Higher Unit 3
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Higher
Outcome 1
Vectors and Scalars
Properties of vectors
Adding / Sub of vectors
Multiplication by a Scalar
Unit Vector
3D Vectors
Properties 3D
Section formula
Scalar Product
Component Form
Position Vector
Collinearity
Section Formula
Exam Type Questions
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Angle between
vectors
Perpendicular
Properties of
Scalar Product
Vectors & Scalars
Outcome 1
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Higher
A vector is a quantity with
BOTH magnitude (length) and direction.
Examples :
Gravity
Velocity
Force
Vectors & Scalars
Outcome 1
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Higher
A scalar is a quantity that has
magnitude ONLY.
Examples :
Time
Speed
Mass
Vectors & Scalars
Outcome 1
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Higher
A vector is named using the letters at the end of
the directed line segment
or
using a lowercase bold / underlined letter
u
This vector is named AB
u
AB
or u
or u
Vectors & Scalars
Outcome 1
Higher
Also known as
column vector
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A vector may also be
represented in component form.
w
 4
CD  w   
 2
 x
AB  d   
 y
 2
FE  z   
 1
z
Magnitude of a Vector
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Higher
Outcome 1
A vector’s magnitude (length) is represented by
PQ
or
u
A vector’s magnitude is calculated
using Pythagoras Theorem.
u
PQ  u  a 2  b2
Vectors & Scalars
Outcome 1
Higher
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Calculate the magnitude of the vector.
w  a2  b2
w
 4
CD  w   
 2
w  42  22
w  20
w  4 5  2 5
Vectors & Scalars
Outcome 1
Higher
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Calculate the magnitude of the vector.
w  a2  b2
w  (4) 2  32
 4 
PQ   
 3
w  16  9
w 5
Equal Vectors
Outcome 1
Higher
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Vectors are equal only if they both have
the same magnitude ( length ) and direction.
Vectors are equal if they have
equal components.
For vectors
a
c
u    and v    , if u  v then a  c and b  d
b
d 
Equal Vectors
Outcome 1
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Higher
By working out the components of each of the
vectors determine which are equal.
a
b
c
d
g
e
f
h
Addition of Vectors
Outcome 1
Higher
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Any two vectors can be added in this way
b
Arrows
must be
nose to tail
 2
a 
 4
 6
 
 3 
8
a + b  
1
Addition of Vectors
Outcome 1
Higher
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Addition of vectors
3
 2
Let AB    and BC   
 4
 5 
B
Then AB + BC  AC
 3   2  5
  +    
 4   5   1
A
C
 5
So AC   
 1
Addition of Vectors
Higher
Outcome 1
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In general we have
For vectors u and v
a
c
If u    and v    then
b 
d 
a
c
a c 
u + v  +   = 

b 
d 
b  d 
Zero Vector
Outcome 1
Higher
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The zero vector
1 
If AB   
 2
 1 
then BA   
 2 
1 
 1 
 0
AB + BA    +   =  
 2
 2 
 0
 0
  is called the zero vector, written 0
 0
Negative Vector
Higher
Outcome 1
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Negative vector
BA is the negative of AB
For any vector u
u   u   u   u   0
a
 a 
If u    then  u   
b 
 b 
Subtraction of Vectors
Outcome 1
Higher
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Any two vectors can be subtracted in this way
Notice arrows
nose to nose
 6
u  
 3
 5
u - v  
 0
1
 
 3
v
Subtraction of Vectors
Outcome 1
Higher
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Subtraction of vectors
6
 2
Let a    and b   
5
 4
Notice arrows
nose to nose
a
a-b
Then a  b  a  (b)
b
 6   2   4 
      
 5   4  1 
6
 
5
 2  4
-    
 4  1
Subtraction of Vectors
Higher
Outcome 1
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In general we have
For vectors u and v
a
c
If u    and v    then
b 
d 
a
c
a c 
u  v     = 

