Math 243: Calculus II

12.2 Vectors in $\mathbb{R}^n$

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12.2.1 Vectors in the Plane

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A vector on the plane has two quantities associated with it; it's magnitude and direction.

The magnitude of a vector in $\mathbb{R}^2$ is the length of the vector, and the direction of such a vector is the angle the vector makes with a vector parallel to $\ihat = \langle 1, 0 \rangle = \begin{bmatrix}\ 1\ \\ \ 0\ \end{bmatrix}$.

Figure 1

Notice that this definition does not include where the vector is located on the plane.

Figure 2

All of the vectors in figure 2 are the same vector! In some applications one says that the vector is applied at a specific point $P = (x_0,y_0)$ if the base of the vector is located at $P$. However the same vector can be applied to any point on the plane.

The coordinates of a vector on the plane are the coordinates of the tip of the vector when the base of the vector is the origin, $(0,0)$. When the base of a vector is the origin, the vector is said to be in standard position. In figure 1, both vectors $\mathbf{x}$ and $\mathbf{y}$ are in standard position.

The standard coordinates of a vector are the coordinates of the tip of the vector in standard position. Often we denotes the coordinates of a vector using the same lower case letter in normal font to indicate it is a scalar, and with a subscript to indicate which coordinate it is.

$$ \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \langle v_1, v_2 \rangle $$

The coordinates are also called the scalar components of the vector since the coordinates are equal to the coefficients of the standard basis vectors,

$$ \ihat = \langle 1, 0 \rangle = \begin{bmatrix}\ 1\ \\ \ 0\ \end{bmatrix}\qquad\text{ and }\qquad \jhat = \langle 0, 1 \rangle = \begin{bmatrix}\ 0\ \\ \ 1\ \end{bmatrix} $$

$$ \mathbf{v} = \begin{bmatrix}\ v_1\ \\ \ v_2\ \end{bmatrix} = v_1\ihat + v_2\jhat $$

Example 1

One uses the coordinates of the points at the base and tip to compute the standard coordinates of a vector in non-standard position

Figure 3

The base is at the point $P_0=(-1,-2)$ and the tip is at the point $P_1=(3,2)$. The same vector in standard position has standard coordinates

$$ \begin{align*} \mathbf{x} &= \begin{bmatrix}\ x_1-x_0\ \\ \ y_1-y_0\ \end{bmatrix} = \begin{bmatrix}\ 3 - (-1)\ \\ \ 2 - (-2)\ \end{bmatrix} = \begin{bmatrix}\ 4\ \\ \ 4\ \end{bmatrix} \\ \\ &= \langle x_1-x_0, y_1-y_0 \rangle = \langle 3 - (-1), 2 - (-2) \rangle = \langle 4, 4 \rangle \end{align*} $$

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12.2.2 What are Vectors?

A vector is an element of a Vector Space. The concept of vectors are central to the study of linear algebra as well as vector calculus (an alternative title used for this course at other universities). As a mathematical concept, vectors are quite common and have several useful applications. For an introduction, students should study the series linear algebra by Grant Sanderson and hosted on his website 3Blue1Brown

Vectors, what even are they?

Please study this second video in this series before continuing with the online notes or reading the textbook.

Throughout this course we will use all three perspectives discussed in the video for our discussion of vectors. We will choose the most useful perspective at each point in the notes. The physicist student's perspective is the most visual, so let us start there with arrows. The visualization of a vector as an arrow in space means that we need an underlying space of points with unique coordinates, and a metric or method of measuring the distance between any two different points. These coordinates are described in the video and discussed below in these notes.

Each vector has an initial point or tail where it begins, and a terminal point or tip where it ends. The magnitude of the vector is the length between these two points and the direction is given in the sense of traveling from the initial to terminal point. This is called a directed line segment, and a directed line segment from $P$ to $Q$ is denoted as $\overrightarrow{PQ}$.

Figure 4

Remember to distinguish between a vector in such an underlying space, and the coordinates of the initial and terminal points of the vector.

In these examples $P$ and $A$ are different points in our underlying coordinate space of points. Likewise $Q$ and $B$ are different points, however $\overrightarrow{PQ}$ and $\overrightarrow{AB}$ are the same vector because they point the same direction and they have the same length.

The two vectors pictured above are equivalent, since they both have the same magnitude and direction. It does not matter where in our underlying coordinate system arrows sit, if two vectors share their magnitude and direction, they represent the same vector.

