Download - Vector Geometry
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CHAPTER ONE 1. Vectors. Examples and Uses.
For example, a wind may be described as being of 10 m.p.h. in a northerly
direction. We say its size (or magnitude) is 10 m.p.h. and its direction is north. Other
examples of vectors are a force, tension in a string, velocity of a car, current in a river,
acceleration, momentum and impulse.
It is understood in vector study that the magnitude and direction of a vector will
completely determine the vector. For example the location, temperature etc. of a wind
will have no relevance with regard to its being considered as a vector.
Many other objects can be thought of as vectors with a little imagination e.g.
i) – 3 has magnitude 3, direction “to the left”. ii) an ordered pair (a,b), its magnitude is
said to be how far from the origin it is,
i.e. 22ba + and its direction is
! where
tan ! =b
a. This means for example, that (3,4)
has magnitude 5 and direction 53.13°.
A Vector is an entity which has both size and direction.
!
(a,b) 22ba +
2
iii) a complex number a + bi can be thought of as in example ii) as having magnitude
22ba + and direction Arctan
a
b (angles being measured as usual counterclockwise
from positive horizontal axis)
Not all objects are vectors. Age has no notion of direction for example (except
maybe up!) and is consequently not a vector. Kinetic energy, work quantity of heat are
other examples which are not vectors.
1.1 Notation
A vector will be written a!
Its magnitude will be written a! . Note that this looks like absolute value. In fact
for real numbers it is. –3 has magnitude 3 (its distance from the origin) because 3! = 3.
Saying (3,4) has magnitude 5 is written 5 |)4,3(| =
!. The arrow over the )4,3(
!tells us we
are thinking of it as a vector. Note that a! is a positive real number and must be distinguished from the vector
a! itself. It makes little sense to talk about a! + a! . After all, what does 5|)4,3(| +
! mean? It
does make sense to talk of 3 + |-3| of course, but this is the exceptional circumstance.
This means, for example that a northerly wind is never equal to an easterly wind,
regardless of their magnitudes.
Two vectors are to be said to be equal if and only if their magnitudes and directions are equal separately.
3
Position of a Vector is irrelevant.
Diagrammatically a vector will be represented by
The length of the arrow represents the magnitude of the vector and
the direction of the arrow is the direction of the vector.
In the diagram the vector has magnitude
|OA|!
and direction north-east.
It is easy to confuse the notation with that of early geometry class. There, OA =
AO but here!!AOOA ! because
!AO and !OA have different directions. In geometry OA
usually means the length of the line segment OA and since the distance from O to A is
the same distance from A to O, then OA = AO. In our vector notation this is the
equivalent of saying |AO| |OA|!!
= which is true.
A 10 m.p.h. northerly wind in Timbuctoo
is identically equal to a 10 m.p.h. northerly wind in Toronto. This is a very important
concept and cannot be overemphasised.
Exercise 1.1
1. Find i) |AC|!
ii) |CA|!
45°
A
O W
N
S
E
A
B C
3
4
4
2. i) Does !!DFDE = ?
ii) Does !!|DF||DE| = ?
3. If !!
ba = does !!
|b||a| = ? Is the converse true?
4. Find all pairs of equal vectors represented in the diagram. ABCD is a parallelogram.
5. Find x so that i)
(x,4)!
= (3,4)!
ii)
| (x,3) |!
= | (3,4) |!
6. Draw a diagram where !!!
|BC| |AC||AB| == but none of !!!BC AC,AB, are equal as
vectors. What is the angle between !!ACand AB ?
Exercise 1.1 Answers
1. i) 5 ii) 5
2. i) No ii) Yes
3. Yes No
4. !!ECAE = ,
!!EBDE = ,
!!BCAD = ,
!!DCAB =
5. i) x = 3 ii) x = 4 or – 4
6. 60° (in an equilateral triangle)
D
E F
2 2
D
A B
C
E
!!EBDE e.g. =
5
150
50
1.2 Addition of Vectors
To add vectors it is clear that we don’t simply add directions and magnitudes
separately. For example, an aeroplane capable of 150 m.p.h. in still air, flying north in an
easterly wind of 50 m.p.h. is certainly not travelling at 200 m.p.h. (or 100 m.p.h. for that
matter) relative to the ground. Addition is best represented diagrammatically.
