This section only expects students to have familiarity with measuring distance and coordinates in three dimensions. If your students are very familiar with vectors and their basic operations, then this section can be covered in one class day. Many instructors present the notation and definitions of vectors and associated operations without a lot of motivation. Activity 9.2.3 offers motivation for why you would want to add, subtract, and rescale vectors using a navigation/mapping motivation.
Quantities like length, speed, area, and mass are all measured using a single number (a scalar). Other quantities, like velocity, force, and displacement, have two attributes: magnitude (or length) and direction. These quantities are represented by vectors and are the focus of this section. For example, we will use vectors to calculate work done by a constant force, calculate torque, determine direction vectors for lines and normal vectors for planes, define curvature, and determine the direction of greatest increase on a surface. For most of these applications, we will be interested in using vectors to measure the direction of some aspect of the application while needing to also pay attention to the magnitude of this measurement. Vectors will be an essential tool for describing the behavior of functions of several variables.
Postscript is a programming language whose primary purpose is to specify how to generate text or graphics. The following is a simple set of Postscript commands that produces the triangle in the plane with vertices \((0,0)\text{,}\)\((1,1)\text{,}\) and \((1,-1)\text{:}\)
Each of these commands encodes two important pieces of information: a direction in which to move and a distance to move. Mathematically, we can capture this information succinctly in a vector. To do so, we record the movement on the map in a pair \(\langle x, y \rangle\text{,}\) where \(x\) is the horizontal displacement and \(y\) the vertical displacement from one point to another. This pair \(\langle x, y \rangle\) is a vector, and we call each number in the pair a component. As an example, the vector from the origin to the point \((1,1)\) is represented by \(\langle 1,1 \rangle\) while the vector from the point \((1,1)\) to the origin is represented by \(\langle -1,-1 \rangle\text{.}\)
Suppose we want to draw the triangle with vertices \(A=(2,3)\text{,}\)\(B=(-3,1)\text{,}\) and \(C=(4,-2)\text{.}\) As a shorthand notation, we will denote the vector from the point \(A\) to the point \(B\) as \(\overrightarrow{AB}\text{.}\) Determine the vectors \(\overrightarrow{AB}\text{,}\)\(\overrightarrow{BC}\text{,}\) and \(\overrightarrow{AC}\text{.}\)
How are the horizontal displacements of \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) related to the horizontal displacement of \(\overrightarrow{AC}\text{?}\) Does the same relationship follow for the vertical displacements? Write a couple of sentences to explain your reasoning in the context of this problem.
Preview Activity 9.2.1 shows how we can record the length and direction of a change in position using a pair of numbers \(\langle x,y\rangle\text{.}\) In particular, these numbers measure the change in each coordinate direction separately. There are many quantities other than displacement, such as force and velocity, that possess the attributes of magnitude and direction. We will call such quantity a vector.
We can represent a vector geometrically as a directed line segment, with the magnitude as the length of the segment and an arrowhead indicating direction, as shown at left in Figure 9.2.2.
According to the definition, a vector possesses the attributes of magnitude (length) and direction; the vector’s position, however, is not mentioned. Consequently, any two vectors having the same magnitude and direction are equal, as shown in Figure 9.2.2.(b). In other words, two vectors are equal provided they have the same magnitude and direction. This means that the same vector may be drawn in the plane in many different ways.
Vectors do not have a particular starting or ending point. However, it will be useful to use directed line segments like \(\overrightarrow{AB}\) as a specific representative of a vector. For instance, suppose that we would like to draw the vector \(\langle 3, 4\rangle\text{,}\) which represents a horizontal change of (positive) three units and a vertical change of (positive) four units. We may place the tail of the vector (the point from which the vector originates) at the origin and the tip (the terminal point of the vector) at \((3,4)\text{,}\) as illustrated in Figure 9.2.3.(a). A vector with its tail at the origin is said to be in standard position.
Alternatively, we may place the tail of the vector \(\langle 3,4\rangle\) at another point, such as \(Q=(1,1)\text{.}\) After a displacement of three units to the right and four units up, the tip of the vector is at the point \(R=(4,5)\) (see the vector in Figure 9.2.3.(b)).
