
⎯→HP.
c⎯→BH and
⎯→BQ are parallel and start at the same point B. Therefore, B, H and Q must lie on
Vectors can be used to prove many results in geometry, as the following examples show.
565
Vectors in geometry25.2
Key wordvector
In this section you will learn how to:● use vectors to solve geometrical problems
EXAMPLE 3
In the diagram,⎯→OA = a,
⎯→OB = b, and
⎯→BC = 1.5a. M is the midpoint of BC, N is the midpoint
of AC and P is the midpoint of OB.
a Find these vectors in terms of a and b.
i⎯→AC ii
⎯→OM iii
⎯→BN
b Prove that ⎯→PN is parallel to
⎯→OA.
a i You have to get from A to C in terms of vectors that you know.⎯→AC =
⎯→AO +
⎯→OB +
⎯→BC
Now⎯→AO = –
⎯→OA, so you can write
⎯→AC = –a + b + 3–2 a
= 1–2 a + b
Note that the letters “connect up” as we go from A to C, and that the negative of a vector represented by any pair of letters is formed by reversing the letters.
⎯→BC = 6b – 1–2 (4b – a) = 6b – 2b + 1–2 a = 1–2 a + 4b
iii⎯→PQ =
⎯→PB +
⎯→BQ =
⎯→PB + 1–2
⎯→BC = b + 1–2 (4b – a) = b + 2b – 1–2 a = 3b – 1–2 a
b⎯→PR =
⎯→PA +
⎯→AO +
⎯→OR = –b – a + 3b = 2b – a
So⎯→PS = 1–2
⎯→PR = b – 1–2 a
⎯→SQ =
⎯→SP +
⎯→PQ = 1–2 a – b + 3b – 1–2 a = 2b
⎯→OC = 6b, so
⎯→SQ is parallel to
⎯→OC
In the triangle OAB, P is the midpoint of AB, X is the
midpoint of OB, OA⎯→
= a and OB⎯→
= b. Q is the pointthat divides OP in the ratio 2 : 1.
a Express these vectors in terms of a and b.
i⎯→AB ii
⎯→AP
iii⎯→OP iv
⎯→OQ
v⎯→AQ vi
⎯→AX
b Deduce that⎯→AX = k
⎯→AQ, where k is a scalar, and
find the value of k.
A
O
BP
X
Qa
b
OABC is a trapezium with AB parallel to OC. P, Q and R are the mid-points of AB, BC and OC respectively. OC is three times the length of AB.⎯→OA = a and
⎯→AP = b
a Express, in terms of a and b, the following vectors.