Draw a diagram so that the plane vector and the wind vector are joined together, as
follows:
The dotted vector represents the addition of the wind
and aeroplane vectors. In fact the plane is flying at
158.1 m.p.h. (Pythagoras) in a direction 18.43° west of
north relative to the ground.
Plane
150 m.p.h. 50 m.p.h.
Wind +
Plane 150
50 Wind
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Example
To find the addition of the two vectors in the diagram it is best to redraw the diagram as
Note that the flow of the arrows is continuous.
The addition of these two vectors produces a vector (called a resultant) whose magnitude
is 10 (Pythagoras). The direction of the resultant depends upon the directions of the
original vectors.
Example
To add redraw as
6 8
6
8 10
The dotted line represents the vector a +b +c
a
b
c
a
b
c
a
b
c
a + b +c
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Note that it is important that arrows of a , b and c on the diagram be continuous, i.e. if a
person were to ‘walk’ from the beginning of a to the end of c he would not be ‘stabbed’
i.e.
Example
An object is being pulled by 3 forces as in the diagram.
To find the resultant of these forces draw a new diagram as follows:
The dotted line represents the addition of the three vectors. It has magnitude 8.02 lbs in
fact.
20
25
10
45°
a
b
c
Is not a representation of a +b +c
because the person would be
‘stabbed’ at the first ‘corner’.
10 45°
20
25
8
Example
This is an important idea and will be used often, later on, where O is considered as the
origin.
Exercise 1.2
1. i) Does BCAB+ always equal toAC?
ii) Does CDBCAB ++ always equal toAD ?
iii) Does BCAB + always equal AC ?
2. Show by diagram that ba + = ab + for all a , b .
3. Show by diagram that ba + rarely equals ba + .
4. Find a condition on a and b so that ba + does equal ba + . Is it good
enough that a and b be parallel?
A
B
O
Note that OBABOA =+ regardless of the
positions of O,A,B. An important way of
thinking of this is OAOBAB !=
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5. In parallelogram ABCD
Represent each of the following as a single vector.
i) BEAB+
ii)
AE + EC
iii) ECBE +
iv) DACDEC ++
v) EABC +
vi) ADAB+
vii) EABADE ++
viii) BCDEAE ++
6. True or False?
i) aba !+
ii) DCAB = means ABCD is a parallelogram
iii) ba + = ca + means cb =
iv) ba + > ba +
v) . )cb(ac)ba( ++=++
D
A B
C
E
10
7.
Express each of the following as a single vector:
i) GCAGOA ++
ii) DEOB +
iii) ADOC +
iv) GFEA +
v) EGGFGA ++
Exercise 1.2 Answers
1. i) Yes ii) Yes iii) No
4. a and b must be in the same direction. No.
5. i) AE ii) AC iii) BC iv)EA v) BE vi) AC
vii) CA viii)AC
6. i) False ii) True iii) True iv) True v) True
7. i) OC ii) OF iii)OF iv)EB v) EB
O
A
B
C
D E
F
G
OAGCFEDB is a cube.
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Example
Parallelogram ABCD
Example
When subtracting vectors it is best to think of adding (-1) times the vector, e.g.
CBAB! means CB)1(AB !+ which is BCAB+ i.e. AC .
Example
Here BA2
1AB2
1AM !==
BC2
1MN =
A
B C
M N
D
A B
C
E
Here CDAB !=
DB2
1DE =
AC2
1AE =
CA2
1EC !=
A
B
C
CBAB! = BCAB+ =AC
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Example
ABCD is a parallelogram.
i) AE.BEABEBABDB2
1AB =+=!=!
BC !DC+ ED = BC + CD + ED
= BD + ED = BD+1
2BD =
3
2BD
Exercise 1.3
1.
.
2.
D
A B
C
E
A
B C
M N
M,N are mid-points of AB, AC respectively.
Express MN in terms of scalar multiples of
AC and AB only.
O
A
B
C
D E
F
G
OAGCFEDB is a cube.
i) Express GO in terms of OC,OB,OA
ii) Express CD in terms of OC,OB,OA
ii)
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3.
4.
