Vectors Part Trois
Vector Geometry
Consider this parallelogram
Q
OR  b  PQ
P
R
a
b
O
OP  a  RQ
Opposite sides are Parallel
OQ  OP  PQ
 a +b
OQ  OR  RQ
 b+a
a +b  b + a
OQ is known as the resultant of a and b
Resultant of Two Vectors
Is the same, no matter which route is
followed
 Use this to find vectors in geometrical
figures

Example
S is the Midpoint of PQ.
Q
.
S
OS
OS  OP  ½ PQ
P
R
a
b
O
Work out the vector
= a + ½b
Alternatively
S is the Midpoint of PQ.
Q
.
S
P
OS
OS  OR  RQ  QS
R
a
b
O
Work out the vector
= b + a - ½b
= ½b + a
= a + ½b
Example
C
AC= p, AB = q
p
A
M
q
Find BC
M is the Midpoint of BC
B
BC = BA + AC
= -q + p
=p-q
Example
C
AC= p, AB = q
p
A
M
q
Find BM
BM = ½BC
= ½(p – q)
M is the Midpoint of BC
B
Example
C
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
AM = AB
B
+ ½BC
= q + ½(p – q)
= q +½p - ½q
= ½q +½p
= ½(q + p)
= ½(p + q)
Alternatively
C
AC= p, AB = q
p
A
M is the Midpoint of BC
M
q
Find AM
AM = AC
B
+ ½CB
= p + ½(q – p)
= p +½q - ½p
= ½p +½q
= ½(p + q)
Why??

We represent objects using
mainly linear primitives:




points
lines, segments
planes, polygons
Need to know how to
compute distances,
transformations…
Basic definitions
Points specify location in space (or in the
plane).
 Vectors have magnitude and direction (like
velocity).

Points  Vectors
Point + vector = point
vector + vector = vector
Parallelogram rule
point - point = vector
Final - Initial
B
A
B
A
point + point: not defined!!
Makes no
sense
Map points to vectors
If we have a coordinate system with
origin at point O
 We can define correspondence between
points and vectors:


P  P  P O
v  Ov
Position Vectors
A position vector for a point X (with respect to the origin 0) is the fixed vector OX
OX
O
X
Components of a Vector
Between Two Points
Given P1  x1, y1  and P2  x2 , y2  , then PP
1 2   x2  x1 , y2  y1 
Given P1  x1, y1, z1  and P2  x2 , y2 , z2  , then PP
1 2   x2  x1 , y2  y1 , z2  z1 
Proof:
P1
PP
 OP2
1 2  PO
1
 OP2  OP1
  x2 , y2 , z2    x1 , y1 , z1 
  x2  x1 , y2  y1 , z2  z1 
O
P2
Understanding
If P1  2, 3, 4  and P2  5, 2,6  then find PP
1 2 and P2 P1
PP
1 2   5, 2,6    2, 3, 4 
  7,1, 2 
P2 P1   2, 3, 4    5, 2,6 
  7, 1, 2 
Understanding
If u  2i  3 j  k and v  3i  5 j  5k find 4u  3v

 
4u  3v  4 2i  3 j  k  3 3i  5 j  5k
 4  2,3, 1  3  3, 5,5 
  8,12, 4    9, 15,15 
 17, 27, 19 

17    27    19 
2
 1379
2
2

Understanding
If u   u1, u2 , u3  and k  R, prove that ku  k u
ku  k  u1 , u2 , u3 
  ku1 , ku2 , ku3 

 ku1    ku2    ku3 
2
2
 k 2  u12  u22  u32 
 k2
k u
u
2
1
 u22  u32 
2
Understanding
Determine the coordinates of the point that divides AB in the ratio 5:3
where A is (2, -1, 4) and B is (3, 1, 7)
If AP:PB = m:n then
OP 
n
m
OA 
OB
mn
mn
3
5
 2, 1, 4  
 3,1,7 
53
53
3
5
  2, 1, 4    3,1,7 
8
8
 21 2 47 
 , , 
 8 8 8 
OP 
Understanding
Consider collinear points P, Q, and R. consider also a reference point O.
Write OQ as a linear combination of OP and OR if:
a) Q divides PR (internally) in the ratio 3:5
b) Q divides PR (externally) in the ratio -2:7
P
3
Q
5
Since:
3
PQ  PR
8
Then:
OQ  OP  PQ
R
3
OQ  OP  PR
8
3
OQ  OP  PO  OR
8
3
OQ  OP  OP  OR
8
3
3
OQ  OP  OP  OR
8
8
5
3
OQ  OP  OR
8
8


