1. Vectors
Revision

2. ( ) Notation 8 5 B The vector AB A … as a column vector 8 across, 5 up

3. Unit Vectors (1) i
i is the unit vector in the x-direction
i = [ ] 1 j
j is the unit vector in the y-direction
j = [ ] 1
All vectors can be expressed as a linear combination of these 2 vectors
[ ] =
e.g. displacement = 25
43.3
[ ] 1

4. [ ] = 25 + 43.3 [ ] = 25 i + 43.3 j Unit Vectors (2) i = [ ] j = [ ]1
j = [ ] 1
All vectors can be expressed as a linear combination of these 2 vectors
[ ] =
e.g. displacement = 25
43.3
[ ] 1
= 25 i j
This is the standard way displacement vectors are presented

5. a = -10 i + 15 j a a Magnitude is notation for magnitude = 18.0
By Pythagoras,
the magnitude = √( )
= √325 = 18.0 (1 d.p.) a a
= 18.0

6. |a| = √(r2 + s2 + t2) Magnitude of a 3D Vector (General) z a y o x t s
“the magnitude is the square root
of the sum of the squares of
the 3 components.” x
[Pythagoras in 3D]

7. Magnitude of a 3D Vector - examplez
By Pythagoras,
OB = √( ) A
√( ) a 10 y
By Pythagoras,
OA2 = AB2 + OB2
AB2 = 102
OB2 = ( )
OA2 =
OA = √( ) 3 o B 4 3 x
|a| = OA
= 11.2

8. Linear Combinations
3a + 6b
2a - b

9. + = (6i + 4j) + (i - 5j) = 7i - j Linear Combination Example 64 1 7 -5
A car drives from A to B with displacement = 6i + 4j
Then he drives from B to C with displacement = i - 5j
6i + 4j B A C
i - 5j 1 -5 64 = 7 -1
The resultant displacement is from A to C
(6i + 4j) + (i - 5j) = 7i - j
The magnitude of the displacement
= √(72 + (-1)2) = √50 = 7.1m (1 d.p.)

10. 3D Vectors
Follow all the same rules of 2D Vectors!

11. Position Vectors R A position vector is r fixed to the origin O
A free vector has magnitude and direction, but is not fixed to the origin s

12. = b - a Distance between 2 points - general case B A
A and B have position vectors x y z B b
What is the distance between them? A a AB
= b - a AB
|AB| = √(x2-x1)2 + (y2-y1)2 + (z2-z1)2

13. a + = b = b - a Distance between 2 points B A |AB| = √(52 + 52 + 82) z
A and B have position vectors x y z b B
What is the distance between them? AB a A
a = b AB AB
= b - a
|AB| = √( )
= √114 = 10.7 AB

14. Parallel Vectors (1) -20 i + 30 j -10 i + 15 j -2 i + 3 j 2 a a
Vectors with a scaler applied are parallel
i.e. with a different magnitude but same direction

15. Parallel Vectors (2) 2c - 3d 4c - 6d 2(2c - 3d) c - 1.5d 0.5(2c - 3d)
With linear combinations,
look to see if you can rearrange them
2c - 3d
Is parallel to ….
4c - 6d
2(2c - 3d)
c d
0.5(2c - 3d)
1000c d
500(2c - 3d)
-6c + 9d
-3(2c - 3d)

16. Parallel 3D Vectors =3 c =1/2 c =-2 c Look for scalers of each other
Is parallel to ….
=3 c
=1/2 c
=-2 c

17. Scaler Product ab = a1 b1+ a2 b2+ a3 b3 cos θ = ab
The angle θ between vectors
is given as:
cos θ = ab
|a||b|

18. Lets see what you are given

19. Parallel Vectors Occur …when cos θ = 1 … so θ = cos-1(1) = 0 degrees
cos θ = ab
|a||b|
Occur
…when cos θ = 1
… so θ = cos-1(1) = 0 degrees
i.e. the lines are Parallel

20. Perpendicular Vectors
cos θ = a.b
|a||b|
ab
If = 0,
…then cos θ = 0
… so θ = cos-1(0) = 90 degrees
i.e. the lines are Perpendicular
So, if ab = 0 then the lines are perpendicular

21. Given: a = 3i + 4j and b = i - 3j Example (2D) - angle between vectors
The scaler product is written as ...
ab
ab = (3 x 1) + (4 x -3)
= = -9
The i components
The j components
|a| = √( ) = √25 = 5
cos θ= ab
|a||b|
|b| = √(12 + (-3)2) = √10
cos θ = =
5√10
θ = cos-1(-0.569) = 124.7o

22. Angle between 3D Vectors
The scaler product is written as ...
ab
ab = (2 x 1) + (3 x -2) + (7 x 5)
= = 31
cos θ = ab
|a||b|
|a| = √( ) = √62
|b| = √(12 + (-2)2 + 52) = √30
cos θ = = 0.719
√62√30
θ = cos-1(0.719) = 44.0o

23. b = pi + qj a = xi + yj E.g. a + tb Vector Equation of a line (2D) y x
A line can be identified by a linear combination
of a position vector and a free vector y
b = pi + qj A
parallel vector to line
a = xi + yj o x
t is a scaler
- it can be any number,
since we only need a parallel vector
E.g. a + tb
= (xi + yj) + t(pi + qj)

24. [ ] [ ] y = 3x - 1 r is the position vector of r Often written …….
Equations of straight lines
y = 3x - 1
….. is a Cartesian Equation
of a straight line x y
[ ]
= t 1 3 2
….. is a Vector Equation
of a straight line
Often written …….
= t 1 3
[ ] 2 r
r is the position vector of
any point R on the line
Any
point
Direction