In this example, the vector was represented by the directed line segment from \(Q\) to \(R\text{,}\) which we denote as \(\overrightarrow{QR}\text{.}\) We may also turn the situation around: given the two points \(Q\) and \(R\text{,}\) we obtain the vector \(\langle 3,4\rangle\) because we move horizontally three units and vertically four units to get from \(Q\) to \(R\text{.}\) In other words, \(\overrightarrow{QR} = \langle 3,4\rangle\text{.}\) In general, the vector \(\overrightarrow{QR}\) from the point \(Q = (q_1, q_2)\) to \(R = (r_1, r_2)\) is found by taking the difference of coordinates, so that
We will use arrows over the name/letters to represent vectors, such as \(\vv = \langle 3, 4 \rangle\text{,}\) to distinguish them from scalars. The entries of a vector are called its components; in the vector \(\langle 3, 4 \rangle\text{,}\) the \(x\)-component is \(3\) and the \(y\)-component is \(4\text{.}\) We use pointed brackets \(\langle\, ,\, \rangle\) and the term components to distinguish a vector from a point \((\, ,\, )\) and its coordinates.
There is, however, a close connection between vectors and points. Given a point \(P\text{,}\) we will frequently consider the vector \(\overrightarrow{OP}\) from the origin \(O\) to \(P\text{.}\) For instance, if \(P=(3,4)\text{,}\) then \(\overrightarrow{OP}=\langle 3,4\rangle\) as in Figure 9.2.4. In this way, we think of a point \(P\) as defining a vector \(\overrightarrow{OP}\) whose components agree with the coordinates of \(P\text{.}\) The vector \(\overrightarrow{OP}\) is called the position vector of \(P\text{.}\)
While we often illustrate vectors in the plane since pictures will be easier to draw and demonstrate the necessary characteristics, many situations will require the use of vectors in three or more dimensions. For instance, a vector \(\vv\) in \(n\)-dimensional space, \(\R^n\text{,}\) has \(n\) components and may be represented as
After being bored in your previous calculus class (and definitely not in multivariable calculus) you count the floor tiles to see how large the room is. You now know that the room is 32 feet wide and 42 feet in length. Additionally, your tall friend Chuck can barely jump and touch the ceiling which means the ceiling is 10 feet high. Your calculus professor notices you doing these measurements and decides to create a classroom coordinate system. Your professor walks to the center of the room and notices that their head is five feet above the ground and sets their head as the origin of the classroom coordinate system. Your friend Alice is sitting at \(A = (9, -6, -2.5)\text{,}\) a projector is located at position \(B = (0,1,3)\text{,}\) and your friend Carlos is working while sitting on the floor at \(C = (-2, 20, -5)\text{.}\) All distances are measured in feet.
Because initial and final locations are not mentioned in the definition of a vector, any two vectors that have the same magnitude and direction are equal. It is helpful to have an algebraic way to determine when this occurs. That is, if we know the components of two vectors \(\vu\) and \(\vv\text{,}\) we will want to be able to determine algebraically when \(\vu\) and \(\vv\) are equal. There is an straightforward set of conditions that we use in terms of components.
Two vectors \(\vu = \langle u_1, u_2 \rangle\) and \(\vv = \langle v_1, v_2 \rangle\) in \(\R^2\) are equal if and only if their corresponding components are equal: \(u_1 = v_1\) and \(u_2 = v_2\text{.}\)
More generally, two vectors \(\vu = \langle u_1, u_2, \ldots, u_n\rangle\) and \(\vv = \langle v_1, v_2, \ldots, v_n \rangle\) in \(\R^n\) are equal if and only if \(u_i = v_i\) for every \(i\) from \(1\) to \(n\text{.}\)
Vectors are not numbers (scalars), but we can now represent them with components that are real numbers. As such, we naturally wonder if it is possible to add two vectors together, multiply two vectors, or combine vectors in any other ways. In this section, we will study two operations on vectors: vector addition and scalar multiplication. To begin, we investigate a natural way to think about combining vectors in the context of measuring the change of location.