C D
E
A
B
M
p
q
r
s
M is the mid-point of AE
Let s,r,q,p be DE,CD,BC,AB respectively
i) Express BM in terms of s,r,q,p
ii) Express MD in terms of s,r,q,p
A
B
D
C a
bO
Let aOA = , let bOB = . C,D are mid-points
of AB, OC respectively.
i) Express AB in terms of a and b
ii) Express OD in terms of a and b
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5.
6.
7. Draw a geometric figure illustrating the idea:
2( ba + ) = b2a2 +
8. True or False?
i) a2a2 =
ii) a3a3 =!
iii) baba !"+
iv) baba !=+ implies b is the zero vector
O
P Q R S
R bisects PQ. PS = 3
1 PQ
Let pOP = , qOQ =
i) Express OR in terms of p and q
ii) ExpressOS in terms of p and q
iii) Express RS in terms of p and q
p q
AD
BA
C
B
D
E
F
p
q
ABCD is a parallelogram
Let pAD = , qDC =
F is the mid-point of BC
Express the following in terms of p and q
i) AC ii) BD iii) DB
vi) ED v) FD vi) FE
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v) am = bn implies bm
na = (assume m ≠ 0)
vi) am + bn = cp implies cm
pb
m
na =+ (assume m ≠ 0)
vii) ba + = ca + implies cb =
viii) ra + s a = (r + s) a for any scalars r and s
9. i) What is the magnitude of a
a
1 ?
ii) What is the angle betweena and b if 2b
b
1a
a
1=+
iii) What is the angle betweena and b if 2b
b
1a
a
1=+
10. Forces of magnitude 5, 10 and 10 ‘act’ along different edges of a rectangular
3-dimensional “brick”. Find the magnitude of their resultant.
11. D,E,F are the mid-points of the sides BC, AC and AB respectively in triangle
ABC. Show that
AF+ BD + EAequals the zero vector, i.e. has magnitude
zero.
12. ABCDEF is a regular hexagon. Show that EFFADE =+ .
13. In quadrilateral ABCD DCsAB = where s is some positive scalar. What can
you say about the figure ABCD? What happens if s = 1?
14. Let
! be the angle between a and b (see diagram)
θ
a!
b
!
i) Express ba + in terms of a , b and
cos
! .
ii) Express ba ! in terms of a , b and
cos
! .
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Exercise 1.3 Answers
1. AB2
1AC2
1!
2. ! OCOBOA,OCOA ++!
3. i) s2
1r2
1q2
1p2
1+++! ii) s
2
1r2
1q2
1p2
1!++
4. i) ab ! ii) b4
1a4
1+
5. i) q2
1p2
1+ ii) q
3
1p3
2+ iii) q
6
1p6
1!
6. i) qp + ii) qp ! iii) pq !
iv) q2
1p2
1! v) qp
2
1! vi) q
2
1!
7.
8. i)T ii) T iii)F iv) F v) T vi) T vii)T viii)T
9. i) 1 ii) 0° iii) 90°
10. 15
13. Trapezoid, parallelogram
14.
a + b =! a 2
+
!
b
2
+ 2a b cos!
a !b =! a 2
+
!
b
2
! 2a b cos"
a!
b
!
2
!b
!a
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1.4 Geometry Proofs Using Vectors
To prove ABCD is a parallelogram it suffices to show DCAB = because this
guarantees that AB is both parallel and equal to DC in length. (see P. 2)
Note that it also guarantees the coplanarity of A, B, C, D.
Example
“To prove that if diagonals of a quadrilateral bisect each other then the quadrilateral is a
parallelogram”.
Proof
We know that AC and DC bisect each other at E.
Hence: ECAE = and DEEB =
ram.parallelog a is ABCD
DCECDEAB
DEECEBAE
!
=+=!
+=+!
Example
“To prove that the line segment joining the mid-points of two sides of a triangle is
parallel to the third side and equal in length to one half of it”.
D
A B
C
E
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!DE is parallel to BC and equal to one half of BC.
Exercise 1.4
1. ABCD is a quadrilateral in which AC = BD. M, N, O, P are mid-points of AB,
BC, CD, DA respectively. Show that MNOP is a rhombus.
2. In triangle ABC, D, E, F are mid-points of AB, BC, AC respectively. Medians
AE, BF, CD are concurrent at point O.
Prove OCOBOAOFOEOD ++=++
3. ABCD is a parallelogram. E, F lie on diagonal BD so that BF = ED. Show, using
vectors, that AFCE is a parallelogram.