In the mathematician picture, we may also refer to the displacement vector $\overrightarrow{PQ} = \overrightarrow{v}$ or $\mathbf{v}$. Vectors are typically represented using particular notation, with several forms being common.

Typical Vector Notations

• Overline $$\overline{v_1},\ \ \overline{v_2}$$
• Arrows over $$\overrightarrow{v_1},\ \ \overrightarrow{v_2}$$
• Underlining $$\underline{v_1},\ \ \underline{v_2}$$
• Bold face $$\mathbf{v}_1,\ \ \mathbf{v}_2$$
• Hats $$\ihat,\ \ \khat$$

Bold face will be preferred in these notes, while arrows over or underlining will be the preferred method for handwritten work. The hatted vectors are typically reserved for special cases, such as the canonical basis vectors $\ihat$, $\jhat$, and $\khat$ or a "changed" or shifted vector $\mathbf{x}\mapsto\mathbf{\hat{x}}$. These will be discussed in more detail when they come up.

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12.2.3 Vector Operations in $\mathbb{R}^n$

Scalar multiplication and vector addition in $\mathbb{R}^3$ is very similar to the same vector operations in $\mathbb{R}^2$.

Definition: Scalar Multiplication in $\mathbb{R}^n$

Let scalar $\alpha\in\mathbb{R}$ (or $\mathbb{C}$) and vector $\mathbf{x}\in\mathbb{R}^n$. Then

$$ \alpha\mathbf{x} = \alpha\begin{bmatrix} x_1 \\ \ddots \\ x_n \end{bmatrix} = \begin{bmatrix} \alpha x_1 \\ \ddots \\ \alpha x_n \end{bmatrix}. $$

Multiplying a vector by a scalar dilates, contracts, and/or changes the vector to point the opposite direction.

Figure 5

Definition: Vector Addition in $\mathbb{R}^n$

Vector addition is computed component-wise from the programmer's perspective. If you have two vectors $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}^n$, their vector sum is

$$ \mathbf{x} + \mathbf{y} = \begin{bmatrix} x_1 \\ \ddots \\ x_n \end{bmatrix} + \begin{bmatrix} y_1 \\ \ddots \\ y_n \end{bmatrix} = \begin{bmatrix} x_1 + y_1 \\ \ddots \\ x_n + y_n \end{bmatrix}. $$

Vector addition is the process to combining two vectors via a sum. To add a pair of vectors, the terminal point of the first vector is used as the initial point to draw the second vector, and the new vector formed from the initial point of the first and terminal point of the second is the vector sum. For two vectors $\mathbf{u}$ and $\mathbf{v}$, this looks like

Figure 6

On the left, we see that added vectors may be thought of as forming a triangle, with the third side being the sum $\mathbf{u}+\mathbf{v}$. On the right, we see that vector addition commutes, that is $\mathbf{u}+\mathbf{v} = \mathbf{v}+\mathbf{u}$. Visually, this forms a parallelogram. In either case, it is typical to think of adding the arrows by drawing them "tip to tail" as in the images.

Exercise 1

Let $\mathbf{u} = \langle -4, -2, 1 \rangle$ and $\mathbf{v} = \langle 0, 3, -4 \rangle$

1. Determine the magnitudes of $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{u} + \mathbf{v}$

2. Perform each vector operation

   (a) $2\mathbf{u} + \mathbf{v}$

   (b) $5\mathbf{u}$

   (c) $3\mathbf{u} - 4\mathbf{v}$
   

View Solution

$$ \begin{align*} 1.\ \ \left\|\langle -4, -2, 1 \rangle\right\| &= \sqrt{ (-4)^2 + (-2)^2 + 1^2 } = \sqrt{16 + 4 + 1} = \sqrt{21} \\ \\ \left\|\langle 0, 3, -4 \rangle\right\| &= \sqrt{ 0^2 + 3^2 + (-4)^2 } = \sqrt{9 + 16} = 5 \\ \\ \left\|\langle -4, -2, 1 \rangle + \langle 0, 3, -4 \rangle\right\| &= \left\|\langle -4, 1, -3 \rangle\right\| \\ &= \sqrt{ (-4)^2 + 1^2 + (-3)^2 } = \sqrt{ 16 + 1 + 9 } = \sqrt{26} \end{align*} $$

$$ \begin{align*} \text{(a)} &\qquad &2\mathbf{u} + \mathbf{v} &= 2\langle -4, -2, 1 \rangle + \langle 0, 3, -4 \rangle = \langle -8, -1, -2 \rangle \\ \\ \text{(b)} &\qquad &5\langle -4, -2, 1 \rangle &= \langle -20, -10, 5 \rangle \\ \\ \text{(c)} &\qquad &3\mathbf{u} - 4\mathbf{v} &= 3\langle -4, -2, 1 \rangle - 4\langle 0, 3, -4 \rangle \\ &\qquad &\ &= \langle -12, -6, 3 \rangle - \langle 0, 12, -16 \rangle \\ &\qquad &\ &= \langle -12, -18, 19 \rangle \end{align*} $$

