Unit 9 Lesson 2
Pre-Calculus Honors
Unit 9 Lesson 2: The Unit Vector and Linear Combinations of Vectors
Objective: ______________________________________________________________
1. Do Now: Represent 3w – 1/2v geometrically. Draw and label each vector.
2. Mark up the following definition for linear combination form of vectors
Definition of Linear Combinations
Component Form: v = < v1, v2 >
Linear combination Form: v = v1i + v2j
The scalars v1 and v2 are called the horizontal and vertical components of v respectively and can be written as,
what is called, a linear combination of vectors i and j.
You can solve vector operation problems by converting u and v from linear combination form to component
form. This, however is not necessary. You can perform the rules you learned yesterday to vectors in linear
combination form.
For example Let u = -3i +8j in linear combination form is equivalent to u = <-3, 8> in component form.
Name the three ways vectors can be represented:

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
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
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Unit 9 Lesson 2
Group Practice 1: Discovering Component Form Formula, Given a Magnitude and Directional
Angle
Diagram
Geometric Representation
Directions: Label the x (the horizontal
component), the y (the vertical
Directions: Mark up the following
Find the component form of a
text and answer the questions below. vector v with a directional angle
of 115 degrees and a magnitude
Direction is measured in different
of 6.
ways and in different contexts,
component), v (the magnitude of the
vector, and  on the diagram below.
v
Example Problem
especially navigation. A precise way to
specify the direction of a vector is the
directional angle (the angle q that v
makes with the positive x-axis.)
1. Using trigonometry, how can
you find the horizontal
component of a vector given
 and v ?
Component Form:
_____________________
x = _________________
2. Using trigonometry, how can
you find the vertical
component of a vector given
 and v ?
y = _________________
3. Using questions #1 and #2,
write a formula that you can
use to find the component
form of a vector given the
directional angle and
magnitude.
V = < ___________ , ___________>
Linear Combination Form
______________________
Unit 9 Lesson 2
Group Practice 2: Discovering How to Find the Direction for a Vector
Proof
Example Problem
Directions: Now that you
Directions: Find the
know
magnitude and directional
v =< v cosq, v sinq >mark up angle of the vector
the steps of the proof below.
v = -2i -5j
It follows that the direction
angle of q for v is determined by
1.) tan  
Geometric Representation
Directions: Draw a diagram that
represents the example problem on the
left. Label your reference angle,
directional angle, and magnitude on the
diagram below.
sin 
cos 
Step 1: The Definition of Tangent
2.) tanq =
v sinq
v cosq
Step 2: Multiply the numerator
and denominator by the
magnitude
3.) tan  
y
x
Step 3: Substitute definition in for
horizontal component and vertical
component
Therefore, you can find the
reference angle of the directional
 y
  .
x
angle by tan 1 
Step4: Following, you have to
calculate the directional angle.
Group Practice 3: Find the vector v with the given magnitude and same direction as u.
v  5, u  5,7 
Unit 9 Lesson 2
Unit 9 Lesson 2 Problem Set
1. Let A = (2, -1), B = (3, 1), C = (-4, 2) and D = (1, -5).
Find the component form and the magnitude of the vector




b) CD 2 AB
a) 3 AB

c) AC  BD
2. Use the figure to sketch a graph of the specified vector. Do each example on a separate
coordinate plane. Label all vectors.
v
u
V
Unit 9 Lesson 2
(a) –3u
(b) u + 2 v
(c) 2u -1/2 v
(d) 1/4v
3. Find the component form of vector v.
Unit 9 Lesson 2
v
47
72◦
4.
Find the magnitude and directional angle of the vector.
a)
5.
< 3, 4 >
b) -3i – 5j
c.) 7(cos135◦I + sin135◦j)
Find the vector v with the given magnitude and same direction as u.
llvll = 2, u = < 3, -3 >
Answer Key
Unit 9 Lesson 2
#1a
<3,6>
Magnitude = 45
#1b
< 3 , -11 >
Magnitude = 130
#1c
<-8 , -3 >
Magnitude = 73
#3
<-14.52, 44.70>
#4a
Magnitude = 5
53.13 degrees
#4b
Magnitude = 34
239.036 degrees
#4c
Magnitude = 7
135 degrees
#5
<1.41, - 1.41>