b 
d 
b  d 
Multiplication by a Scalar
Higher
Outcome 1
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Multiplication by a scalar ( a number)
x
 kx 
If a vector v    then kv   
 y
 ky 
The vector kv is parallel to vector v ( different size )
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv
Multiplication by a Scalar
Outcome 1
Higher
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Multiplication by a scalar
Write down a vector
b
a
parallel to a
Write down a vector
parallel to b
Multiplication by a Scalar
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Higher
Outcome 1
Show that the two vectors are parallel.
 6
 18 
w    then z   
9
 27 
If z = kw then z is parallel to w
18 
 6
z    = 3 
 27 
9
z  3w
Multiplication by a Scalar
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Higher
Outcome 1
Alternative method.
 6
 18 
w    then z   
9
 27 
If w = kz then w is parallel to z
 6
1 18 
w   =  
3  27 
9
1
w  z
3
Unit Vectors
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Higher
Outcome 1
For ANY vector v there exists
a parallel vector u of magnitude 1 unit.
This is called the unit vector.
Unit Vectors
u
Outcome 1
Higher
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v
Find the components of the unit vector, u ,
parallel to vector v , if
3
v  
 4
v  3 4
2
v 5
2
So the unit vector is u
1 3
u   
5  4
3
5
 
4
 
5
A
Position Vectors
Outcome 1
Higher
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B
A is the point (3,4) and B is the point (5,2).
Write down the components of
3
OA   4 
 
OB
OB  OA 
5
  2
 
AB 
5  3   2
     
 2   4   2 
 2
 
 2 
Answers
the same !
A
Position Vectors
Higher
a
Outcome 1
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0
B
b
OA is called the position vector of the point A
relative to the origin O, written as a
OB is called the position vector of the point B
relative to the origin O, written as b
AB  AO  OB  a  b  b  a
A
Position Vectors
Higher
a
Outcome 1
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0
B
b
AB  b  a
where a and b are the position vectors of A and B.
Position Vectors
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Higher
Outcome 1
If P and Q have coordinates (4,8) and (2,3)
respectively, find the components of PQ
 2   4   2 
PQ  q  p         
 3   8   5 
Position Vectors
Outcome 1
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Higher
P
Graphically
P (4,8)
Q (2,3)
p
 2 
PQ =  
 5 
q-p
Q
q
O
Collinearity
Outcome 1
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Higher
Reminder from chapter 1
Points are said to be collinear if they lie
on the same straight line.
For vectors
If AB  kBC , where k is a scalar, then AB is parallel to BC.
If B is also a point common to both AB and BC then
A,B and C are collinear.
Collinearity
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Higher
Outcome 1
Prove that the points A(2,4), B(8,6) and C(11,7)
are collinear.
8   2  6 
AB  b  a         
 6  4  2
11  8   3 
BC  c  b         
 7   6  1 
 6   3
AB     2    2BC
 2  1 
Collinearity
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Higher
Outcome 1
Since AB  2BC , AB is parallel to BC.
B is a point common to both AB and BC
so A, B and C are collinear.
Section Formula
Outcome 1
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Higher
B
OS  OA  AS
3
1
OS  OA  AB
3
1
s  a  (b  a)
3
2
1
s  a b
3
3
1
2
S
s
A
a
b
O
General Section Formula
Outcome 1
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Higher
B
OP  OA  AP
m+n
m
OP  OA 
AB
mn
m
p a
(b  a )
mn
n
m
p
a
b
mn
mn
m
n
P
p
A
a
b
O
General Section Formula
Outcome 1
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Higher
Summarising we have
B
n
If p is a position vector
of the point P that divides
AB in the ratio m : n
then
n
m
p
a
b
mn
mn
A
m
P
General Section Formula
Higher
Outcome 1
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A and B have coordinates (-1,5)
and (4,10) respectively.
P
Find P if AP : PB is 3:2
2
3
p  a b
5
5
2  1 3  4 
p      
5  5  5 10 
2
3
A
 2   12 
 5  5  
   
 2  6
 10 
 5    2
   8 
 8
B
3D Coordinates
Outcome 1
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Higher
In the real world points in space can be located using a
3D coordinate system.
For example, air traffic controllers find the location a
plane by its height and grid reference.
z
y
O
(x, y, z)
x
3D Coordinates
Outcome 1
Higher
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Write down the coordinates for the 7 vertices
z
(0, 0, 2) F
(0,0, 0) G
E(0, 1, 2)
H
O
y
(6, 0, 2)B
6
A(6, 1, 2)
2
1
D(6, 1, 0)
C
(6, 0, 0)
What is the coordinates of the vertex H
so that it makes a cuboid shape.
x
H(0, 1, 0 )
3D Vectors
Outcome 1
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Higher
3D vectors are defined by 3 components.
For example, the velocity of an aircraft taking off can
be illustrated by the vector v.
z
(7, 3, 2)
2
v
y
2
3
O
3
7
7
x
3D Vectors
Outcome 1
Higher
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Any vector can be represented in terms of the
i , j and k
Where i, j and k are unit vectors
z
in the x, y and z directions.
 1
 