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12.2.4 Vector Components

Consider the vector with initial point $P(-2,2)$ and terminal point $Q(3,0)$, plotted below

Figure 7

We see that this vector is the same as the other vector $\mathbf{v}$ plotted from $(0,0)$ to $(5,-2)$. It is generally inconvenient to track a pair of points to define a vector, and we would rather work with the version with the tail (initial point) at the origin. This is the component form of a vector. To find the component form for a vector, you take the difference, in order, of the final and initial coordinates of the points defining the vector. So,

$$ \overrightarrow{PQ} = \begin{bmatrix} 3 - (-2) \\ - 2 \end{bmatrix} = \begin{bmatrix} 5 \\ -2 \end{bmatrix} $$

Here I have written the results as a column vector, one of the few notations available. There are a few and it is extremely important that you pay attention to the context of the problem because it is assumed in many texts that the reader will know if the discussion concerns a point in space or a vector. The use of one of the following notations is typically present, but in some places other authors are not so careful.

Component Form of a Vector

Given a vector $\mathbf{v}$ with components $v_1,v_2,\ldots,v_n$, there are several possible notations

• Column vector $$ \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} $$
• Transpose coordinates $$ \mathbf{v} = (v_1,v_2,\ldots,v_n)^T $$
• Angle brackets $$ \mathbf{v} = \langle\,v_1,v_2,\ldots,v_n\,\rangle $$
• Just plain lazy $$ \mathbf{v} = (v_1,v_2,\ldots,v_n) $$

The column vector representation is the "most correct," as it fits best with how we represent vectors in linear algebra. However, it is also the one that takes the most real estate when writing. The other notations mostly exist for convenience for either handwriting or printing. (Paper is expensive and our printing presses had limitations in the past, which is why some of these notations were originally necessary.) The transpose coordinate version is just a compact form letting you know what a column vector is meant. The angle brackets are an uncommon vector notation employed by the Stewart text. It is used exclusively to represent component forms, so that parentheses will always represent an ordered pair or triple.

This is not typical. This notation is used primarily to save space in Stewart's textbooks. Angle brackets tend to be reserved for inner products, something we will discuss next section. You instructor will accept angle bracket notation, however student's should use column vector notation or transposed tuples.

The just plain lazy notation is exactly that. It uses the same notation for a point and expects you to know from context that a vector is being implied. In practice, this is the second most common way you are will see vectors used in other classes. For this class, use column vectors or the angle bracket notation while you are getting familiar with vectors. Do not shortcut anything until you are certain you know what you are doing.

Use parentheses only when you intend to communicate a point. Never use the lazy notation and expect your instructor to understand that you meant something else. Misuse of the notation will be interpreted as a lack of understanding. Practice your skills.

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12.2.5 Vector Spaces

Vectors live in vector spaces, and it is typical to refer to a generic $n$-dimensional vector space as $V_n$. Both $\mathbb{R}^2$ and $\mathbb{R}^3$ are vector spaces, so these properties specifically work there, but also in any vector space, as you find if you later study linear algebra. For our purposes, we take the following as given.

Closure Properties

If $\mathbf{u}$ and $\mathbf{v}$ are vectors in the same vector space $V_n$, and $\alpha$ is a scalar, then we have the following properties:

1. Addition

   $\mathbf{u} + \mathbf{v}$ is also a vector in the same Vector Space as $\mathbf{u}$ and $\mathbf{v}$.

2. Scalar Multiplication

   $\alpha\mathbf{u}$ is also a vector in the same Vector Space as $\mathbf{u}$.