4. Let A, B, C, D be four non-coplanar points. Let M, N, O, P be mid-points of AB,
BC, CD, DA respectively. Show MNOP is a parallelogram.
5. ABCD is a quadrilateral. M, N are mid-points of diagonals BD and AC
respectively.
Show NM4CDCBADAB =+++
(Hint: - expressNM in four different ways and add)
A
B C
D E
Given ABC! with D,E mid-points of
AB and AC respectively.
BC2
1
)ACBA(2
1
AC2
1BA2
1
AEDADE
=
+=
+=
+=
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1.5 Geometric Applications of Vectors
Example
Let P, Q, R be three collinear points and O be any point not on the line.
Let Q divide PR in the ratio
! :1- !
For example if Q divides PR into the ratio 5
2:5
3 then PQ = 5
3 PR (i.e.
! =3
5)and hence
.OP5
2OR5
3OQ +=
Note that the scalar 5
3multiplies OR (notOP as one might at first think) i.e. there
is a cross-over effect of the scalars. A good way to think of this is to imagine
! as being
nearly equal to 1 e.g. let
! =7
8. Q is then close to R.
P Q R
O
i.e.
PQ
PR= ! . This means that
PQ = !PR
Note : OQ = OP+ PQ
= OP+ !PR
= OP+ !(OR "OP)
= !OR + (1" !) OP
P Q R
O
8
1
8
7
i.e. OQ is “nearly equal” to OR hence
OQ = OR8
7OP8
1+
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Note that the converse is also true; namely that if OQ = ORnOPm + where m + n = 1
then P, Q, R are collinear. Furthermore if m or n is negative the converse statement is still
true. For example, if OQ = OR2OP3 ! then this can be easily rearranged to
OR3
2OQ3
1OP += . It is best to do this because this rearrangement avoids all difficulties
related to external division and immediately tells you that P is the point which is between
the other two.
Exercise 1.5
1. If AB5
3DA5
2AC =+ find out which three points are collinear and which point is
between the other two.
2.
3. 3OP! "!!
= mOR! "!!
! 2OQ! "!!. Find m, a scalar, so that P, Q, R are collinear. Determine
which point is between the other two.
In particular if Q is the mid-point of PR then OQ = OR2
1OP2
1+
X Z
O
Y 7
4
7
3
Write OY in terms of OX and OZ
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4.
Express in terms of a and b .
i) BD ii) DE Make a geometric deduction concerning DE.
5. ORnOQmOA and OR2
1OA2OP +=+=
P, Q, R are collinear. Find m + n.
6. If OR6OQ5OP +!= explain why P, Q, R are collinear and find the ratio into
which R divides PQ.
7. Show that if G is the centroid of triangle ABC and O is any point that
OC3
1OB3
1OA3
1OG ++=
(Hint: medians of a triangle intersect at a point dividing medians in the ratio
)3
1:3
2
8. Convince yourself that if O,P,Q, R are four points such that P, Q, R are collinear,
then OORpOQnOPm =++ where m + n + p = 0.
A
F
B
D
E C
F,D are mid-points of AB, CF
respectively. CE: EB = .4
3:4
1
Let aAC = and bAB =
22
9.
Hint: - G is intersection of four line segments joining vertices to centroids of opposite
faces but it turns out to be better to use the fact that G is also the intersection of the three
line segments joining mid-points of opposite edges.
You might like to show the truth of the hint using vector methods.
10. Let ABCD be a square whose centroid is G. Let O be any external point. Show
that
OG! "!!
=1
4(OA! "!!
+OB! "!!
+OC! "!!
+OD! "!!)
11. i) Show that in any triangle ABC
OGCGBGA =++ where G is the centroid.
ii) Show that in any tetrahedron ABCD
OGDGCGBGA =+++ where G is the centroid.
12. Investigate whether )OA....OAOA(n
1OG n21 ++= where A1, A2, A3… An is a
regular n-gon whose centroid is G and where O is any point. Does it make any
difference if the n-gon is not regular?
D
C
B A
G
Let ABCD be a tetrahedron whose centroid
is G. If O is any point show that
OD4
1OC4
1OB4
1OA4
1OG +++=
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Exercise 1.5 Answers
1. B – C – D
2. OZ7
4OX7
3OY +=
3. m = 5. P – R – Q
4. i) b4
3a2
1! ii) a
4
1 . DE is parallel to AC.
