Vector Arithmetic
These notes cover vector arithmetic, focusing on parts of a vector (magnitude and direction) and operations of vectors (addition, subtraction, and scalar multiplication). Readers will learn how to manipulate vectors to solve problems in physics, engineering, computer science, and mathematics. Through practical exercises and theoretical discussions, they will develop a strong understanding of vector arithmetic and its applications.
What is a vector?
Vectors are used to represent mathematical quantities that hold both a magnitude and a direction. You may have seen them before in physics when working with force or velocity.
For example, a force vector would be described by something like: a person is pushing a toy car forward with a power of 5 Newtons. We would depict this vector with an arrow starting at the point of impact and ending some distance away in the described direction.
Notice that the point of impact does not define the vector, only the magnitude and the direction.
Consider this sketch of vectors in a 2-Dimensional coordinate plane.
While all of these directed line segments originate from different points, each of them depicts the same vector. In particular, they depict a vector that moves $X$ units left and $Y$ units up. We denote this vector by $\boldsymbol{v}= \langle X, Y \rangle$ with the bolded $\boldsymbol{v}$ and angled brackets to differentiate between a point and a vector.
Notice that we obtain the vector $\boldsymbol{v}$ from a directed line segment $\overrightarrow{AB}$ from the point $A=(x_1, y_1)$ to the point $B=(x_2, y_2)$ by using the equation $\boldsymbol{v}=\langle x_2-x_1, y_2-y_1 \rangle $. Also, note that the vector from point $A$ to point $B$ is different from the reverse vector going from point $B$ to point $A$.
Since the initial point of the vector is not important, we often denote all vectors as originating from the origin. We call these vectors position vectors.
These constructions are not unique to 2-dimensional space. In fact, while we will only go up to 3-dimensional space, you can extend the terminology for a vector up to $n$-dimensional space by including $n$-components
Example: Find the directed line segments between the points $A = (1, 2, 3)$ and $B = (4, 5, 6)$ in 3-dimensional space.
- Subtract the coordinates of the initial point, $A$, from the coordinates of the terminal point, $B$, to find the vector representation of the line segment: $ \overrightarrow{AB} = \langle b_1 - a_1, b_2 - a_2, b_3 - a_3 \rangle $.
- The directed line segment from $A$ to $B$ is given by $ \overrightarrow{AB} = \langle 4 - 1, 5 - 2, 6 - 3 \rangle = \langle 3, 3, 3 \rangle $.
- Subtract the coordinates of the initial point, $B$, from the coordinates of the terminal point, $A$, to find the vector representation of the line segment: $ \overrightarrow{BA} = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle $.
- The directed line segment from $B$ to $A$ is given by $ \overrightarrow{BA} = \langle 1 - 4, 2 - 5, 3 - 6 \rangle = \langle -3, -3, -3 \rangle $.
Answer: $ \overrightarrow{AB} = \langle 3, 3, 3 \rangle $ and $ \overrightarrow{BA} = \langle -3, -3, -3 \rangle $.
Magnitude and Direction
Magnitude
Recall that a vector encodes both a magnitude and a direction. Sometimes we want to just know the magnitude of the vector, i.e. the distance traveled by the position vector.
Definition: The magnitude, length, of the vector $\boldsymbol{v}=\langle x_1,x_2,x_3 \rangle $ is given by $||\boldsymbol{v}||=\sqrt{ x_1^2 + x_2^2 + x_3^2 }$.
Or more generally, the magnitude of the vector $\boldsymbol{v}=\langle x_1,...,x_n \rangle$ is $||\boldsymbol{x}||=\sqrt{ x_1^2 + ... + x_n^2 }$
Example: Calculate the magnitude of the following vectors:
- $\boldsymbol{a} = \langle 3, -5, 10 \rangle$
- $\boldsymbol{b} = \langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\rangle$
- $\boldsymbol{c} = \langle 0, 0 \rangle$
- $\boldsymbol{d} = \langle 1, 0, 0 \rangle$
- Use the formula for the magnitude of a vector: $||\boldsymbol{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
- Substitute the given values: $||\boldsymbol{a}|| = \sqrt{3^2 + (-5)^2 + 10^2} = \sqrt{9 + 25 + 100} = \sqrt{134}$.