i=  0 
0
 
y
k
O
j
i
x
0
 
j =  1
0
 
0
 
k=0
 1
 
3D Vectors
Outcome 1
Higher
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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.
z
(7, 3, 2)
v
y
3
O
7
2
v = ( 7i+ 3j + 2k )
x
7
 
v =  3
 2
 
3D Vectors
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Higher
Outcome 1
Good News
All the rules for 2D vectors
apply in the same way for 3D.
Magnitude of a Vector
Outcome 1
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Higher
A vector’s magnitude (length) is represented by v
A 3D vector’s magnitude is calculated using
Pythagoras Theorem twice.
v  x2  y 2  z 2
v  3  2 1
2
2
v  14
2
z
v
y
1
2
O
3
x
Addition of Vectors
Higher
Outcome 1
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Addition of vectors
3
 2
 
 
Let u   4  and v   5 
1 
 2
 
 
Then u + v
 3   2  5
     
 4  +  5    1
1   2   3 
     
Addition of Vectors
Higher
Outcome 1
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In general we have
For vectors u and v
a
d 
 
If u   b  and v   e  then
f
c 
 
 
a
d 
a d 
 
 


u + v  b  + e  = b  e 
c 
f
c  f 
 
 


Negative Vector
Higher
Outcome 1
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Negative vector
BA is the negative of AB
For any vector u
u   u   u   u   0
a
 a 
 
 
If u   b  then  u   b 
c 
 c 
 
 
Subtraction of Vectors
Higher
Outcome 1
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Subtraction of vectors
6
 2
 
Let a   5  and b   4 
 2
3
 
 
Then a  b
6
 
 5
3
 
 2  4
-  4    1 
 2 1
   
Subtraction of Vectors
Higher
Outcome 1
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For vectors u and v
a
d 
 
 
If u   b  and v   e  then
c 
f
 
 
a
d 
a d 
 
 


u  v  b   e  = b  e 
c 
f
c  f 
 
 


Multiplication by a Scalar
Higher
Outcome 1
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Multiplication by a scalar ( a number)
x
 kx 
 
If a vector v   y  then kv   ky 
 kz 
z 
 
 
The vector kv is parallel to vector v ( different size )
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv
Multiplication by a Scalar
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Higher
Outcome 1
Show that the two vectors are parallel.
6 
 12 
 
w   9  then z   18 
 24 
 12 
 
 
If z = kw then z is parallel to w
12 
6 
 
 
z  18  = 2  9 
 24 
 12 
 
 
z  2w
Position Vectors
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Higher
Outcome 1
The position vector of a 3D point A is OA,
usually written as a
3
 
OA = a =  2 
1 
 
A (3,2,1)
z
a
y
1
2
O
3
x
Position Vectors
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Higher
Outcome 1
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 
     
General Section Formula
Outcome 1
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Higher
Summarising we have
B
n
If p is a position vector
of the point P that divides
AB in the ratio m : n
then
n
m
p
a
b
mn
mn
A
m
P
The scalar product
Higher
Outcome 1
Must be
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The scalar product is defined
being:
tail as
to tail
a
a  b  a b cos
0    180
0
θ
b
The Scalar Product
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Higher
Outcome 1
Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
a  b  a b cos
= 4  5cos 45
20 10  2  2
a b 

