Module 8 Lesson 2: Vectors in 2D
Two basic vector operations are scalar multiplication, vector addition, and the dot
product. Scalar multiplication involves multiplying a vector by a real number. Vector
addition involves adding two vectors. We will work with all of these operations
separately with a few more details. Hopefully you’ll find that operations with vectors
are quite simple and straightforward. 
Then, we will discuss how to find the unit vector, how to determine direction angles of
vectors, and how to find the angle between two vectors.
Let’s begin!
Scalar Multiplication
If k is any real number and 𝐯 = ⟨𝑣1 , 𝑣2 ⟩, then 𝑘𝐯 = ⟨𝑘𝑣1 , 𝑘𝑣2 ⟩.
Geometrically, the product of a vector v and its scalar k is the vector that is |k| times as
long as v. For example, if v is 3 units long, then the vector 2v is 2 ∙ 3 = 6 units long.


If k is positive, then kv has the same direction as v. See below.
If k is negative, then kv has the opposite direction as v. See below.
Ex. Let 𝐯 = ⟨−2, 5⟩ and 𝐰 = ⟨3, 4⟩. Find (a) 3𝐯 and (b) −2𝐰.
a. Solution: 3𝐯 = 3⟨−2, 5⟩ = ⟨3(−2), 3(5)⟩ = ⟨−6, 15⟩
b. Solution: −2𝐰 = −2⟨3, 4⟩ = ⟨−2(3), −2(4)⟩ = ⟨−6, −8⟩
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Vector Addition
Let 𝐯 = ⟨𝑣1 , 𝑣2 ⟩ and 𝐰 = ⟨𝑤1 , 𝑤2 ⟩ be vectors (as shown below).
Then the sum of v and w is
𝐯 + 𝐰 = ⟨𝑣1 + 𝑤1 , 𝑣2 + 𝑤2 ⟩
Notice the new vector that is created by their sum shown above. Notice how the
second vector w is simply added on to the first vector v by connecting the initial point of
w with the terminal point of v.
Note: The difference of u and v can be found by
𝐯 − 𝐰 = 𝐯 + (−𝐰) = ⟨𝑣1 − 𝑤1 , 𝑣2 − 𝑤2 ⟩
For this course, you will only need to know how to operate with vectors algebraically
and not necessarily how to sketch them; however, you should be familiar with and
understand what vector addition looks like in the coordinate plane (as shown in the
figures above).
Now, let’s use basic addition properties to find the sum of two vectors algebraically.
Ex. Let 𝐯 = ⟨−2, 5⟩ and 𝐰 = ⟨3, 4⟩. Find 𝐯 + 𝐰.
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Solution: 𝐯 + 𝐰 = ⟨−2, 5⟩ + ⟨3, 4⟩ = ⟨−2 + 3, 5 + 4⟩ = ⟨1, 9⟩
Note: Addition of vectors is commutative (the order of vectors being added does not
matter).
For example, 𝐰 + 𝐯 = ⟨3, 4⟩ + ⟨−2, 5⟩ = ⟨3 + (−2), 4 + 5⟩ = ⟨1, 9⟩.
We get the same vector sum. Therefore, 𝐯 + 𝐰 = 𝐰 + 𝐯.
Combining Scalar Multiplication & Addition
Ex. Given 𝐯 = ⟨7, −3⟩ and 𝐰 = ⟨2, 8⟩, find (a) 3𝐯 + 𝐰 and (b) 𝐰 − 4𝐯.
a. Solution: 3𝐯 + 𝐰 = 3⟨7, −3⟩ + ⟨2, 8⟩
= ⟨21, −9⟩ + ⟨2, 8⟩
= ⟨21 + 2, −9 + 8⟩
= ⟨23, −1⟩
b. Solution: 𝐰 − 4𝐯 = ⟨2, 8⟩ − 4⟨7, −3⟩
= ⟨2, 8⟩ − ⟨28, −12⟩
= ⟨2 − 28, 8— 12)
= ⟨−26, 20⟩
The Linear Combination Form of a Vector
Now that you are a little more familiar with how to write and operate with vectors, let’s
introduce another notation.
You will also see vectors represented with an “i” and a “j”. For example, the vector
𝐯 = ⟨3, 6⟩ can also be represented as 𝐯 = 3𝐢 + 6𝐣.
Here, i represents the horizontal component of the vector v and j
represents the vertical component of v. Recall that the coefficients of
i and j are the directed differences in the x-values and y-values of the
initial and terminal points of the vector (as found in the previous
Component Form Examples).
Ex. Let 𝐮 = 2𝐢 − 4𝐣 and 𝐰 = −6𝐢 + 𝐣. Find (a) 𝐮 + 𝐰 and (b) 6𝐮 + 2𝐰.
(a) Solution: 𝐮 + 𝐰 = (2𝐢 − 4𝐣) + (−6𝐢 + 𝐣)
= 2𝐢 − 6𝐢 − 4𝐣 + 𝐣
= −4𝐢 − 3𝐣.
