higher unit 2 ef 1.4 higher unit 3 vectors and scalars properties of vectors adding / sub of...
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Higher Unit 2 EF 1.4
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Higher Unit 3Higher Unit 3
Vectors and ScalarsProperties of vectorsAdding / Sub of vectorsMultiplication by a Scalar
Position VectorCollinearity
Unit Vector
Exam Type Questions
Section Formula
3D Vectors
Properties 3DSection formulaScalar Product
Angle between vectorsPerpendicular
Component Form
Properties of Scalar Product
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Higher Unit 2 EF 1.4
Vectors & Scalars
A vector is a quantity with
BOTH magnitude (length) and direction.
Examples : Gravity
Velocity
Force
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Higher Unit 2 EF 1.4
Vectors & Scalars
A scalar is a quantity that has
magnitude ONLY.
Examples : Time
Speed
Mass
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Higher Unit 2 EF 1.4
Vectors & Scalars
A vector is named using the letters at the end of the directed line segment
or
using a lowercase bold / underlined letter
This vector is named
AB,,,,,,,,,,,,,,
AB,,,,,,,,,,,,,,
or u
or u
uu
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Higher Unit 2 EF 1.4
Vectors & Scalars
A vector may also be
represented in component form.
4
2CD w
,,,,,,,,,,,,,,
w
xAB d
y
,,,,,,,,,,,,,,
z
2
1FE z
,,,,,,,,,,,,,,
Also known as column vector
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Higher Unit 2 EF 1.4
Magnitude of a Vector
A vector’s magnitude (length) is represented by
or PQ u,,,,,,,,,,,,,,
A vector’s magnitude is calculated using Pythagoras
Theorem.
2 2PQ u a b ,,,,,,,,,,,,,, u
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Higher Unit 2 EF 1.4
Vectors & Scalars
Calculate the magnitude of the vector.
4
2CD w
,,,,,,,,,,,,,,w
2 2w a b
2 24 2w
20w
4 5 2 5w
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Higher Unit 2 EF 1.4
Vectors & Scalars
Calculate the magnitude of the vector.
4
3PQ
,,,,,,,,,,,,,,
2 2w a b
2 2( 4) 3w
16 9w
5w
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Higher Unit 2 EF 1.4
Equal Vectors
Vectors are equal only if they both have
the same magnitude ( length ) and direction.
Vectors are equal if they have
equal components.
For vectors
and , if then and a c
u v u v a c b db d
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Higher Unit 2 EF 1.4
Equal Vectors
By working out the components of each of the vectors determine which are equal.
a
b c
d
ef
g
h
a
g
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Higher Unit 2 EF 1.4
Addition of Vectors
Any two vectors can be added in this way
a
b
a + b
b
2
4
6
3
8
1
Arrows must be
nose to tail
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Higher Unit 2 EF 1.4
Addition of Vectors
Addition of vectors
3 2Let and
4 5AB BC
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
A
B
C
Then + AB BC AC,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
5So
1AC
,,,,,,,,,,,,,,
3 2 5 +
4 5 1
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Higher Unit 2 EF 1.4
Addition of Vectors
In general we have
If and then a c
u vb d
For vectors u and v
+ + = a c a c
u vb d b d
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Higher Unit 2 EF 1.4
Zero Vector
The zero vector
1 If AB
2
,,,,,,,,,,,,,,
1 1 0 AB + BA + =
2 2 0
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
0 is called the zero vector, written 0
0
1 then BA
2
,,,,,,,,,,,,,,
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Higher Unit 2 EF 1.4
Negative Vector
Negative vector
is the negative of BA AB,,,,,,,,,,,,,,,,,,,,,,,,,,,,
0 u u u u
For any vector u
If then a a
u ub b
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Higher Unit 2 EF 1.4
Subtraction of Vectors
Any two vectors can be subtracted in this way
u
v
u - v
6
3
5
0
Notice arrows nose
to nose
1
3
v
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Higher Unit 2 EF 1.4
Subtraction of Vectors
Subtraction of vectors
6 2Let and
5 4a b
Then ( )a b a b
6 2 4 -
5 4 1
6 2 4
5 4 1
a b
a - b
Notice arrows nose
to nose
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Higher Unit 2 EF 1.4
Subtraction of Vectors
In general we have
If and then a c
u vb d
For vectors u and v
= a c a c
u vb d b d
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Multiplication by a scalar ( a number)
If a vector then x kx
v kvy ky
The vector is parallel to vector ( different size )kv v
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Multiplication by a scalar
Write down a vector
parallel to a
Write down a vector
parallel to ba
b
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Show that the two vectors are parallel. 6 18
then z 9 27
w
If z = kw then z is parallel to w
18 6 = 3
27 9z
3z w
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Alternative method.