Vector Properties

If $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are vectors in the same vector space $V_n$, and $\alpha$ and $\beta$ are scalars, then we have the following properties:

1. Vector Addition Commutes $$ \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} $$
2. Vector Addition Associates $$ \mathbf{u} + \left(\mathbf{v} + \mathbf{w} \right) = \left(\mathbf{u} + \mathbf{v}\right) + \mathbf{w} $$
3. The Zero Vector is the Additive Identity $$ \mathbf{v} + \mathbf{0} = \mathbf{u} $$
4. Each Vector has an Additive Inverse $$ \mathbf{v} + (-\mathbf{v}) = \mathbf{0} $$
5. Scalar Multiplication Distributes Over Vector Addition $$ \alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v} $$
6. Scalar Multiplication Distributes Over Scalar Addition $$ (\alpha + \beta)\mathbf{v} = \alpha\mathbf{v} + \beta\mathbf{v} $$
7. Scalar Multiplication Associates $$ (\alpha\beta)\mathbf{v} = \alpha(\beta\mathbf{v}) $$
8. (The Scalar) One is a Multiplicative Identity $$ 1\mathbf{v} = \mathbf{v} $$

Using these properties, we can perform vector algebra in much the same way as you are familiar with algebra involving formulas and functions by doing the appropriate symbolic manipulations. Also note that we have not said anything about vector multiplication, just scalar multiplication. We will see in the next couple of sections that there are peculiarities with the idea of multiplying vectors and that it does not always just work like we are used to with real numbers.

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12.2.6 Properties of Vector Space $\mathbb{R}^n$

In Euclidean vector space $\mathbb{R}^n$ scalar multiplication of a vector and addition of vectors is defined in terms of coordinates.

Let $\mathbf{u}$, $\mathbf{v}$, and $\boldsymbol{\phi}$ be vectors in $\mathbb{R}^n$, then for indices $1 \le i \le n$,

$$ \begin{align*} \mathbf{u} = \overrightarrow{u} = \overline{u} = \langle u_1, u_2, \dots, u_n \rangle &= \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} = \begin{bmatrix} u_i \end{bmatrix} \\ \mathbf{v} = \overrightarrow{u} = \overline{v} = \langle v_1, v_2, \dots, v_n \rangle &= \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} = \begin{bmatrix} v_i \end{bmatrix} \\ \boldsymbol{\phi} = \overrightarrow{\phi} = \overline{\phi} = \langle \phi_1, \phi_2, \dots, \phi_n \rangle &= \begin{bmatrix} \phi_1 \\ \phi_2 \\ \vdots \\ \phi_n \end{bmatrix} = \begin{bmatrix} \phi_i \end{bmatrix} \end{align*} $$

Example 2

Consider the vector $\boldsymbol{\psi} = \langle 2, 1, -3 \rangle\in\mathbb{R}^3$

The coordinates of vector $\boldsymbol{\psi}$ are

$$ \begin{align*} \psi_1 &= 2 \\ \psi_2 &= 1 \\ \psi_3 &= -3 \end{align*} $$

This notation allows us to describe the second axiom of a vector space (commutativity) using the fact that the coordinates (n-tuple) of a vector in $\mathbb{R}^n$ is a list of $n$ scalars.

If $\mathbf{x}$ and $\mathbf{y}$ are vectors in $\mathbb{R}^n$, then

$$ \begin{align*} \mathbf{x} + \mathbf{y} &= \begin{bmatrix} x_i \end{bmatrix} + \begin{bmatrix} y_i \end{bmatrix} \\ &= \begin{bmatrix} x_i + y_i \end{bmatrix} \\ &= \begin{bmatrix} y_i + x_i \end{bmatrix}\qquad\text{because we have $n$ sums of two scalar components!} \\ &= \begin{bmatrix} y_i \end{bmatrix} + \begin{bmatrix} x_i \end{bmatrix} \\ &= \mathbf{y} + \mathbf{x} \end{align*} $$

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12.2.7 The Standard Basis

We represent points and vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$ using ordered pairs or triples. For example, in 2D we would have $(x,y)$ and $\langle\, x,y\, \rangle$ for a point and its associated coordinate vector. The reason this works is because we have a meaning assigned to the ordering of the coordinates $x$ and $y$. This ordering is associated with the basis that we are using for a coordinate system. A basis is composed of vectors that define the key directions, and hence coordinate system, within a vector space.