0


Understanding
Consider collinear points P, Q, and R. consider also a reference point O.
Write OQ as a linear combination of OP and OR if:
a) Q divides PR (internally) in the ratio 3:5
b) Q divides PR (externally) in the ratio -2:7
Since: PQ 
Then:
7
Q
-2
P
R
5
2
PR
5
OQ  OP  PQ
2
OQ  OP  PR
5
2
OQ  OP  PO  OR
5
2
OQ  OP  OP  OR
5
2
2
OQ  OP  OP  OR
5
5
7
2
OQ  OP  OR
5
5


0


Understanding
Determine the point Q that divides line segment PR in the ratio 5:8 if P is (1,5,2)
and R is (-4,1,2).
8
5
1,5, 2    4,1, 2 
13
13
 8 20 40 5 16 10 
  ,  ,  
 13 13 13 13 13 13 
OQ 
 12 45 26 
  , , 
 13 13 13 
Inner (dot) product

Defined for vectors: v  w  | v || w |  cosθ
w·cos(θ) is the scalar projection of w
on vector v, which we will call L
w
the dot product can be understood
geometrically as the product of the length
of this projection and the length of v.

L
v
 
v  w   L | v |
The dot product or scalar product allows us to determine a product the
magnitude of one vector by the magnitude of the component of the
other vector that’s parallel to the first. (The dot product was invented
specifically for this purpose.)
Dot product in coordinates
v  ( xv , yv )
w  ( xw , yw )
y
v  w  xv xw  yv yw
yw
yv
An interesting proof
w
wc  v
v
c vw
2
O
xw
xv
x