6 18 then z
9 27w
If w = kz then w is parallel to z
6 181 =
9 273w
1
3w z
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Higher Unit 2 EF 1.4
Unit Vectors
For ANY vector v there exists
a parallel vector u of magnitude 1 unit.
This is called the unit vector.
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Higher Unit 2 EF 1.4
Unit Vectors
Find the components of the unit vector, u , parallel to vector v , if
So the unit vector is u
3
4v
2 23 4v
5v 31
45u
3
5 4
5
v
u
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Higher Unit 2 EF 1.4
Position Vectors
A is the point (3,4) and B is the point (5,2).
Write down the components of
OA,,,,,,,,,,,,,,
OB OA ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
3
4
B
A
OB ,,,,,,,,,,,,,, 5
2
AB ,,,,,,,,,,,,,, 2
2
5 3 2
2 4 2
Answers
the same !
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Higher Unit 2 EF 1.4
Position Vectors
is called the position vector of the point A
relative to the origin , written as
OA,,,,,,,,,,,,,,
O a
is called the position vector of the point B
relative to the origin , written as
OB,,,,,,,,,,,,,,
O b
AB ,,,,,,,,,,,,,,
b a AO OB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
a b
B
A
a
b0
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Higher Unit 2 EF 1.4
Position Vectors
where and are the position vectors of and .
AB b a
a b A B
,,,,,,,,,,,,,,
B
A
a
b0
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Higher Unit 2 EF 1.4
Position Vectors
If P and Q have coordinates (4,8) and (2,3)
respectively, find the components of
2 4
3 8
PQ,,,,,,,,,,,,,,
2
5
PQ q p ,,,,,,,,,,,,,,
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Higher Unit 2 EF 1.4
P
Q
O
Position Vectors
Graphically
P (4,8)
Q (2,3)
= PQ,,,,,,,,,,,,,, 2
5
p
q
q - p
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Higher Unit 2 EF 1.4
Collinearity
Reminder !
If , where is a scalar, then is parallel to .AB kBC k AB BC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Points are said to be collinear if they lie on the same straight line.
For vectors
If B is also a point common to both AB and BC then
A,B and C are collinear.
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Higher Unit 2 EF 1.4
BC c b ,,,,,,,,,,,,,,
Prove that the points A(2,4), B(8,6) and C(11,7) are collinear.
8 2
6 4
6
2
AB b a ,,,,,,,,,,,,,,
11 8
7 6
3
1
6 32
2 1AB
,,,,,,,,,,,,,,2BC,,,,,,,,,,,,,,
Collinearity
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Higher Unit 2 EF 1.4
Collinearity
Since 2 , is parallel to .AB BC AB BC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
is a point common to both and
so , and are collinear.
B AB BC
A B C
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13I
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Higher Unit 2 EF 1.4
Section Formula
OS OA AS ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
1
3OS OA AB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
1 ( )
3s a b a
2 1
3 3s a b O
A
B
1
2
S
a
b
3
s
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Higher Unit 2 EF 1.4
General Section Formula
OP OA AP ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
mOP OA AB
m n
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
( )m
p a b am n
n mp a b
m n m n
O
A
B
m
n
P
a
b
m + n
p
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Higher Unit 2 EF 1.4
If p is a position vector
of the point P that divides
AB in the ratio m : n
then
General Section Formula
n mp a b
m n m n
A
B
m
n
P
Summarising we have
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Higher Unit 2 EF 1.4
General Section Formula
2 3
5 5p a b
A
B
3
2P
A and B have coordinates (-1,5)
and (4,10) respectively.
Find P if AP : PB is 3:2
1 42 3
5 105 5p
2 12 5 5
2 6
102
58
8
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13K
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Higher Unit 2 EF 1.4
O
3D Coordinates
In the real world points in space can be located using a 3D coordinate system.
x
y
z
For example, air traffic controllers find the location a plane by its height and grid reference.