In rectangular coordinates (our typical choice) we use something called the standard basis or canonical basis. This basis is composed of a unit vector (a vector of length one) that points in each of the $x$, $y$, and $z$ (if 3D) directions. We call these vectors $\ihat$, $\jhat$, and $\khat$, where in 2D we have

$$ \ihat = \langle\, 1,0\,\rangle\qquad\qquad \jhat = \langle\, 0,1\,\rangle $$

and in 3D

$$ \ihat = \langle\, 1,0,0\,\rangle\qquad\qquad \jhat = \langle\, 0,1,0\,\rangle\qquad\qquad \khat = \langle\, 0,0,1\,\rangle $$

Figure 7

Using the vector properties, we can see that it is possible to represent a vector $\langle\, x,y\,\rangle$ or $\langle\, x,y,z\,\rangle$ in terms of these vectors.

$$ \begin{align*} \langle\, x,y\,\rangle &= x\langle\, 1,0\,\rangle + y\langle\, 0,1\,\rangle = x\,\ihat + y\,\jhat \\ \\ \langle\, x,y,z\,\rangle &= x\langle\, 1,0,0\,\rangle + y\langle\, 0,1,0\,\rangle + z\langle\, 0,0,1\,\rangle = x\,\ihat + y\,\jhat + z\,\khat \end{align*} $$

You may be given information about vectors in either component form or in terms of the standard basis vectors $\ihat$, $\jhat$, and $\khat$. Typically, you report answers using the same notation the information in the exercise is given in unless instructed otherwise.

Exercise 2

Let $\mathbf{a} = 3\ihat - 2\jhat + \khat$ and $\mathbf{b} = 5\jhat - 2\khat$. Determine $3\mathbf{a} + 4\mathbf{b}$ in terms of the standard basis vectors.

Check Your Work

$$ \begin{align*} 3\mathbf{a} + 4\mathbf{b} &= 3\left(3\ihat - 2\jhat + \khat\right) + 4\left(5\jhat - 2\khat\right) \\ \\ &= 9\ihat - 6\jhat + 3\khat + 20\jhat - 10\khat\\ \\ &= 9\ihat + 14\jhat - 7\khat \end{align*} $$

Any vector that is multiplied by a positive scalar points in the same direction as the original vector. So, if we want to find a unit vector that points in the same direction as a given nonzero vector, all we need to is find the magnitude of that vector and multiply the original by the reciprocal of the magnitude.

A Unit Vector

If $\mathbf{v} \neq 0$, then

$$ \mathbf{u} = \dfrac{1}{|\mathbf{v}|}\mathbf{v} = \dfrac{\mathbf{v}}{|\mathbf{v}|} $$

is a unit vector in the same direction as $\mathbf{v}$.

Exercise 3

Find a unit vector that points in the same direction as $\mathbf{a}$ from the previous exercise.

Check Your Work

Begin by finding the magnitude of $\mathbf{a}$,

$$ |\mathbf{a}| = \sqrt{(3)^2 + (-2)^2 + (1)^2} = \sqrt{14} $$
then divide $\mathbf{a}$ by that value $$ \begin{align*} \mathbf{u} &= \dfrac{\mathbf{a}}{|\mathbf{a}|} \\ \\ &= \dfrac{1}{\sqrt{14}}\left(3\ihat - 2\jhat + \khat\right) \\ \\ &= \frac{3}{\sqrt{14}}\ihat - \dfrac{1}{7}\jhat + \frac{1}{\sqrt{14}}\khat \end{align*} $$

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12.2.8 Properties of Additive Identity and Inverse

Theorem 12.2.1 Properties of Additive Identity and Additive Inverse

Let $\mathbf{v}$ be vectors in the plane, and let $c$ be scalars.

1. If $\mathbf{v} + \mathbf{u} = \mathbf{v}$, then $\mathbf{u} = \mathbf{0}$
2. If $\mathbf{v} + \mathbf{u} = \mathbf{0}$, then $\mathbf{u} = -\mathbf{v}$
3. $0\mathbf{v} = \mathbf{0}$
4. $c\mathbf{0} = \mathbf{0}$
5. If $c\mathbf{v} = \mathbf{0}$, then $c=0$ or $\mathbf{v}=\mathbf{0}$
6. $-(-\mathbf{v})= \mathbf{v}$

Proof (of some of the properties)

Let $\mathbf{u}$, and $\mathbf{v}$ be vectors $\mathbb{R}^n$, and let $c$ a scalar.