In this problem, we will be navigating to help find a friend who is lost at a local state park. We will be navigating using traditional map coordinates with east going in the positive \(x\)-direction and north going in the positive \(y\)-direction.
Your friend told you they would be staying 3 km east and 4 km north of the main parking lot. You drive to the parking lot and park next to your friend’s car, then hike 3 km east and 4 km north. You get to the location you expected your friend to be at, but you don’t find your friend and call the ranger station. The ranger station says they think your friend is 1 km west and 4 km north of your current location.
Let \(\vu\) be the vector that represents your hike from your car to the expected location of your friend, \(\vv\) be the vector that represents the vector from your current location to the ranger’s suggested location, and \(\vw\) be the vector from your car to the ranger’s suggested location. Compute the components of \(\vu\text{,}\)\(\vv\text{,}\) and \(\vw\text{.}\)
Draw a picture that represents \(\vu\text{,}\)\(\vv\text{,}\) and \(\vw\) with the reference points \(C\) being the location of your car, \(P\) being your current position, and \(S\) being the ranger’s suggested location.
In the context of this problem, explain why you can add the horizontal components of \(\vu\) and \(\vv\) to get the horizontal component of \(\vw\text{.}\) Write a sentence or two about why this argument should work for the vertical components as well.
After hiking to the location suggested by the ranger, you still don’t see your friend and call the ranger station again. The regional manager of rangers answers this time. She says the first ranger made a mistake in their navigation. A drone spotted your friend along the same direction from your car to point \(P\text{,}\) but your friend is three times as far away from your car as you were when you were at point \(P\text{.}\) In order to avoid confusion or any other mistakes, you want to compute the vector from your car to the drone’s suggested location, which we will call \(D\text{.}\) What are the components of \(\overrightarrow{CD}\text{?}\) How does that compare to \(\vw\text{?}\)
After hiking to the drone’s suggested location, you find your friend. Yay! On the walk back to your car with your friend, you decide to help make sure the first ranger understands vectors. In particular, you want to let them know what direction they should have told you to go from \(P\text{.}\) Use the components of \(\overrightarrow{CD}\) and \(\vu\) to figure out what the first ranger should have told you.
In Activity 9.2.3 we saw how adding two vectors componentwise gave a vector that represented the total change of traveling along one vector followed by doing the change described by the other vector. Additionally, we saw that to go three times as far in a direction given by a vector, we can describe the total change by multiplying each component by three. Finally, we saw that we can find the difference between two vectors by componentwise subtraction. This difference vector completed a triangle formed by positioning the two subtracted vectors with the same initial point. The ideas of adding vectors or multiplying a scalar by a vector generalizes beyond the context of Activity 9.2.3.
In general, we can add vectors \(\vv=\langle v_1,\ldots,v_n \rangle\) and \(\vw=\langle w_1,\ldots,w_n \rangle\) componentwise to get \(\vv+\vw=\langle v_1+w_1,\ldots,v_n+w_n\rangle\text{,}\) where \(\vv+\vw\) gives the total change described by \(\vv\) then \(\vw\text{.}\) We define the scalar multiplication of vectors componentwise as well. If \(k\) is a real number, then \(k\vv = \langle k v_1,\ldots, k v_n \rangle \text{.}\) Geometrically, the scalar multiplication of \(k\) by \(\vv\) will scale (stretch/shrink) \(\vv\) by a factor of \(k\text{.}\) We will discuss the way different values of \(k\) scale vectors in Subsection 9.2.6.
We can now add vectors and multiply vectors by scalars, and thus we can add together scalar multiples of vectors. This allows us to define vector subtraction, \(\vv - \vu\text{,}\) as the sum of \(\vv\) and \(-1\) times \(\vu\text{,}\) so that
You can alternatively think of of \(\vv-\vu\) as the vector you need to add to \(\vu\) to get a result of \(\vv\) (as you did in part e of Activity 9.2.3).