(x, y, z)
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Higher Unit 2 EF 1.4
O
3D Coordinates
Write down the coordinates for the 7 vertices
x
yzA
B
C
D
E
F
G
What is the coordinates of the vertex H so that it makes a cuboid shape.
6
2
1
(6, 1, 2)
(6, 0, 2)
(6, 0, 0)
(6, 1, 0)
(0, 1, 2)
(0, 0, 2)
(0,0, 0)
H(0, 1, 0 )
H
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Higher Unit 2 EF 1.4
3D Vectors
3D vectors are defined by 3 components.
z
For example, the velocity of an aircraft taking off can be illustrated by the vector v.
(7, 3, 2)
v
3
7
2
O x
y2
7
3
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Higher Unit 2 EF 1.4
O
3D Vectors
z
j
i x
y
k
Any vector can be represented in terms of the i , j and k
Where i, j and k are unit vectors
in the x, y and z directions.
1
= 0
0
i
0
= 1
0
j
0
= 0
1
k
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Higher Unit 2 EF 1.4
3D Vectors
z
Any vector can be represented in terms of the i , j and k
Where i, j and k are unit vectors
in the x, y and z directions.
(7, 3, 2)
v
3
7
2
O x
y v = ( 7i+ 3j + 2k )
7
= 3
2
v
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Higher Unit 2 EF 1.4
3D Vectors
Good NewsAll the rules for 2D vectors apply in the same way for
3D.
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Higher Unit 2 EF 1.4
Magnitude of a Vector
A vector’s magnitude (length) is represented by
v
A 3D vector’s magnitude is calculated using Pythagoras Theorem twice.
2 2 2v x y z
z v
2
3
1
O x
y2 2 23 2 1v
14v
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Higher Unit 2 EF 1.4
Addition of Vectors
Addition of vectors
3 2
Let 4 and 5
1 2
u v
Then + u v
3 2 5
4 + 5 1
1 2 3
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Higher Unit 2 EF 1.4
Addition of Vectors
In general we have
If and then
a d
u b v e
c f
For vectors u and v
+ + =
a d a d
u v b e b e
c f c f
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Higher Unit 2 EF 1.4
Negative Vector
Negative vector
is the negative of BA AB,,,,,,,,,,,,,,,,,,,,,,,,,,,,
0 u u u u
For any vector u
If then
a a
u b u b
c c
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Higher Unit 2 EF 1.4
Subtraction of Vectors
Subtraction of vectors
6 2
Let 5 and 4
3 2
a b
Then a b
6 2 4
5 - 4 1
3 2 1
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Higher Unit 2 EF 1.4
Subtraction of Vectors
If and then
a d
u b v e
c f
For vectors u and v
=
a d a d
u v b e b e
c f c f
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Multiplication by a scalar ( a number)
If a vector then
x kx
v y kv ky
z kz
The vector is parallel to vector ( different size )kv v
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv
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Higher Unit 2 EF 1.4
Multiplication by a Scalar
Show that the two vectors are parallel.
6 12
9 then z 18
12 24
w
If z = kw then z is parallel to w12 6
18 = 2 9
24 12
z
2z w
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Higher Unit 2 EF 1.4
Position Vectors
The position vector of a 3D point A is ,
usually written as
OA,,,,,,,,,,,,,,
a
3
= = 2
1
OA
,,,,,,,,,,,,,,a
z a
2
3
1
O x
y
A (3,2,1)
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Higher Unit 2 EF 1.4
Position Vectors
If R is (2,-5,1) and S is (4,1,-3) then RS s r ,,,,,,,,,,,,,,
4 2 2
1 5 6
3 1 4
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13M
HHM Ex13L
HHM Ex13N (Q1 - 14)
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Higher Unit 2 EF 1.4
Collinearity in 3D
Reminder !
If , where is a scalar, then is parallel to .AB kBC k AB BC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Points are said to be collinear if they lie on the same straight line.
For vectors
If B is also a point common to both AB and BC then
A,B and C are collinear.