Magnitude of Vector $a$: $||\boldsymbol{a}|| = \sqrt{134}$.
- Use the formula for the magnitude of a vector: $||\boldsymbol{b}|| = \sqrt{b_1^2 + b_2^2}$.
- Substitute the given values: $||\boldsymbol{b}|| = \sqrt{(\frac{1}{\sqrt{5}})^2 + (\frac{-2}{\sqrt{5}})^2 } = \sqrt{\frac{1}{5} + \frac{4}{5}} = 1$.
Magnitude of Vector $b$: $||\boldsymbol{b}|| = 1$.
- Use the formula for the magnitude of a vector: $||\boldsymbol{c}|| = \sqrt{c_1^2 + c_2^2}$.
- Substitute the given values: $||\boldsymbol{c}|| = \sqrt{0^2 + 0^2} = \sqrt{0+0} = 0$.
Magnitude of Vector $c$: $||\boldsymbol{c}|| = 0$.
- Use the formula for the magnitude of a vector: $||\boldsymbol{d}|| = \sqrt{d_1^2 + d_2^2 + d_3^2}$.
- Substitute the given values: $||\boldsymbol{d}|| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1 + 0 + 0} = 1$.
Magnitude of Vector $d$: $||\boldsymbol{d}|| = 1$.
Direction
Similiarly, one might want to know the direction a vector points in. This information is not immediately obvious; for example, the position vectors $\langle 1,2 \rangle$ and $\langle 3,6 \rangle$ both point in the same direction. In fact, there are ifninitely many vectors that point in one direction (one for each magnitude possible).
So, how do we remove magnitude from the vector in order to see the direction more clearly?
Definition: The unit vector, or direction, of the vector $\boldsymbol{v}=\langle x_1,x_2,x_3 \rangle $ is given by $\boldsymbol{u}=\frac{\boldsymbol{v}}{||\boldsymbol{v}||}$.
Notice, that a unit vector always has magnitude 1. To see why, click proof.
Proof
Given a vector $\boldsymbol{v}=\langle x_1,x_2,x_3 \rangle $ with magnitude $ ||\boldsymbol{v}||=\sqrt{ x_1^2 + x_2^2 + x_3^2 } $ , we have the unit vector $$ \boldsymbol{u} = \frac{\boldsymbol{v}}{||\boldsymbol{v}||} = \langle \frac{x_1}{\sqrt{x_1^2 + x_2^2 + x_3^2}} , \frac{x_2}{\sqrt{x_1^2 + x_2^2 + x_3^2}} , \frac{x_3}{\sqrt{x_1^2 + x_2^2 + x_3^2}} \rangle. $$ Therefore, the magnitude of $\boldsymbol{u}$ is $$\begin{align*} ||\boldsymbol{u}|| & = \sqrt{ \left( \frac{x_1}{ \sqrt{x_1^2 + x_2^2 + x_3^2} } \right)^2 + \left( \frac{x_2}{ \sqrt{x_1^2 + x_2^2 + x_3^2} } \right)^2 + \left( \frac{x_3}{ \sqrt{x_1^2 + x_2^2 + x_3^2} } \right)^2 } \\ & = \sqrt{ \frac{x_1^2}{\left( \sqrt{x_1^2 + x_2^2 + x_3^2} \right)^2} + \frac{x_2^2}{ \left( \sqrt{x_1^2 + x_2^2 + x_3^2} \right)^2} + \frac{x_3^2}{ \left( \sqrt{x_1^2 + x_2^2 + x_3^2} \right)^2} }\\ & = \sqrt{ \frac{x_1^2}{x_1^2 + x_2^2 + x_3^2} + \frac{x_2^2}{x_1^2 + x_2^2 + x_3^2} + \frac{x_3^2}{x_1^2 + x_2^2 + x_3^2} } \\ & = \sqrt{ \frac{x_1^2 + x_2^2 +x_3^2}{x_1^2 + x_2^2 + x_3^2} } \\ & = 1 \end{align*}$$Example: Find the unit vector of the following vectors:
- $\boldsymbol{a} = \langle 3, -5, 10 \rangle$
- $\boldsymbol{b} = \langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\rangle$
- $\boldsymbol{c} = \langle 0, 0 \rangle$
- $\boldsymbol{d} = \langle 1, 0, 0 \rangle$
- Use the formula for the magnitude of a vector: $||\boldsymbol{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
- Substitute the given values: $||\boldsymbol{a}|| = \sqrt{3^2 + (-5)^2 + 10^2} = \sqrt{9 + 25 + 100} = \sqrt{134}$.