2. If $\mathbf{v} + \mathbf{u} = \mathbf{0}$, then using our axioms we have

$$ \begin{align*} \mathbf{v} + \mathbf{u} &= \mathbf{0} &\qquad\text{Given} \\ -\mathbf{v} + \left( \mathbf{v} + \mathbf{u} \right) &= -\mathbf{v} + \mathbf{0} &\qquad\text{Axiom 1} \\ -\mathbf{v} + \left( \mathbf{v} + \mathbf{u} \right) &= -\mathbf{v} &\qquad\text{Axiom 4} \\ \left( -\mathbf{v} + \mathbf{v} \right) + \mathbf{u} &= -\mathbf{v} &\qquad\text{Axiom 3} \\ \mathbf{0} + \mathbf{u} &= -\mathbf{v} &\qquad\text{Axiom 5} \\ \mathbf{u} + \mathbf{0} &= -\mathbf{v} &\qquad\text{Axiom 2} \\ \mathbf{u} &= -\mathbf{v} &\qquad\text{Axiom 4} \\ \end{align*} $$

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12.2.9 Magnitude and Length

The magnitude or length of a vector in $\mathbb{R}^n$ is the distance from the base of the vector to its tip. Therefore

Definition

The magnitude, or length of a vector in $\mathbf{a}\in\mathbb{R}^n$ in standard position is the distance from the origin to the tip of the vector,

$$ \left|\mathbf{a}\right| = \sqrt{a_1^2 + \dots + a_n^2} $$

Example 3

This problem is taken from Open Stax Calculus, Volume 3

A force $\mathbf{F}$ or $40 N$ acts on a box in the direction of the vector $\overrightarrow{OP}$, where $P(1,0,2)$. Express the force as a vector using standard unit vectors.

Solution

Vector $\overrightarrow{OP}$ is in standard position, however it is not a unit vector. A unit vector is a vector with magnitude $1$. To find a unit vector that points the same direction as $\overrightarrow{OP}$ one must divide the vector by its length.

$$ \left|\overrightarrow{OP}\right| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5} $$

Now one can determine a unit vector in the direction of both our force $\mathbf{F}$ and $\overrightarrow{OP}$. One knows this because or problem states that force $\mathbf{F}$ acts on the box in the direction of the vector $\overrightarrow{OP}$. The appropriate unit vector $\mathbf{u}$ is

$$ \mathbf{u} = \frac{\overrightarrow{OP}}{\left|\overrightarrow{OP}\right|} = \frac{\langle 1, 0, 2 \rangle}{\sqrt{5}} = \left\langle \frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}} \right\rangle $$

Finally one computes the standard coordinates of force $\mathbf{F}$

$$ \mathbf{F} = 40\mathbf{u} = 40\left\langle \frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}} \right\rangle = \left\langle \frac{40}{\sqrt{5}}, 0, \frac{80}{\sqrt{5}} \right\rangle = \begin{bmatrix} 8\sqrt{5} \\ 0 \\ 16\sqrt{5} \end{bmatrix} $$

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12.2.10 Applications

One of the reasons that vectors are so important is their usefulness in solving applied problems. If there is a physical quantity such as a force or a velocity that we can interpret as a vector, then we can use vector arithmetic to determine aspects related to that quantity. To do this, we will need to put the vector into component form prior to performing our calculations. Unfortunately, the real world does not come equipped with a fixed coordinate system, so we will need to measure the length (magnitude) and angle (direction) of a vector and translate it to a coordinate system that we choose for a particular problem.

The nice thing about the freedom to choose a coordinate system is that we can both place the origin somewhere convenient and orient the axes in the simplest way for the problem. For most problems there is typically one (or maybe two) choices for the coordinate system that simplify the problem the most. Setting things up properly by smartly choosing how to place the coordinate axes is one of most important critical thinking skills to develop as you solve problems. We will use this skill consistently throughout the course in different forms.

Another thing to keep in mind is that vectors are translation invariant. This is another way of saying that their magnitude and direction are the only things that matter. Where a vector is drawn does not. This means that if we are trying to figure out information about a particular vector, we can move the coordinate system for that particular vector, determine its properties, then move to any other vectors we need to deal with.

Exercise 4

Consider a 200 kg beam that is suspended by cables as shown in the image on the right, where the cables are connected at the ends of the beam in a way so that they are both at an angle of $60^\circ$ with the beam. This system is in static equilibrium which means that the sum of all forces acting on it are zero. The three forces acting on the beam are the force of gravity and the tension forces from the two cables in contact with the beam. Determine the tension vector in each of the cables connected to the beam and the magnitude of those tensions.

Video Solution

Suspended Beam

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