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13N (Q15 - 18)
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Higher Unit 2 EF 1.4
If p is a position vector
of the point P that divides
AB in the ratio m : n
then
General Section Formula
n mp a b
m n m n
A
B
m
n
P
Summarising we have
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13N (Q19-24)
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Higher Unit 2 EF 1.4
The scalar product
a
b
The scalar product is defined as being:
θ
cos a b a b
0 0 180
Must be
tail to tail
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Higher Unit 2 EF 1.4
The Scalar Product
Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
cos a b a b = 4 5cos 45 o
20 10 2 2 10 2
2 2a b
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Higher Unit 2 EF 1.4
The Scalar Product
Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
cos a b a b = 4 5cos90 o
20 0 0a b
Important : If a and b are perpendicular then
a . b = 0
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13O
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13O
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Higher Unit 2 EF 1.4
Component Form Scalar Product
If
1 1 2 2 3 3 a b a b a b a b
1 1
2 2
3 3
and
a b
a a b b
a b
then
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Higher Unit 2 EF 1.4
Angle between Vectors
To find the angle between two vectors we simply use the scalar product formula
rearranged
cos = a b
a b 1 1 2 2 3 3 cos =
a b a b a b
a b
or
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Higher Unit 2 EF 1.4
Angle between Vectors
Find the angle between the two vectors below.
cos = a b
a b
3 4
2 and 1
5 3
p q
2 2 23 2 5 38p 2 2 24 1 3 26q
3 +2j+5k and 4 + j+3kp i q i
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Higher Unit 2 EF 1.4
Angle between Vectors
Find the angle between the two vectors below.
cos = a b
a b
2 2 23 2 5 38p
2 2 24 1 3 26q 3 4 2 1 5 3 29p q
29 =
38 26 = 0.923
-1 o = cos (0.923) = 22.7
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Higher Unit 2 EF 1.4
Show that for
1 1 2 2 3 3 a b a b a b a b
3 1
2 and 2
1 7
a b
a and b are perpendicular
3 1 2 2 7 ( 1)a b
3 4 (-7) 0 a b
a . b =0
Perpendicular Vectors
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Higher Unit 2 EF 1.4
Perpendicular Vectors
0 cos = 0
a b
a b a b
Given a 0 and b 0 and a 0b
-1 o = cos (0) = 90
If a . b = 0 then a and b are perpendicular
Then
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13P
HHM Ex13Q
HHM Ex13R
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Higher Unit 2 EF 1.4
Properties of a Scalar Product
( )a b c a b a c
a ab b
Two properties that you need to be aware of
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Higher Unit 2 EF 1.4
Extra Practice
HHM Ex13U
Applications of Vectors
HHM Ex13S
HHM Ex13T
HHM Ex13T
Vectors
Strategies
Higher Maths
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Vectors Higher
Vectors
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Vectors Higher
The questions are in groups
Angles between vectors (5)
Points dividing lines in ratios
Collinear points (8)
General vector questions (15)
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General Vector Questions
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Vectors u and v are defined by and
Determine whether or not u and v are perpendicular to each other.
3 2 u i j 2 3 4 v i j k
Is Scalar product = 0
3 2
2 3
0 4
u.v
3 2 2 3 0 4 u.v 6 6 0 u.v
0u.v Hence vectors are perpendicular
Hint
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Vectors Higher
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For what value of t are the vectors and perpendicular ?2
3
t
u2
10
t
v
Put Scalar product = 0
2
2 10
3
t
t
u.v
2 2 10 3t t u.v 5 20t u.v
Perpendicular u.v = 0 0 5 20t
4t
Hint
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Vectors Higher
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VABCD is a pyramid with rectangular base ABCD.
The vectors are given by
Express in component form.
, andAB AD AV,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
8 2 2AB ,,,,,,,,,,,,,,
i j k 2 10 2AD ,,,,,,,,,,,,,,
i j k
7 7AV ,,,,,,,,,,,,,,
i j k
CV,,,,,,,,,,,,,,
AC CV AV ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
CV AV AC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
BC AD,,,,,,,,,,,,,,,,,,,,,,,,,,,,
AB BC AC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Ttriangle rule ACV Re-arrange
Triangle rule ABC also
CV AV AB AD ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1 8 2
7 2 10
7 2 2
CV
,,,,,,,,,,,,,,5
5
7
CV
,,,,,,,,,,,,,,
9 5 7CV ,,,,,,,,,,,,,,
i j k
Hint
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The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.
These vectors are inclined at an angle of 45° to each other.
a) Evaluate
i) a.a ii) b.b iii) a.b
b) Another vector p is defined by
Evaluate p.p and hence write down | p |.
