Transcript 6.3 Vectors in the Plane

6.3 Vectors in the Plane
Many quantities in geometry and physics, such as area, time,
and temperature, can be represented by a single real number.
Other quantities, such as force and velocity, involve both
magnitude and direction and cannot be completely
characterized by a single real number.
To represent such a quantity, we use a directed line segment.
The directed line segment PQ has initial point P and terminal
point Q and we denote its magnitude (length) by PQ .
Q
Terminal Point
P
Initial Point
PQ
Vector Representation by Directed Line Segments
Let u be represented by the directed line segment from
P = (0,0) to Q = (3,2), and let v be represented by the directed
line segment from R = (1,2) to S = (4,4). Show that u = v.
S
4
v
3
2
Q
R
u
1
P
1
Using the distance formula, show
that u and v have the same length.
Show that their slopes are equal.
u 
3  02  2  02
 13
v 
4  12  4  22
 13
2
Slopes of u and v are both
3
2
3
4
Component Form of a Vector
The component form of the vector with initial point P = (p1, p2)
and terminal point Q = (q1, q2) is
PQ
 q1  p1,q2  p2  v1, v2  v
The magnitude (or length) of v is given by
v 
q1  p1   q2  p2 
2
2
 v1  v2
2
2
Find the component form and length of the vector v that has
initial point (4,-7) and terminal point (-1,5)
6
Let P = (4, -7) = (p1, p2) and
Q = (-1, 5) = (q1, q2).
4
Then, the components of v = v1 ,v2
are given by
2
v1 = q1 – p1 = -1 – 4 = -5
-2
-2
2
4
v2 = q2 – p2 = 5 – (-7) = 12
 5,12
-4
Thus, v =
-6
and the length of v is
-8
v  (5) 2  12 2  169  13
Vector Operations
The two basic operations are scalar multiplication and vector
addition. Geometrically, the product of a vector v and a scalar
k is the vector that is k times as long as v. If k is positive,
then kv has the same direction as v, and if k is negative, then
kv has the opposite direction of v.
v
½v
2v
-v
3
 v
2
Definition of Vector Addition & Scalar Multiplication
Let u = u1 ,u2 and v = v1 ,v2 be vectors and let k be a
scalar (real number). Then the sum of u and v is
u + v = u1  v1 , u2  v2
and scalar multiplication of k times u is the vector
ku  k u1, u2  ku1, ku2
Vector Operations
Ex. Let v =  2,5 and w = 3,4 . Find the following vectors.
a. 2v
b. w – v
2v   4,10
w v  3  (2),4  5  5,1
10
2v
v
8
4
6
3
4
2
2
-4
-2
-2
w
-v
1
2
-1
1
2
3
w-v
4
5
Writing a Linear Combination of Unit Vectors
Let u be the vector with initial point (2, -5) and terminal point
(-1, 3). Write u as a linear combination of the standard unit
vectors of i and j.
6
Solution
10
(-1, 3) 4
8
u  1  2,3  5
2
-2
-2
-4
-6
-8
u 2
4
(2, -5)
  3,8
6
 3i  8 j
4
Graphically,
it looks like…
8j
2
-3i
-4
-2
-2
2
Writing a Linear Combination of Unit Vectors
Let u be the vector with initial point (2, -5) and terminal point
(-1, 3).Write u as a linear combination of the standard unit
vectors i and j.
Begin by writing the component form of the vector u.
u  1 2,3 5
u  3,8



u  3i  8 j
Unit Vectors
v  1 
  v
u = unit vector 
v  v 
Find a unit vector in the direction of v =
v 

v
2,5
2  5
2

2
1

2,5 
29


2,5
2 5
,
29 29
Vector Operations
Let u = -3i + 8j and let v = 2i - j. Find 2u - 3v.
2u - 3v = 2(-3i + 8j) - 3(2i - j)
= -6i + 16j - 6i + 3j
= -12i + 19 j