 10 2
2
2
o
The Scalar Product
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Higher
Outcome 1
Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
a  b  a b cos
= 4  5cos90
o
a  b  20  0  0
Important : If a and b are perpendicular then
a.b=0
Component Form Scalar Product
Outcome 1
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Higher
If
 a1 
 b1 
 
 
a   a2  and b   b2  then
a 
b 
 3
 3
a  b  a1b1  a2b2  a3b3
Angle between Vectors
Outcome 1
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Higher
To find the angle between two vectors we simply
use the scalar product formula rearranged
a b
cos =
a b
or
a1b1  a2b2  a3b3
cos =
ab
Angle between Vectors
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Higher
Outcome 1
Find the angle between the two vectors below.
p  3i +2j+5k and q  4i +j+3k
 3
 4
 
 
p   2  and q  1 
 5
3
 
 
a b
cos =
a b
p  32  22  52  38
q  42  12  32  26
Angle between Vectors
Outcome 1
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Higher
Find the angle between the two vectors below.
p  q  3 4  2 1  5 3  29
a b
cos =
=
a b
q  42  12  32  26
29
= 0.923
38  26
 = cos (0.923) = 22.7
-1
p  32  22  52  38
o
Perpendicular Vectors
Outcome 1
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Higher
Show that
for
a.b
=0
 3
1 
 
 
a   2  and b   2 
 1
7
 
 
a and b are perpendicular
a  b  a1b1  a2b2  a3b3
a  b  31  2  2  7  (1)
a  b  3  4  (-7)  0
Perpendicular Vectors
Outcome 1
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Higher
Given a  0 and b  0 and a  b  0
Then
a b
0
cos =

0
a b a b
 = cos (0) = 90
-1
o
If a . b = 0 then a and b are perpendicular
Properties of a Scalar Product
Higher
Outcome 1
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Two properties that you need to be aware of
a b  ba
a  (b  c)  a  b  a  c
Higher Maths
Vectors
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Vectors
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Vectors
Higher
The questions are in groups
General vector questions (15)
Points dividing lines in ratios
Collinear points (8)
Angles between vectors (5)
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Vectors
Higher
Vectors u and v are defined by u  3i  2 j and v  2i  3 j  4k
Determine whether or not u and v are perpendicular to each other.
Is Scalar product = 0
 3
 
u.v   2 
 0
 
u.v  3  2  2  3  0  4
u.v  0
2
 
 3 
4
 
 u.v  6   6   0
Hence vectors are perpendicular
Hint
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Vectors
Higher
For what value of t are the vectors
Put Scalar product = 0
 t 
 
u   2 
 3
 
 t 
 
u.v   2 
 3
 
u.v  2t   2 10  3t
Perpendicular  u.v = 0
and v
2
 
 10 
t 
 
perpendicular ?
2
 
10 
t 
 
 u.v  5t  20
 0  5t  20
 t4
Hint
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Vectors
Higher
VABCD is a pyramid with rectangular base ABCD.
The vectors AB, AD and AV are given by
AB  8i  2 j  2k
AD  2i  10 j  2k
AV  i  7 j  7k
Express CV in component form.
Ttriangle rule  ACV
AC  CV  AV
Triangle rule  ABC
AB  BC  AC also

 CV  AV  AB  AD

Re-arrange
 CV  AV  AC
BC  AD
 1   8   2 
     
 CV   7    2    10 
 7   2   2 
     
 5 
 
 CV   5 
7
 
 CV  9i  5 j  7k
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Vectors
Higher
The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.
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 p  2a  3b
Evaluate p.p and hence write down | p |.
i)
a  a  a a cos0  3  3 1  9
iii) a  b  a b cos 45  3  2 2 
b)
p  p   2a  3b   2a  3b 
 36  72  72  180
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8
1
6
2
 4a.a  12a.b  9b.b
Since p.p = p2
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bb  2 2 2 2
ii)
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p  180  6 5
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Hint
Vectors
Higher
Vectors p, q and r are defined by
p  i  j - k,
a)
Express p  q  2r in component form
b)
Calculate p.r
c)
Find |r|
q  i  4k ,
  i  j - k    i  4k   2  4i  3 j 
and
r  4i  3 j
 8i  5 j - 5k
a)
p  q  2r
b)
p.r   i  j - k  .  4i  3 j   p.r  1 4  1 (3)  (1)  0  p.r  1
c)
r 
42  (3) 2
 r  16  9 
r 5
Hint
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Vectors
Higher
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
S (a, 0, 0)
PS 2  49  (4  a)2  22  62
 a  43
T (b, 0, 0)
 49  (4  a)2  40
 9  (4  a)2
 a  7 or a  1
hence there are 2 points on the x axis that are 7 units from P
S (1, 0, 0)
and
i.e. S and T
T (7, 0, 0)
Hint
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Vectors
Higher
The position vectors of the points P and Q are
p = –i +3j+4k and q = 7 i – j + 5 k respectively.
a) Express PQ in component form.
b) Find the length of PQ.
a)
b)
PQ  q - p
 PQ
 7   1
   