2 3 p a b
cos0 a a a a 3 3 1 9 2 2 2 2 b b 8
cos 45 a b a b1
3 2 2 62
i) ii)
iii)
b) 2 3 2 3 p p a b a b 4 . 12 9 a a a.b b.b
36 72 72 180 Since p.p = p2 180 6 5 p
Hint
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Vectors p, q and r are defined by
a) Express in component form
b) Calculate p.r
c) Find |r|
- , 4 , and 4 3 p i j k q i k r i j
2 p q r
2 p q r - 4 2 4 3 i j k i k i j 8 5 -5 i j ka)
b) . - . 4 3 p r i j k i j . 1 4 1 ( 3) ( 1) 0 p r . 1 p r
c)2 24 ( 3) r 16 9 5 r r
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The diagram shows a point P with co-ordinates
(4, 2, 6) and two points S and T which lie on the x-axis.
If P is 7 units from S and 7 units from T,
find the co-ordinates of S and T.
Use distance formula ( , 0, 0)S a ( , 0, 0)T b
2 2 2 249 (4 ) 2 6PS a 249 (4 ) 40a 29 (4 )a
4 3a 7 1a or a
hence there are 2 points on the x axis that are 7 units from P
(1, 0, 0)S (7, 0, 0)T
i.e. S and T
and
The position vectors of the points P and Q are
p = –i +3j+4k and q = 7 i – j + 5 k respectively.
a) Express in component form.
b) Find the length of PQ.
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PQ,,,,,,,,,,,,,,
PQ ,,,,,,,,,,,,,,
q - p7 1
1 3
5 4
PQ
,,,,,,,,,,,,,,
-a)8
4
1
PQ
,,,,,,,,,,,,,,
8 4 i j k
2 2 28 ( 4) 1PQ ,,,,,,,,,,,,,,
b) 64 16 1 81 9
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PQR is an equilateral triangle of side 2 units.
Evaluate a.(b + c) and hence identify
two vectors which are perpendicular.
, , andPQ PR QR ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
a b c
( ) a. b c a.b a.c
cos60 a.b a b1
22 2 a.b 2 a.b
Diagram
P
RQ60° 60°
60°a b
c
NB for a.c vectors must point OUT of the vertex ( so angle is 120° )
cos120 a.c a c1
22 2
a.c 2 a.c
Hence ( ) 0 a. b c so, a is perpendicular to b + c
Table of
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Calculate the length of the vector 2i – 3j + 3k
22 22 ( 3) 3 Length 4 9 3
16
4
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Find the value of k for which the vectors and are perpendicular 1
2
1
4
3
1k
Put Scalar product = 0
1 4
2 3
1 1
0k
0 4 6 ( 1)k
3k
0 2 1k
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A is the point (2, –1, 4), B is (7, 1, 3) and C is (–6, 4, 2).
If ABCD is a parallelogram, find the co-ordinates of D.
AD BC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
c b6 7
4 1
2 3
BC
,,,,,,,,,,,,,, 13
3
1
BC
,,,,,,,,,,,,,,
D is the displacement AD,,,,,,,,,,,,,,
from A
hence2 13
1 3
4 1
d11
2
3
d 11, 2, 3D
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If and write down the components of u + v and u – v
Hence show that u + v and u – v are perpendicular.
3
3
3
u1
5
1
v
2
8
2
u v4
2
4
u v
2 4
8 2
2 4
.
u v u v
look at scalar product
.
( 2) ( 4) 8 ( 2) 2 4
u v u v
8 16 8 0
Hence vectors are perpendicular
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The vectors a, b and c are defined as follows:
a = 2i – k, b = i + 2j + k, c = –j + k
a) Evaluate a.b + a.c
b) From your answer to part (a), make a deduction about the vector b + c 2 1
0 2
1 1
a.ba) 2 0 1 a.b 1a.b
2 0
0 1
1 1
a.c
b)
0 0 1 a.c 1a.c 0 a.b a.c
b + c is perpendicular to a
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A is the point ( –3, 2, 4 ) and B is ( –1, 3, 2 )
Find:
a) the components of
b) the length of AB
AB,,,,,,,,,,,,,,
AB ,,,,,,,,,,,,,,
b aa)1 3
3 2
2 4
AB
,,,,,,,,,,,,,, 2
1
2
AB
,,,,,,,,,,,,,,
2 2 22 1 ( 2)AB b) 4 1 4AB
9AB 3AB
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In the square based pyramid,
all the eight edges are of length 3 units.