Using vector addition and scalar multiplication, we will often represent vectors in terms of the special vectors \(\vi = \langle 1, 0 \rangle\) and \(\vj = \langle 0,1 \rangle\text{.}\) For instance, we can write the vector \(\langle a, b \rangle\) in \(\R^2\) as
\begin{equation*}
\langle a, b \rangle = a\langle 1, 0 \rangle + b\langle 0, 1 \rangle = a \vi + b \vj,
\end{equation*}
In \(\R^3\text{,}\) we let \(\vi = \langle 1, 0, 0 \rangle\text{,}\)\(\vj = \langle 0,1,0 \rangle\text{,}\) and \(\vk = \langle 0,0,1 \rangle\text{,}\) and we can write the vector \(\langle a, b, c \rangle\) in \(\R^3\) as
\begin{equation*}
\langle a, b,c \rangle = a\langle 1, 0,0 \rangle + b\langle 0, 1,0 \rangle + c\langle 0,0,1 \rangle = a \vi + b \vj + c \vk.
\end{equation*}
We know that the sum (of scalars) \(1+2\) is equal to the sum \(2+1\text{.}\) This is called the commutative property of scalar addition. Any time we define operations on objects (like addition of vectors) we usually want to know what kinds of properties the operations have. For example, is addition of vectors a commutative operation? To answer this question we take two arbitrary vectors \(\vv\) and \(\vu\) and add them together and see what happens. Let \(\vv = \langle v_1, v_2 \rangle\) and \(\vu = \langle u_1, u_2 \rangle\text{.}\) Now we use the fact that \(v_1\text{,}\)\(v_2\text{,}\)\(u_1\text{,}\) and \(u_2\) are scalars, and that the addition of scalars is commutative to see that
So the vector sum is a commutative operation. Similar arguments can be used to show the following properties of vector addition and scalar multiplication.
The vector \(\vzero = \langle 0, 0, \ldots, 0 \rangle\) has the property that \(\vv + \vzero = \vv\text{.}\) The vector \(\vzero\) is called the zero vector.
We verified the first property for vectors in \(\R^2\text{.}\) The other properties can be verified to hold for all vectors in \(\R^n\) in a similar manner because the operations are done componentwise on the vectors.
In this activity, use Properties of vector operations to simplify each of the following expressions into a single vector of the form \(\langle a,b,c \rangle\text{.}\) For this activity, let \(\vu=\langle 1,-2,3 \rangle\text{,}\)\(\vv=\langle 2,-2,5\rangle\text{,}\) and \(\vw=\langle -3,0,3\rangle\text{.}\)
Subsection9.2.6Geometric Interpretation of Vector Operations
Next, we explore the geometric representations of vector addition and scalar multiplication which will allow us to visualize these operations. Let \(\vu = \langle 4, 6 \rangle\) and \(\vv = \langle 3, -2 \rangle\text{.}\) Then \(\vw = \vu + \vv = \langle 7, 4 \rangle\text{,}\) as shown in Figure 9.2.5.(a).
If we think of these vectors as displacements measuring change in position in the plane, we can see a geometric way to envision vector addition. For instance, the vector \(\vu + \vv\) will represent the displacement obtained by following the displacement \(\vu\) with the displacement \(\vv\text{.}\) We may picture this by placing the tail of \(\vv\) at the tip of \(\vu\text{,}\) as seen in Figure 9.2.5.(b). In other words, the change described by \(\vu+\vv\) is given by going along \(\vu\) then immediately along \(\vv\text{.}\)
Vector addition is commutative. This means we obtain the same sum if we place the tail of \(\vu\) at the tip of \(\vv\text{.}\) We therefore see that \(\vu+\vv\) appears as the diagonal of the parallelogram determined by \(\vu\) and \(\vv\text{,}\) as shown in Figure 9.2.5.(c).