  1 -  3 
5 4
   
PQ  82  (4)2  12
 PQ
 64  16  1
8
 
  4 
1
 
 81
 8i  4 j  k
9
Hint
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Vectors
Higher
PQR is an equilateral triangle of side 2 units.
P
PQ  a, PR  b, and QR  c
Evaluate a.(b + c) and hence identify
a
b
Diagram
two vectors which are perpendicular.
Q
a.(b  c)  a.b  a.c
a.b  a b cos60
60°
 a.b  2  2 
1
2
60°
60°
c
R
 a.b  2
NB for a.c vectors must point OUT of the vertex ( so angle is 120° )
 1

a.c

2

2

 
a.c  a c cos120
 2
Hence
a.(b  c)  0
 a.c   2
so, a is perpendicular to b + c
Hint
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Vectors
Higher
Calculate the length of the vector 2i – 3j + 3k
Length
 2  (3) 
2
2
 3
2
 493
 16
4
Hint
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Vectors
Higher
Find the value of k for which the vectors
Put Scalar product = 0
1 
 
 2
 1 
 
1
 
0  2 
 1
 
and
 4 


 3 
 k 1


are perpendicular
 4 


3


 k 1


0  4  6  (k  1)
 0  2  k 1
 k 3
Hint
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Vectors
Higher
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  b
D is the displacement
hence
d
BC
AD
 2   13 
  

  1   3 
 4   1 
  

 6   7 
   
  4   1
 2   3
   
BC
 13 


 3 
 1 


from A
d
 11


 2 
 3 


 D  11, 2, 3
Hint
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Vectors
If u
 3 
 
 3
3
 
Higher
and v
1
 
5
 1
 
write down the components of u + v and u – v
Hence show that u + v and u – v are perpendicular.
uv
 2 
 
 8 
2
 
 u  v  . u  v 
uv
 2   4 
   
  8    2 
2 4
   
 4 
 
  2 
4
 
look at scalar product
 u  v  . u  v 
 (2)  (4)  8  (2)  2  4
 8  16  8
0
Hence vectors are perpendicular
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Vectors
Higher
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
a)
 2  1
   
a.b   0    2 
 1  1 
   
a.b  2  0  1
a.b  1
2 0
   
  0    1
 1  1 
   
a.c  0  0  1
a.c  1
a.c
b)
a.b  a.c  0
b + c is perpendicular to a
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Vectors
Higher
A is the point ( –3, 2, 4 ) and B is ( –1, 3, 2 )
Find:
a) the components of AB
b) the length of AB
a)
b)
AB  b  a
AB
 1  3 
   
  3  2 
2 4
   
AB  22  12  (2) 2
AB
2
 
 1 
 2 
 
AB  4  1  4
AB  3
AB  9
Hint
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Vectors
Higher
In the square based pyramid,
all the eight edges are of length 3 units.
AV  p, AD  q, AB  r ,
Evaluate
p.(q + r)
Triangular faces are all equilateral
p.(q  r )  p.q  p.r
p.q  p q cos60
p.r  p r cos60
1
2
p.(q  r )  4  4
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1
2
Table of
p.q
1
 3 3
2
p.r
1
 3 3
2
p.q  4
p.q
1
2
1
4
2
p.(q  r )  9
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Vectors
Higher
You have completed all 15 questions in this section
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Vectors
Higher
Points dividing lines in ratios
Collinear Points
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Vectors
Higher
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
AB 1

AD 3
 3AB  AD
 3b  3a  d  a
 d
2
 1
 
 
 3  1  2  3 
1
2
 
 
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 3 b  a   d  a
 d  3b  2a
 d
8
 
 3
 1
 
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 D(8, 3, 1)
Hint
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Vectors
Higher
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.
Diagram
PQ 2

QR 1
 3q
P
1
R
 q  p  2r  2q
 PQ  2QR
 5   1
   
 2  2    1
 3   0 
   
2
Q
 3q  2r  p




1 9 
 q 3
3  6 
 Q(3, 1,  2)
Hint
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Vectors
a) Roadmakers look along the tops of a set of T-rods to ensure
Higher
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. a)
AB  b  a AB
 6
 2
 