Evaluate p.(q + r)
, , ,AV AD AB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
p q r
Triangular faces are all equilateral
( ) p. q r p.q p.r
cos60 p.q p q1
23 3 p.q
1
24p.q
cos60 p.r p r1
23 3 p.r
1
24p.q
1 1
2 2( ) 4 4 p. q r ( ) 9 p. q r
Table of
Exact Values
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Points dividing lines in ratios
Collinear Points
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A and B are the points (-1, -3, 2)
and (2, -1, 1) respectively.
B and C are the points of trisection of AD.
That is, AB = BC = CD.
Find the coordinates of D 1
3
AB
AD
,,,,,,,,,,,,,,
,,,,,,,,,,,,,, 3AB AD ,,,,,,,,,,,,,,,,,,,,,,,,,,,, 3 b a d a
3 3 b a d a 3 2 d b a
2 1
1 3
1 2
3 2
d8
3
1
d (8, 3, 1)D
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The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1.
Find the co-ordinates of Q.
2
1
PQ
QR
,,,,,,,,,,,,,,
,,,,,,,,,,,,,, 2PQ QR ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
2 2 q p r q
3 2 q r p
5 1
2 1
3 0
3 2
q9
3
6
1
3
q (3, 1, 2)Q
Diagram P
QR
21
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a) Roadmakers look along the tops of a set of T-rods to ensure
that straight sections of road are being created.
Relative to suitable axes the top left corners of the T-rods are
the points A(–8, –10, –2), B(–2, –1, 1) and C(6, 11, 5).Determine whether or not the section of road ABC has been
built in a straight line.
b) A further T-rod is placed such that D has co-ordinates (1, –4, 4).
Show that DB is perpendicular to AB. AB ,,,,,,,,,,,,,,
b aa)6 2
9 3 3
3 1
AB
,,,,,,,,,,,,,, 14 2
21 7 3
7 1
AC
,,,,,,,,,,,,,,
andAB AC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
are scalar multiples, so are parallel. A is common. A, B, C are collinear
b) Use scalar product6 3
9 3
3 3
. .AB BD
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
. 18 27 9 0AB BD ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Hence, DB is perpendicular to AB
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VABCD is a pyramid with rectangular base ABCD.
Relative to some appropriate axis, represents – 7i – 13j – 11k
represents 6i + 6j – 6k
represents 8i – 4j – 4k
K divides BC in the ratio 1:3
Find in component form.
VA,,,,,,,,,,,,,,
AB,,,,,,,,,,,,,,
AD,,,,,,,,,,,,,,
VK,,,,,,,,,,,,,,
VA AB VB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
VK KB VB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1 1 1
4 4 4KB CB DA AD ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
VK VB KB ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, 1
4VK VA AB AD ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
7 6 81
13 6 44
11 6 4
VK
,,,,,,,,,,,,,, 1
8
18
VK
,,,,,,,,,,,,,,
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The line AB is divided into 3 equal parts by
the points C and D, as shown.
A and B have co-ordinates (3, –1, 2) and (9, 2, –4).
a) Find the components of and
b) Find the co-ordinates of C and D.
AB,,,,,,,,,,,,,,
AC,,,,,,,,,,,,,,
AB ,,,,,,,,,,,,,,
b a6
3
6
AB
,,,,,,,,,,,,,, 2
1
2
1
3AC AB
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
a)
b) C is a displacement of from AAC,,,,,,,,,,,,,, 3 2
1 1
2 2
c (5, 0, 0)C
similarly5 2
0 1
0 2
d (7, 1, 2)D
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Relative to a suitable set of axes, the tops of three chimneys have
co-ordinates given by A(1, 3, 2), B(2, –1, 4) and C(4, –9, 8).
Show that A, B and C are collinear
AB ,,,,,,,,,,,,,,
b a1
4
2
AB
,,,,,,,,,,,,,, 3 1
12 3 4
6 2
AC
,,,,,,,,,,,,,,
andAB AC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
are scalar multiples, so are parallel. A is common. A, B, C are collinear
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A is the point (2, –5, 6), B is (6, –3, 4) and C is (12, 0, 1).