- Use the formula for the unit vector: $\boldsymbol{u}=\frac{\boldsymbol{v}}{||\boldsymbol{v}||}$.
- Subsitude the given values: $\boldsymbol{u}=\frac{\langle 3, -5, 10 \rangle}{\sqrt{134}}$.
- Distribute for clarity: $\boldsymbol{u}= \langle \frac{3}{\sqrt{134}} , \frac{-5}{\sqrt{134}} , \frac{10}{\sqrt{134}} \rangle$.
Magnitude of Vector $a$: $\boldsymbol{u} = \langle \frac{3}{\sqrt{134}} , \frac{-5}{\sqrt{134}} , \frac{10}{\sqrt{134}} \rangle$.
- Use the formula for the magnitude of a vector: $||\boldsymbol{b}|| = \sqrt{b_1^2 + b_2^2}$.
- Substitute the given values: $||\boldsymbol{b}|| = \sqrt{(\frac{1}{\sqrt{5}})^2 + (\frac{-2}{\sqrt{5}})^2 } = \sqrt{\frac{1}{5} + \frac{4}{5}} = 1$.
- Use the formula for the unit vector: $\boldsymbol{u}=\frac{\boldsymbol{v}}{||\boldsymbol{v}||}$.
- Subsitude the given values: $\boldsymbol{u}=\frac{\\langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\rangle}{1}$.
- Distribute for clarity: $\boldsymbol{u}= \langle \frac{\frac{1}{\sqrt{5}}}{1},\frac{\frac{-2}{\sqrt{5}}}{1} \rangle$.
Magnitude of Vector $b$: $\boldsymbol{u} = \langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}} \rangle$.
- Use the formula for the magnitude of a vector: $||\boldsymbol{c}|| = \sqrt{c_1^2 + c_2^2}$.
- Substitute the given values: $||\boldsymbol{c}|| = \sqrt{0^2 + 0^2} = \sqrt{0+0} = 0$.
- Use the formula for the unit vector: $\boldsymbol{u}=\frac{\boldsymbol{v}}{||\boldsymbol{v}||}$.
- Subsitude the given values: $\boldsymbol{u}=\frac{\langle 0, 0 \rangle}{0}$.
- Distribute for clarity: $\boldsymbol{u}= \langle \frac{0}{0} , \frac{0}{0} \rangle$.
Magnitude of Vector $c$: $\boldsymbol{u} = \langle 0, 0 \rangle $.
- Use the formula for the magnitude of a vector: $||\boldsymbol{d}|| = \sqrt{d_1^2 + d_2^2 + d_3^2}$.
- Substitute the given values: $||\boldsymbol{d}|| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1 + 0 + 0} = 1$.
- Use the formula for the unit vector: $\boldsymbol{u}=\frac{\boldsymbol{v}}{||\boldsymbol{v}||}$.
- Subsitude the given values: $\boldsymbol{u}=\frac{\langle 1, 0, 0 \rangle}{1}}$.
- Distribute for clarity: $\boldsymbol{u}= \langle \frac{1}{1} , \frac{0}{1} , \frac{0}{1} \rangle$.
Magnitude of Vector $d$: $\boldsymbol{u} = \langle 1,0,0 \rangle $.
Special Vectors
There are a couple special vectors that we use to simplify notation. The first is the only vector with no direction:
Definition: The zero vector, $\langle 0,0,0 \rangle $, often denoted $\boldsymbol{0}$, has an undfined direction and a magnitude of zero.