c  vw

2
2
 v  2v  w  w
2
 v  w  2 v w cos  
2
2
Proof of Geometric and
Coordinate Definitions Equality
2
u  u12 iˆ 2  u22 ˆj 2
2
 u12 12  02   u22  02  12 
 u12  u22
v  v12 iˆ 2  v22 ˆj 2
 v12 12  02   v22  02  12 
 v12  v22
c  v u
c   v1  u1    v2  u2 
2
2
 v12  2v1u1  u12  v22  2v2u2  u22
  v  v    u  u   2  v1u1  v2u2 
2
1
2
2
2
1
2
2
 u  v  2  v1u1  v2u2 
2
2
From the Cosine Law
2
c  u  v  2 u v cos  
2
2
2
therefore
u1v1  v2u2  u v cos  
Vector Properties
u v  v u
2
u u  u
 ku   v  k  u  v   u   kv 
u   v  w  u  v  u  w
Understanding
Given u  8.1 and v  4.7 and   126, find u  v
u  v  u v cos 126
u  v  8.1 4.7 cos 126
u  v  22.4
Understanding
Prove that for any vector
Proof:
u  u  u u cos  0
u
2
u
2
1
u
,
u u  u
2
Understanding
a) If u  4 and v  2, what values can u  v take?
b) Prove the Cauchy-Schwarz Inequality ; u  v  u v
u  v  u v cos  
u  v  u v cos  
u  v   4 2 cos  
u  v  u v cos  
u  v  8cos  
u  v  u v cos  
8  u  v  8
Cosine has the
range of -1 to 1
u v  u v
1
Understanding
Given that work is defined as W  f  s (the dot product of force, f, and
distance travelled, s. A crate on a ramp is hauled 8 m up the ramp under a
constant force of 20 N applied at an angle of 300 to the ramp. Determine the
work done.
W  f s
f
W  f s cos  
30
s
W   20N 8m cos 30
W  140J
Understanding
The angle between two vectors u and v is 1100, if the magnitude of vector u is
12, then determine: proj u
v
The projection of vector u onto
vector v
projv u  12cos 110 
u
 4.1
110
projv u
v
The Cross Product A  B
Some physical concepts (torque, angular momentum, magnetic force)
require that we multiply the magnitude of one vector by the magnitude of
the component of the other vector that is perpendicular to the first (recall
in the dot product we used the parallel component).
The cross product was invented for this specific purpose
The magnitude
(length) of the
cross product
θ
A  B  AB sin  
The Cross Product
Unlike the dot product, the cross product is a vector quantity.
For vectors x and y, we define the cross product of x and y as:
x  y  x y sin   nˆ
Where θ is the angle between x and y, and n̂ is a unit vector
perpendicular to both x and y such that x , y and n̂ form a right
handed triangle.
Take your right hand, and point all 4 fingers (not the
thumb) in the direction of the first vector x. Next, rotate
your arm or wrist so that you can curl your fingers in
the direction of y. Then extend your thumb, which is
perpendicular to both x and y. The thumb’s direction is
the direction the cross product will point.
Understanding
If u  8, v  5, and the angle between the vectors is 30, determine:
u
a) u  v
b) v  u
30
u  v  u v sin   nˆ
v
v  u  v u sin   nˆ
  8  5  sin  30  nˆ
  5  8  sin  30   nˆ 
 20nˆ
 20nˆ
u
30
Into board
v
u
Out off board
30
v
Understanding
Prove that for non-zero vectors u and v ,
u  v  0 if and only if u and v are collinear
First suppose u and v are collinear, then the angle, θ, between u and v is 00.
Since sin(00)=0, then:
u  v  u v sin  0  nˆ
0
Conversely suppose that u  v  0 then u v sin  0  nˆ  0 . Since u
and v are non-zero vectors, then sin(θ)=0, and hence θ=0 or θ=π.
Therefore u and v are collinear.
Properties
1. u   v  w   u  v  u  w
2.  u  v   w  u  w  v  w
3.  ku   v  k  u  v   u   kv 
Understanding
Determine the area of the parallelogram below, using the cross product.
8
300
13
area  u  v
 u v sin  
  8 13 sin  30 
 52
Understanding
State whether each expression has meaning. If not, explain why, if so, state
whether it is a vector or a scalar quantity.
a) u   v  w 
scalar
b) u   v  w
Non sense
c) u   v  w 
vector
d ) u  v   w
Non sense
e)  u  v    u  w
vector
f )  u  v    u  w
Non sense
g ) u  v   u  w
scalar
Understanding
A right-threaded bolt is tightened by applying a 50 N force to a 0.20 m
wrench as shown in the diagram. Find the moment of the force about
the centre of the bolt.
M  r F
70
M  r F
 r F sin   nˆ
  50  0.20  sin 110  nˆ
 9.4nˆ
The moment of the bolt is
9.4 Nm and is directed
down
Vector Product in Component
Form
Recall:
u  v  u v sin   nˆ
But how do we obtain the answer in component form?
The easiest technique is called the Sarrus’ Scheme A   ax , a y , az  , B   bx , by , bz 
iˆ
A  B  ax
bx
a b
y z
ˆj
ay
by
kˆ
az
bz
iˆ
ax
bx
ˆj
ay
by
kˆ
az
bz
 azby  iˆ   az bx  axbz  ˆj   axby  a y bx  kˆ
Understanding
Find  6, 1,3   2,5,4 
ˆj kˆ iˆ
ˆj kˆ
iˆ
A  B  6 1 3 6 1 3
2 5 4 2 5 4
 1 4  35  iˆ  3 2   6 4  ĵ   6 5   1 2 kˆ
 19, 30,28
Understanding
The points A 1,1,1 , B  2,0, 4  , C 1, 2, 3 , and D  4,3, 2  are vertices of a parallelogram.
Find the area of each of the following:
a) ABCD
b) triangle ABC
D
A
AB   2,0, 4   1,1,1   3, 1, 5
a. Area  AB  AD
AD   4,3, 2   1,1,1   3, 2,1
Area  AB  AD
  3, 1, 5    3, 2,1
  9, 12, 3
 92   12    3
2
 234
B
C
ˆj kˆ iˆ
ˆj kˆ
iˆ
A  B  3 1 5 3 1 5
3 2 1 3 2 1
2
 11   5 2 ,  53   31 ,  3 2   13  9, 12, 3
Understanding
The points A 1,1,1 , B  2,0, 4  , C 1, 2, 3 , and D  4,3, 2  are vertices of a parallelogram.
Find the area of each of the following:
a) ABCD
b) triangle ABC
b.
Area of triangle is one-half the area of the parallelogram
Area 
1
234
2
Understanding
If u  1, 2,1 , v   2,1, 1 , and w   1, 1,3 then evaluate  u  v   w
iˆ ˆj kˆ
u  v  1 2 1
2 1 1
iˆ ˆj kˆ
1 2 1
2 1 1
u  v    2  1  11 , 1 2   1 1 , 11   2  2  
 1,3,5
ˆj kˆ iˆ
ˆj kˆ
iˆ
A w  1 3 5 1 3 5
1 1 3 1 1 3
A  w    3 3   5 1 ,  5 1  1 3 , 1 1   3 1 
 14, 8, 2 
Understanding
Find a unit vector that is perpendicular to both: u  1, 2,3
v   2,3, 5 
We need to apply the cross product to determine a vector perpendicular to
both u and v. Then we divide by the length of this vector to make it a unit
vector
nˆ 
a b
a b
nˆ 

1, 2,3   2,3, 5 
1, 2,3   2,3, 5 
 19, 1, 7 
2
2
2
 19    1   7 
1
7 
 19

,
,

 411 411 411 
Useful Properties
i i  0
j k  i
j  i  k
j j 0
i  j k
k  j  i
k k  0
k i  j
i k   j
u  v  v  u
 u  v   w  u   v  w