Show that A, B and C are collinear
and determine the ratio in which B divides AC
AB ,,,,,,,,,,,,,,
b a4 2
2 2 1
2 1
AB
,,,,,,,,,,,,,, 6 2
3 3 1
3 1
BC
,,,,,,,,,,,,,,
andAB BC,,,,,,,,,,,,,,,,,,,,,,,,,,,,
are scalar multiples, so are parallel. B is common. A, B, C are collinear
2
3
AB
BC
,,,,,,,,,,,,,,
,,,,,,,,,,,,,,A
BC
23
B divides AB in ratio 2 : 3
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Relative to the top of a hill, three gliders
have positions given by
R(–1, –8, –2), S(2, –5, 4) and T(3, –4, 6).
Prove that R, S and T are collinear
RS ,,,,,,,,,,,,,,
s r3 1
3 3 1
6 2
RS
,,,,,,,,,,,,,, 4 1
4 4 1
8 2
RT
,,,,,,,,,,,,,,
andRS RT,,,,,,,,,,,,,,,,,,,,,,,,,,,,
are scalar multiples, so are parallel. R is common. R, S, T are collinear
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Angle between two vectors
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The diagram shows vectors a and b.
If |a| = 5, |b| = 4 and a.(a + b) = 36
Find the size of the acute angle
between a and b.
cos a.b
a b( ) 36 36 a. a b a.a a.b
25 a.a a a 25 36 a.b 11 a.b
11cos
5 4
1 11
cos20
56.6
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The diagram shows a square based pyramid of height 8 units.
Square OABC has a side length of 6 units.
The co-ordinates of A and D are (6, 0, 0) and (3, 3, 8).
C lies on the y-axis.
a) Write down the co-ordinates of B
b) Determine the components of
c) Calculate the size of angle ADB.
andDA DB,,,,,,,,,,,,,,,,,,,,,,,,,,,,
a) B(6, 6, 0) b)3
3
8
DA
,,,,,,,,,,,,,, 3
3
8
DB
,,,,,,,,,,,,,,
c).
cosDADB
DA DB
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,,,,,,,,,3 3
3 . 3 64
8 8
.DADB
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
64cos
82 82 38.7
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A box in the shape of a cuboid designed with circles of different
sizes on each face.
The diagram shows three of the circles, where the origin represents
one of the corners of the cuboid.
The centres of the circles are A(6, 0, 7), B(0, 5, 6) and C(4, 5, 0)
Find the size of angle ABC 6
5
1
BA
,,,,,,,,,,,,,,4
0
6
BC
,,,,,,,,,,,,,,Vectors to point
away from vertex . 24 0 6 18BA BC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
36 25 1 62BA ,,,,,,,,,,,,,,
16 36 52BC ,,,,,,,,,,,,,,
18cos
62 52 71.5
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A cuboid measuring 11cm by 5 cm by 7 cm is placed centrally on top of another cuboid measuring 17 cm by 9 cm by 8 cm.
Co-ordinate axes are taken as shown.
a) The point A has co-ordinates (0, 9, 8) and C has
co-ordinates (17, 0, 8).
Write down the co-ordinates of B
b) Calculate the size of angle ABC. (3, 2, 15)Ba) b)
3
7
7
BA
,,,,,,,,,,,,,,15
2
7
BC
,,,,,,,,,,,,,,
. 45 14 49 10BA BC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
225 4 49 278BC ,,,,,,,,,,,,,,
9 49 49 107BA ,,,,,,,,,,,,,, 10
cos278 107
93.3
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A triangle ABC has vertices A(2, –1, 3), B(3, 6, 5) and C(6, 6, –2).
a) Find and
b) Calculate the size of angle BAC.
c) Hence find the area of the triangle.
AB,,,,,,,,,,,,,,
AC,,,,,,,,,,,,,,
1
7
2
AB
,,,,,,,,,,,,,,b aa)
4
7
5
AC
,,,,,,,,,,,,,,c a
b) 2 2 21 7 2 54AB ,,,,,,,,,,,,,,
90AC ,,,,,,,,,,,,,,
. 4 49 10 43AB AC ,,,,,,,,,,,,,,,,,,,,,,,,,,,,
43cos 0.6168
54 90 1cos 0.6168 51.9 51.9 BAC =
c) Area of ABC = 1
2sinab C
190 54
2sin 51.9 2
unit27.43
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