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Note: This bold
lowercase i is not to
be confused with the
imaginary number 𝑖 =
√−1.
(b) Solution: 6𝐮 + 2𝐰 = 𝟔(2𝐢 − 4𝐣) + 𝟐(−6𝐢 + 𝐣)
= 12𝐢 − 24𝐣 − 12𝐢 + 2𝐣
= 0𝐢 − 22𝐣
= −22𝐣
Note: Scalar Multiplication and Vector Addition work the same in both component
form or linear combination form. We will use both interchangeably from here on out.
The Dot Product
The dot product is a way of multiplying two vectors. Say we have two vectors, 𝐮 =
⟨𝑢1 , 𝑢2 ⟩ and 𝐯 = ⟨𝑣1 , 𝑣2 ⟩. The product or 𝐮 ∙ 𝐯 is the product of the horizontal
components plus the product of the vertical components. In other words:
𝐮 ∙ 𝐯 = 𝑢1 𝑣1 + 𝑢2 𝑣2
Notice that we use the “dot” multiplication symbol here.
Ex. Let 𝐮 = ⟨3, −4⟩ and 𝐯 = ⟨5, 2⟩. Find 𝐮 ∙ 𝐯.
Solution: 𝐮 ∙ 𝐯 = (3)(5) + (−4)(2) = 15 − 8 = 7
So notice what’s a little different here. When we multiply two vectors, we get just a
plain number… a scalar. Sometimes this is also called the scalar product (if you happen
to find additional resources for your work here in Module 8).
Here is an example of two vectors in linear combination form.
Ex. Let 𝐯 = 6𝐢 − 2𝐣 and 𝐰 = −4𝐢 + 5𝐣. Find the dot product of (a) 𝐯 ∙ 𝐰, (b) 𝐯 ∙ 𝐯 and
(c) 𝐰 ∙ 𝐰.
(a) Solution: 𝐯 ∙ 𝐰 = (6)(−4) + (−2)(5) = −24 − 10 = −34
(b) Solution: 𝐯 ∙ 𝐯 = (6)(6) + (−2)(−2) = 36 + 4 = 40
(c) Solution: 𝐰 ∙ 𝐰 = (−4)(−4) + (5)(5) = 16 + 25 = 41
Notice in (b) and (c) above how we were asked to find a vector multiplied by itself.
One property of vectors is that 𝐯 ∙ 𝐯 = ‖𝐯‖𝟐 . So a vector squared is equal to its
magnitude or length squared.
Refer again to example (b) where 𝐯 = 6𝐢 − 2𝐣 or in component form, 𝐯 = ⟨6, −2⟩.
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Let’s check to see if 𝐯 ∙ 𝐯 = ‖𝐯‖𝟐 by finding the magnitude of v first.
‖𝐯‖ = √62 + (−2)2 = √36 + 4 = √40
So 𝐯 ∙ 𝐯 = ‖𝐯‖𝟐 holds true:
‖𝐯‖𝟐
𝐯 ∙ 𝐯 = 40
= (√40)2 = 40

Properties of the Dot Product
Let’s look at a few more properties of the Dot Product. Most of these are pretty
straightforward.
Let u, v, and w be vectors and let c be a scalar.
1.
2.
3.
4.
5.
𝐮∙𝐯= 𝐯∙𝐮
0∙𝐯=0
𝐮 ∙ (𝐯 + 𝐰) = 𝐮 ∙ 𝐯 + 𝐮 ∙ 𝐰
𝐯 ∙ 𝐯 = ‖𝐯‖𝟐
c(𝐮 ∙ 𝐯) = c𝐮 ∙ 𝐯 = 𝐮 ∙ c𝐯
Let’s try a few more examples.
Ex. Let 𝐮 = ⟨−1, 3⟩, 𝐯 = ⟨2, −4⟩, and 𝐰 = ⟨1, −2⟩.
(a) Find (𝐮 ∙ 𝐯)𝐰
Solution: 𝐮 ∙ 𝐯 = (−1)(2) + (3)(−4) = −2 − 12 = −14
Then, (𝐮 ∙ 𝐯)𝐰 = −14⟨1, −2⟩ = ⟨−14, 28⟩.
(b) Find 𝐮 ∙ 2𝐯
Solution: Using Property #5 we have…
𝐮 ∙ 2𝐯 = 2(𝐮 ∙ 𝐯) = 2(−14) = −28
Or, we can find 2𝐯 = 2⟨2, −4⟩ = ⟨4, −8⟩
Then, 𝐮 ∙ 2𝐯 = (−1)(4) + (3)(−8) = −4 − 24 = −28.
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(c) Find 𝐮 ∙ (𝐯 + 𝐰).
From Property #3 we know that 𝐮 ∙ (𝐯 + 𝐰) = 𝐮 ∙ 𝐯 + 𝐮 ∙ 𝐰.