25. [ ] [ ] [ ] r y = 4x + 3 y - 3 = 4x = t = + t = + t
Convert this Cartesian equation into a Vector equation
Easiest Method
y = 4x + 3
y - 3 = 4x = t
Write:
t = 4x
t = y - 3
x = /4 t
y = t x y
[ ]
= t 3
1/4 1
[ ]
Can replace
with a parallel
vector
= t 1 4
[ ] 3 r

26. Equations of form y-b=m(x-a)
Summary
A line can be identified
by a linear combination
of a position vector
and a free [direction] vector
= t 1 m
[ ] a b r
the
direction
vector
Any
point
Equations of form y-b=m(x-a)
Line goes through (a,b) with gradient m

27. r a + = b = b - a Equation given 2 points B A z b AB a AB x AB ABy z
A and B have position vectors b B AB
What is the vector equation of the
line passing through them a A
a = b AB AB
= b - a
= t r
The equation of the line AB
This gives the direction vector

28. r r = (1+3t)i + (2+2t)j + (-4+10t) = + t 13i + 10j + 36k
Points on line
A line has the vector equation
(3i + 2j + 10k)
(i + 2j - 4k)
= t r
13i + 10j + 36k
Show that the point with position vector
is on the line
Need to find a value of t that fit all points
1. Rearrange equation
A single value of t exists for all points.
Therefore the point is on the line
r = (1+3t)i + (2+2t)j + (-4+10t)
13i + 10j + 36k
2. Has to equal
i coefficients
j coefficients
k coefficients
3. Equate
coefficients
1+3t = 13
2+2t = 10
-4+10t = 36
t = 4
t = 4
t = 4

29. Example = + A line has the equation ...
The point P (11, a, b) is on the line.
a and b are constants. =
Find a and b
P has position vector
Therefore
for some value of
Substitute = 2
= a
= b
Equating the x values:
= 11
-3+10 = a
-4+16 = b
5 = 10
a = 7
b = 12
= 2

30. Intersect of 2D lines in vector form - 1
For example x y
[ ] 1 3
[ ] 2
[ ] x y
[ ]
[ ]
[ ] 6 -2 -1 4
= s
and
= t
If the lines intersect, there must be values of
s and t that give the position vector of the point of intersection.
x part: s = 6 - t
y part: s = t x y
[ ] = 1 3 2 x y
[ ] = 1 3
Subtract x from y : 2 = t
5t = 10
t = 2
[ ] x y = 4 6
position vector of the point of intersection
Substitute: s = 6 - 2
2s = 3
s = 1.5

31. Example The lines r and s have the equations ...
Show they intersect and
find the point of intersection
and
If the lines intersect, there must be values of
and  that give the position vector of the point of intersection.
x : = 4 +2
y : = 7 - 2
z : = 2 +3
x+y : = 11
3 = 6
= 2
Check the values in the 3rd equation
z : x2 = 2 +3x3
11 = 11
Satisfied!
Substitute x : x2 = 4 +2
= 3
Hence a point exists
common to both lines

32. Example (continued) The lines r and s have the equations ...
Show they intersect and
find the point of intersection
and
If the lines intersect, there must be values of
and that give the position vector of the point of intersection.
= 2
= 3
Values satisfy all equations
Hence a point exists
common to both lines
Substitute
The 2 lines intersect at (10, 1, 11)

33. Skew lines z b a x In 3D lines can be that are not parallel and
do not intersect are called skew lines x y z
Don’t meet b a

34. Skew Example r = (2i + 3j + 6k) + t(4i - j + 6k)
2 lines have the equations ...
r = (2i + 3j + 6k) + t(4i - j + 6k)
Show they are skew
and
r = (4i + 7j + 8k) + s(2i - 2j + k)
If the lines intersect, there must be values of
s and t that give the position vector of the point of intersection.
i : t = 4 +2s
Check the values in the 3rd equation
j : 3 - t = 7 - 2s
k : x2 = 8 + 3
18 = 11
k : t = 8 + s
Not Satisfied!
i+j : t = 11
3t = 6
t = 2
Direction vectors:
(4i - j + 6k) and (2i - 2j + k)
are not parallel
Therefore lines are skew
Substitute i : x2 = 4 +2s
s = 3

35. Angles Between Skew Lines
Skew lines do not meet!
However you can work out
angle between them by
‘transposing’ one to the other -
keeping the direction the same.
E.g. the angle between
and
You just need to look at the angle
between the direction vectors:
and

36. cos θ = ab Skew Angle Example |a||b|
2 lines have the equations … find the angle between them.
r = (2i + 3j + 6k) + t(4i - j + 6k)
and
r = (4i + 7j + 8k) + s(2i - 2j + k)
cos θ = ab
|a||b|
Direction Vectors are:
a = 4i - j + 6k
b = 2i - 2j + k
ab
= 4x2 + -1x-2 + 6x1 = 16
|a|
= √( ) = √53
|b|
= √( ) = √9 = 3
cos θ = = 0.733
θ = cos-1(0.733)
= 42.9o
3√53

37. Angles Between Skew Lines - you find the angle!
and
The direction vectors:
and
ab
= 4x2 + -1x-2 + 3x3 = 19
|a|
= √( ) = √26
|b|
= √( ) = √17
cos θ = = 0.904
√26√17
θ = cos-1(0.904)
= 25.3o