It’s time for Chapter 6…
Section 6.1a
Vectors in the Plane
Scalar – a single real number, has magnitude
 Examples???
Vector – has magnitude and direction
 Examples???
Vectors are denoted by lowercase boldface
letters (e.g., u, v, w), and can be defined
by directed line segments
Our first example:
Notation!
Consider directed line segment PQ, with P(– 4, 2) and Q(1, 1):
Initial
Point
P
y
We can find the length, or magnitude
of the segment, |PQ|, using the
distance formula:
Terminal
Q Point
x
PQ 
 4  1   2  1
2
PQ  25  1
|PQ| = 26
2
Another example:
Let u be the vector represented by the directed line segment
from R(– 4, 2) to S(–1, 6) and v the vector represented by the
directed line segment from O(0, 0) to P(3, 4). Prove that u = v.
How ‘bout a graph?
Lengths?
Slopes?
The two vectors are equivalent  a vector is defined by its
direction and magnitude, not by its location
Definition: Component Form of a Vector
If v is a vector in the plane equal to the vector with initial
point (0, 0) and terminal point (v1 , v 2), then the component
form of v is:
v = v1 , v2
Components
This vector is called the
position vector of the
point (v1 , v 2 )
The vector 0, 0 with length 0 and no direction
is called the zero vector, and is denoted 0
We need some more practice examples…
Find the component form and magnitude of the vector
v = PQ, where P(–3, 4) and Q(–5, 2).
Let’s see a graph:
v = –2, –2
|v|= 2 2
We need some more practice examples…
Find the component form and magnitude of the vector
u = FG, where G(1, –2) and F(7, 1).
Let’s see a graph:
u = – 6, –3
|u|= 3 5
Vector Operations
Sum of vectors:
u+v=
u1 , u2  v1 , v2  u1  v1,u2  v2
Would this work for subtraction as well???
Product of a scalar and a vector:
ku =
k u1 , u2  ku1 , ku2
Would this work for a product of two vectors???
Let’s see these graphically
y
(– 2, 8)
(– 6, 6)
What is u + v?
u+v
v
u
The parallelogram law…
(4, 2)
x
Let’s see these graphically
y
If k = 3, what is ku?
(6, 9)
3u
(2, 3)
u
x
Practice Problems
Let u = –2, 8 and v = 3, – 5 . Find the
component form of:
u + 2v
4, –2
Unit Vectors
Unit vector – a vector u with length |u| = 1
Unit vector in the direction of v:
v
1 v
u=
=
|v| |v|
Unit Vectors
Standard unit vectors: i = 1, 0
and j = 0, 1
Any vector w can be written as an expression
using the standard unit vectors:
In a graph?
w = a, b
= a, 0 + 0, b
= a 1, 0 + b 0, 1
= ai + bj
Horizontal and vertical components of w
How ‘bout some examples?
Find a unit vector in the direction of u = 7, 1 , and
verify that it has length 1.
u  7  1 = 50  5 2
2
Unit Vector:
2
1
1
7
1
u
7,1 
,
u
5 2
5 2 5 2
Magnitude:
2
2
 7   1   49  1  50  1

 

50 50
50
5 2  5 2 
Whiteboard Problem…
Let P = (–2,2), Q = (3,4), R = (–2,5), and S = (2,–8).
Find the component form and magnitude of
PS  3PQ
PS  2  2, 8  2  4, 10
PQ  3  2, 4  2  5, 2
PS  3PQ  4, 10  3 5, 2  11, 16
PS  3PQ 
 11   16 
2
2
 377
Whiteboard Problems
Let u = –2, 8 and v = 3, – 5 . Find the
component form of:
3u – 4v
–18, 44
Whiteboard Problem…
Find the unit vector in the direction of v = 4, –2 .
Write your answer in both component form and
as a combination of the standard unit vectors.
v  4   2   20  2 5
2
Unit Vector:
2
2
4, 2 
,
2 5
2 5 2 5
1
Component Form:
2
1
,
5
5
4
Standard Unit Vectors:
2
1
i
j
5
5