𝐮 ∙ 𝐯 = −14
𝐮 ∙ 𝐰 = (−1)(1) + (3)(−2) = −1 − 6 = −7
Therefore 𝐮 ∙ (𝐯 + 𝐰) = 𝐮 ∙ 𝐯 + 𝐮 ∙ 𝐰 = −14 − 7 = −21.
Notice that the product in (a) was a vector and the products in (b) and (c) were
scalars.
The Unit Vector Review
Let’s again review how to find the unit vector, but now we will use the linear
combination notation instead of component notation as we did in Lesson 1.
Ex. Find the unit vector, u, for 𝐯 = 3𝐢 + 6𝐣.
From the linear combination notation, we know that the horizontal component
of the vector is 3 and the vertical component of the vector is 6. Therefore, we
can determine that
‖𝐯‖ = √32 + 62 = √45 = 3√5
Next, simply use our formula for the unit vector.
𝐯
𝐮 = ‖𝐯‖ =
3𝐢+6𝐣
3√5
3
6
= 3√5 𝐢 + 3√5 𝐣
Separate
=
Simplify
=
1
𝐢+
√5
√5
𝐢
5
+
2
𝐣
√5
2√5
5
𝐣
Simplify/Rationalize
Remember, the unit vector is a vector with the same direction, but it has a length of 1.
Unit vectors will be helpful in the following Lesson. 
Direction Angles
If u is a unit vector such that 𝜃 is the angle (measured counterclockwise) from the
positive x-axis to u, the terminal point of u lies on the unit circle and you have
𝐮 = ⟨𝑥, 𝑦⟩ = ⟨cos 𝜃 , sin 𝜃⟩ = (cos 𝜃) 𝐢 + (sin 𝜃) 𝐣
as shown in the figure to the right. The angle 𝜃 is called the directional angle of the
vector u.
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If v is any vector that makes an angle 𝜃 with the positive x-axis, then it has the same
direction as the unit vector u and we can write
𝐯 = ‖𝐯‖⟨(cos 𝜃 , sin 𝜃⟩ = ‖𝐯‖ cos 𝜃 𝐢 + ‖𝐯‖sin 𝜃 𝐣
Because 𝐯 = 𝑎𝐢 + 𝑏𝐣 = ‖𝐯‖ cos 𝜃 𝐢 + ‖𝐯‖sin 𝜃 𝐣, it follows that the directional angle 𝜃
can be found by using the formula
tan 𝜃 =
𝑏
𝑎
Ex. Find the direction angle of the vector 𝐮 = 3𝐢 + 3𝐣.
Here, 𝑎 = 3 and 𝑏 = 3. The direction angle s
3
tan 𝜃 =
3
tan 𝜃 = 1
𝜃 = tan−1
3
= 45°
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Let’s draw a picture to help us determine the directional angle. Since the horizontal and
vertical components of this vector are both positive 3, we know the vector lies in
Quadrant I.
Therefore, the directional angle of u is 𝜃 = 45°.
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Ex. Find the direction angle of 𝐯 = 3𝐢 − 4𝐣.
Find 𝜃 by
4
tan 𝜃 = − 3
4
𝜃 = tan−1 (− ) ≈ −53.130°
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Note: Stay in
degree mode.
Let’s draw a picture to help us determine the directional angle. Since the horizontal
component is positive and the vertical component is negative, we know the vector lies i
n Quadrant IV.
The angle 𝜃 = −53.130° is the angle made moving clockwise from the positive x-axis…
not counterclockwise! So in order to find the directional angle, we can subtract this
‘reference angle” of 53.130° from 360°.
The direction angle is 𝜃 = 306.870°.
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The Angle Between Two Vectors
Before we get started with our second method of multiplying two vectors, we must first
learn how to find the angle between two vectors.
The angle between two nonzero vectors is the angle 𝜃, 0 ≤ 𝜃 ≤ 𝜋, between its
respective standard position vectors. A vector is in standard position when its initial
point is at the origin. See the figure below.
If 𝜃 is the angle between two nonzero vectors u and v, then
cos 𝜃 =
𝐮∙𝐯
‖𝐮‖‖𝐯‖
Ex. Find the angle between 𝐮 = ⟨4, 3⟩ and 𝐯 = ⟨3, 5⟩.
cos 𝜃 =
=
⟨4, 3⟩ ∙ ⟨3, 5⟩
‖⟨4, 3⟩‖‖⟨3, 5⟩‖
(4)(3) + (3)(5)
(√42 + 32 )(√32 + 52 )
=
27
(√25)(√34)
=
27
5√34
This implies that the angle between the two vectors is
𝜃 = arccos
9
27
5√34
= 22.166°
Make sure your calculator is in Degree Mode and use 𝑐𝑜𝑠 −1 for 𝑎𝑟𝑐𝑐𝑜𝑠.
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