 
 9  3  3
 3
1
 
 
AC
 14 
 2
 
 
  21  7  3 
7
1
 
 
AB and AC are scalar multiples, so are parallel. A is common. A, B, C are collinear
b)
Use scalar product
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AB.BD
 6  3 
   
  9  .  3 
 3  3 
   
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AB.BD  18  27  9  0
Hence, DB is perpendicular to AB
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Vectors
VABCD is a pyramid with rectangular base ABCD.
Higher
Relative to some appropriate axis,
VA represents – 7i – 13j – 11k
6i + 6j – 6k
AB represents
8i – 4j – 4k
AD represents
K divides BC in the ratio 1:3
Find VK in component form.
VA  AB  VB
VK  VB  KB
 VK
1
4
1
4
1
4
 VK  VA  AB  AD
 7   6 
8

   1 
  13    6    4 
 11  6  4  4 

  
 
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1
4
KB  CB  DA   AD
VK  KB  VB
 VK
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 1 


  8 
 18 


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Vectors
Higher
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 AB and AC
b) Find the co-ordinates of C and D.
a)
b)
AB  b  a
AB
6
 
 3 
 6 
 
C is a displacement of AC from A
similarly
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d
 5  2 
   
  0   1 
 0   2 
   




2
1
 
AC  AB   1 
3
 2 
c
3  2
   
  1   1 
 2   2 
   
 C (5, 0, 0)
 D(7, 1,  2)
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Vectors
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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  a
AB
1
 
  4 
2
 
AC
 3 
1


 
  12   3  4 
 6 
2


 
AB and AC are scalar multiples, so are parallel. A is common. A, B, C are collinear
Hint
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Vectors
Higher
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  a
AB and BC
AB 2

BC 3
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AB
4
2
 
 
  2   2 1 
 2 
 1
 
 
BC
6
2
 
 
  3   3 1 
 3 
 1
 
 
are scalar multiples, so are parallel. B is common. A, B, C are collinear
A
2
B
3
C
B divides AB in ratio 2 : 3
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Vectors
Higher
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  r
RS
and
RT
RS
 3
1
 
 
  3  3  1 
 6
 2
 
 
RT
 4
1
 
 
  4  4  1
8
 2
 
 
are scalar multiples, so are parallel. R is common. R, S, T are collinear
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Vectors
Higher
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Vectors
Higher
Angle between two vectors
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Vectors
Higher
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
a.a  a a  25
cos  
11
5 4
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a.(a  b)  36  a.a  a.b  36
 25  a.b  36
 11 
  cos  
 20 
  56.6
1
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 a.b  11
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Vectors
The diagram shows a square based pyramid of height 8 units.
Higher
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 DA and DB
c) Calculate the size of angle ADB.
a)
c)
B(6, 6, 0)
cos  
DA.DB
DA DB
cos  
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b)
3
 
DA   3 
 8 
 
64
82 82
DA.DB
DB
3
 
 3
 8 
 
3 3
   
  3  .  3   64
 8   8 
   
  38.7
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Vectors
A box in the shape of a cuboid designed with circles of different
Higher
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
 
Vectors to point
BA   5 
1
away from vertex
 
 4
 
BC   0 
 6 
 
BA  36  25  1  62
cos  
18
62 52
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BA.BC  24  0  6  18
BC  16  36  52
  71.5
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Vectors
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.
Higher
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.
a)
B(3, 2, 15)
b)
 15 
 
BC   2 
 7 
 
BA.BC  45 14  49  10
BA  9  49  49  107
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 3 
 
BA   7 
 7 
 
BC  225  4  49  278
10
cos  
278 107
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  93.3
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Vectors
Higher
A triangle ABC has vertices A(2, –1, 3), B(3, 6, 5) and C(6, 6, –2).
a) Find
AB
and
AC
b) Calculate the size of angle BAC.
c) Hence find the area of the triangle.
1
 4
 
AC  c  a   7 
a) AB  b  a   7 
 2
 5 
 
 
b)
AB  12  72  22  54
cos  
c)
43
 0.6168
54 90
Area of  ABC =
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AC  90
AB.AC  4  49 10  43
  cos1 0.6168  51.9
1
ab sin C
2
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
1
90 54 sin 51.9
2
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 BAC =

27.43
51.9
unit
2
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Vectors
Higher
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