5. m + n = 4
1
6. Because 6
1:
6
5 .OQ
6
5OP
6
1OR +=
12. i) True ii) Yes, conjecture is now false. (You might like to think about the
possibility of an irregular n-gon in n – 1 space having the required property.)
1.6 More Geometric Applications with Vectors
Example
In the diagram below BD and CE are drawn so that
A
B C
D E
F
6
5:6
1DA:CD =
and
BE: EA = 5
3:5
2
Find BF: FD and CF : FE 6
1
6
5
5
2
5
3
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Method
Let b.AC and aAB == Join AF.
Let BF: FD = m:1 – m and CF: FE = n: 1 – n
Since B – F – D are collinear and using A as an external point
am)1()b6
5m(AF !+=
Similarly since E- F- C are collinear
)a5
3(nbn)-(1AF +=
In equations and we equate the scalars of b and a separately then we would have a
solution for m and n. Later on we shall see that this is the only solution.
i.e. m6
5 = 1 – n (equating scalars of b in and )
and 1 – m = n5
3 (equating scalars of a in and )
Solving these equations by elementary algebraic methods yields n = 3
1 and m = 5
4 .
i.e. F divides BD into ratio 5
1:5
4
F divides CD into ratio 3
2:3
1
25
Exercise 1.6
1.
Find i) BF:FD ii) CF: FE
2.
Find i) BF:FD and ii) CF:FE
3. 2 ORmOQOP =+ and P, Q, R are collinear.
Find m. Which point is between the other two and into what ratio does it divide
the line segment?
A
B C
D E
F
In triangle ABC, D and E are on AC
and AB respectively so that D is the
mid-point of AC and BE:EA = 3
2:3
1
A
B C
D E
F
In triangle ABC, D and E are on AC
and AB respectively so that AD: DC =
5
2:5
3 and AE:EB = 7
3:7
4
7
3
7
4
5
3
5
2
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4. If Ocpbnam =++ and m + n + p = 0 show that ac ! and cb ! are parallel
vectors.
5.
i) Prove that CF: FE = 3
1:3
2 and BF: FD =3
1:3
2
ii) Join AF and extend to point G on BC. Show that AG is a median and hence
deduce medians of a triangle are concurrent.
6. True or False? ( means “implies”)
i) ba = ba =
ii) ba + = ab +
iii) ba = ba =
iv) ba ! = ab !
v) ba + = ba !
vi) bma = a and b are parallel
vii) ( )BCAD and BCAB == ABCD is a rhombus
viii) m( ba ! ) = bmam !
ix) c)ba()cb(a +!=!!
A
B C
D E
F
In triangle ABC, BD and CE are
medians intersecting at F.
27
x) If a = 3 and b = 4 and ba + = 5 then a is perpendicular tob .
xi) If a = 3 and b = 4 then ba + < 7
xii) a > 0 for all a
xiii) bmbm = for all scalars m.
xiv) CDAB = BDAC =
7.
8. Three forces act on a body so that it is in equilibrium (i.e. resultant is zero vector).
The magnitude of the forces are 7, 3, and 8 units. Find the angle between lines of
directions of the smallest and largest force.
9.
A B
C D
E
F
P
ABCD is a square. E, F are mid-points
of AB and AD respectively.
Find DP: PE.
A
B C
D E
F
In the diagram, F is the mid-point of
EC and BE: EA = 3
2:3
1
Find BF: FD and AD: DC 3
2
3
1
2
1
2
1
28
10.
Exercise 1.6 Answers
1. i) 23
8:
23
15 ii) 23
9:
23
14
2. i) 2
1:2
1 ii) 4
1:4
3
3. m = 3, P – R – Q. R divides PQ into ratio of 3
2:3
1
6. i)F ii) T iii) T iv) T v) F vi) T vii) T viii) T ix) T x) T xi) T
xii) T xiii) False (m < 0)
7. 5
3:5
2
8. 120°
9. 3
1:3
2 and 4
1:4
3
10. y2x2 !
A B
C F
E D
x
y
ABCDEF is a regular hexagon.
xCD = and yDE =
Find AD in terms of y and x