Be careful to distinguish between 0 the origin and $\boldsymbol{0}$ the vector with no direction and zero magnitude.
The second type of special vector is called a standard basis vector:
Definition: A standard basis vector is a unit vector that moves in the direction of an axis.
For example, in 3-dimensional space, there are 3 standard basis vectors: $\boldsymbol{i}= \langle 1,0,0 \rangle$, $\boldsymbol{j}= \langle 0,1,0 \rangle$, and $\boldsymbol{k}= \langle 0,0,1 \rangle$. In 2-D space there are only two standard basis vectors, $\boldsymbol{i}$ and $\boldsymbol{j}$. In $n$-dimensions, there are $n$.
Addition and Subtraction
Now that we know what a vector is and how to break it down into its parts, we ask a new question: how do you work with vectors?
Given two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$, we can define additional coordinate-wise:
Definition: The additiona of two vectors, $\boldsymbol{a}= \langle a_1,a_2a_3 \rangle $ and $\boldsymbol{b}= \langle b_1,b_2,b_3 \rangle $, is defined by the formula $\boldsymbol{a}+\boldsymbol{b}= \langle a_1+b_1,a_2+b_2,a_3+b_3 \rangle $.
Note that substraction is just the addition of the negative form of the second vector, i.e. $\boldsymbol{a}-\boldsymbol{b}= \langle a_1-b_1,a_2-b_2,a_3-b_3 \rangle $
.
Example: Let $\boldsymbol{a}= \langle 1,0,-1 \rangle $, $\boldsymbol{b}= \langle 1,2,3 \rangle$, and $\boldsymbol{j}= \langle 0,1,0 \rangle $. Find the following:
- $\boldsymbol{a+b}$
- $\boldsymbol{a+j}$
- $\boldsymbol{b+a}$
- $\boldsymbol{j-a}$
- Use the formula for addition of two vectors: $\boldsymbol{a+b}= \langle a_1+b_1,a_2+b_2,a_3+b_3 \rangle $.
- Substitute the given values: $\boldsymbol{a+b} = \langle 1+1,0+2,(-1)+3 \rangle $.
- Simplify: $\boldsymbol{a+b}= \langle 2,2,2 \rangle $.
Addition of Vectors $\boldsymbol{a}$ and $\boldsymbol{b}$: $\boldsymbol{a+b}= \langle 2,2,2 \rangle $.
- Use the formula for addition of two vectors: $\boldsymbol{a+j}= \langle a_1+j_1,a_2+j_2,a_3+j_3 \rangle $.
- Substitute the given values: $\boldsymbol{a+b} = \langle 1+0,0+1,(-1)+0 \rangle $.
- Simplify: $\boldsymbol{a+b}= \langle 1,1,-1 \rangle $.
Addition of Vectors $\boldsymbol{a}$ and $\boldsymbol{j}$: $\boldsymbol{a+j}= \langle 1,1,-1 \rangle $.
- Use the formula for addition of two vectors: $\boldsymbol{b+a}= \langle b_1+a_1,b_2+a_2,b_3+a_3 \rangle $.
- Substitute the given values: $\boldsymbol{b+a} = \langle 1+1,2+0,3+(-1) \rangle $.
- Simplify: $\boldsymbol{a+b}= \langle 2,2,2 \rangle $.
Addition of Vectors $\boldsymbol{b}$ and $\boldsymbol{a}$: $\boldsymbol{a+b}= \langle 2,2,2 \rangle $.
- Use the formula for addition of two vectors: $\boldsymbol{j-a}= \langle j_1-a_1,j_2-a_2,j_3-a_3 \rangle $.
- Substitute the given values: $\boldsymbol{j-a} = \langle 0-1,1-0,0-(-1) \rangle $.
- Simplify: $\boldsymbol{a+b}= \langle -1,1,1 \rangle $.
Subtraction of Vector $\boldsymbol{a}$ from $\boldsymbol{j}$: $\boldsymbol{j-1}= \langle -1,1,1 \rangle $.
Scalar Multiplication
This is the transcript for Video 4...
Self-assessment Questions
1. Recall: What is the formula for adding two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in component form?
The correct answer is A. $\boldsymbol{a}+\boldsymbol{b}=\langle a_x + b_x, a_y + b_y \rangle$. This is the formula for adding two vectors in component form given in the section titled "Addition and Subtraction."
2. Mimic: Given the vector $\boldsymbol{a}=\langle 1,2\rangle $, find the magnitude and direction.
The correct answer is B. The magnitude is $\sqrt{5}$ and the direction is $\boldsymbol{u}=\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle$. See the section "Magnitude and Direction" for their respective formulas.
| $$ \begin{align*} ||\boldsymbol{v}|| = \sqrt{(v_1)^2+(v_2)^2} \\ ||\boldsymbol{a}|| = \sqrt{1^2+2^2} = \sqrt{5} \end{align*}$$ | $$ \begin{align*} \boldsymbol{u} = \frac{\boldsymbol{u}}{||\boldsymbol{u}||} \\ \boldsymbol{u} = \langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle \end{align*}$$ |
3. Apply: If $\boldsymbol{a}=\langle 3,2 \rangle$ and $\boldsymbol{b}=\langle -1,4 \rangle$, find the magnitude of $\boldsymbol{a}+\boldsymbol{b}$.
The correct answer is C.$2\sqrt{3}$. After adding the two vectors, you can find the magnitude of the resulting vector using the magnitude formula found in the "Magnitude and Direction" section.
Add the two vectors: $$\begin{align*} \boldsymbol{u}+\boldsymbol{v} = \langle u_1+v_1,u_2+v_2 \rangle \\ \langle 3,2 \rangle + \langle -1,4 \rangle = \langle 2,2 \rangle \\ \end{align*}$$ Compute the magnitude: $$\begin{align*} ||\boldsymbol{u}||= \sqrt{(u_1)^2+(u_2)^2} \\ ||\boldsymbol{a}+\boldsymbol{b}|| = ||\langle 2,2 \rangle || \\ = \sqrt{2^2+2^2} = \sqrt{8} = 2 \sqrt{2} \end{align*}$$
4. Expand: Which of the following statements is true regarding the addition of two vectors?
The correct answer is D. The magnitude of the sum can be equal to or less than, but not greater than, the sum of the magnitudes of the individual vectors, depending on their directions. The two sums will be equal when the vectors point in the same direction. Otherwise, the magnitude of the sum will be shorter due to the triangle inequality theorem. To see a visual of this, go to the "Magnitude and Direction" section of this page.
| Example of adding two vectors in the same direction: $$ \begin{align*} \boldsymbol{a}=\langle 2,0 \rangle,\boldsymbol{b}=\langle 2,0 \rangle \\ ||\boldsymbol{a}|| = \sqrt{2^2+0^2} = \sqrt{4} = 2 \\ ||\boldsymbol{b}|| = \sqrt{2^2+0^2} = \sqrt{4} = 2 \\ ||\boldsymbol{a}|| + ||\boldsymbol{b}|| = 2 + 2 = 4 \\ \boldsymbol{a}+\boldsymbol{b}=\langle 4,0 \rangle \\ ||\boldsymbol{a}+\boldsymbol{b}|| = \sqrt{4^2+0^2} = \sqrt{16} = 4 \end{align*}$$ | Example of adding two vectors in different directions: $$ \begin{align*} \boldsymbol{a}=\langle 2,0 \rangle ,\boldsymbol{b}=\langle 0,2 \rangle \\ ||\boldsymbol{a}|| = \sqrt{2^2+0^2} = \sqrt{4} = 2 \\ ||\boldsymbol{b}|| = \sqrt{0^2+2^2} = \sqrt{4} = 2 \\ ||\boldsymbol{a}|| + ||\boldsymbol{b}|| = 2 + 2 = 4 \\ \boldsymbol{a}+\boldsymbol{b}=\langle 2,2 \rangle \\ ||\boldsymbol{a}+\boldsymbol{b}|| = \sqrt{2^2+2^2} = \sqrt{8} = 2\sqrt{2} \end{align*}$$ |