Vector subtraction has a similar interpretation. In Figure 9.2.6.(a) we see vectors \(\vu\text{,}\)\(\vv\text{,}\) and \(\vw =\vu + \vv\text{.}\) If we rewrite \(\vv = \vw - \vu\text{,}\) we have the arrangement shown at right in Figure 9.2.6.(b). In other words, to form the difference \(\vw-\vu\text{,}\) we draw a vector from the tip of \(\vu\) to the tip of \(\vw\text{.}\) This is consistent with our other interpretation of vector subtraction as well: the vector \(\vw-\vu\) is what we need to add to \(\vu\) to get \(\vw\text{.}\) This is stated algebraically as \(\vu + \underline{\vw-\vu} = \vw\text{.}\)
In a similar way, we may geometrically represent a scalar multiple of a vector. For instance, if \(\vv=\langle 2,3\rangle\text{,}\) then \(2\vv = \langle 4,6\rangle\text{.}\) As shown in Figure 9.2.7, multiplying \(\vv\) by 2 leaves the direction unchanged, but stretches \(\vv\) by 2. Remember that \(2\vv = \vv +\vv\text{,}\) which describes the total change of doing \(\vv\) then \(\vv\) again and thus gives a result that will be in same direction as \(\vv\text{.}\) Also, \(-2\vv = \langle -4, -6\rangle\text{,}\) which shows that multiplying by a negative scalar gives a vector pointing in the opposite direction of \(\vv\text{.}\)
Figure 9.2.7 motivates the following definition of parallel vectors: vectors \(\vv\) and \(\vw\) in \(\R^n\) are parallel provided that there exists a non-zero scalar \(k\in\R\) such that \(\vv = k \vw\text{.}\) In other words, parallel vectors are (non-zero) scalar multiples of each other. There are a couple of things to note that are important about parallel vectors. First, the zero vector \(\vzero = \langle 0, \ldots, 0\rangle \in \R^n\) is not parallel to any other vector. Geometrically, this should make sense because the zero vector is the only vector that has no change and that will not correspond to any direction. Second, parallel vectors do not have to be in the same direction; in fact, parallel vectors can be in exactly opposite directions, as shown by \(\vv\) and \(-2\vv\) in Figure 9.2.7. Parallel vectors are important when trying to classify the orientation of lines, planes, and changes on surfaces.
On the set of axes in Figure 9.2.8.(b), sketch the vectors \(\vu-3\vv\text{,}\)\(\vu-2\vv\text{,}\)\(\vu-\vv\text{,}\)\(\vu + \vv\text{,}\) and \(\vu + 2\vv\text{.}\)
Before we move on to the magnitude of a vector, we will take a moment to point out a concept that permeates most the mathematics you will see for awhile. Our definitions of vector addition and scalar multiplication give us a way to take a few vectors and generate entire spaces with both familiar algebraic and geometric properties. We say that \(a\vv + b\vw\) is a linear combination of the vectors \(\vv\) and \(\vw\) (where \(a\) and \(b\) are scalars). Earlier in this section we mentioned how every vector in \(\R^3\) can be written as a linear combination of \(\vi\text{,}\)\(\vj\text{,}\) and \(\vk\text{.}\) Important questions in the next few math courses include “Can you write any vector in \(\R^n\) as a linear combination of \(\vu\text{,}\)\(\vv\text{,}\) and \(\vw\text{?}\)” For now, we will point out that much of what has been done in this section is setting up the algebraic and geometric properties of linear combinations as a function of the vectors used.
By definition, vectors have both direction and magnitude (or length). We will now investigate how to calculate the magnitude of a vector. Since a vector \(\vv\) can be represented by a directed line segment, we can use the distance formula to calculate the length of the segment. This length is the magnitude of the vector \(\vv\) and is denoted \(\vecmag{\vv}\text{.}\)
The magnitude of a vector is measured as a the length of the directed line segments that represent the vector. For example, the directed line segment \(\overrightarrow{AB}\) in Figure 9.2.10.(a) will represent a vector with components of \(2\) and \(4\text{,}\) which represent the horizontal and vertical changes from \(A\) to \(B\text{.}\) Notice that when we apply the distance formula to find the length of the segment from \(A\) to \(B\text{,}\) we compute \((x_B-x_A)\) and \((y_B-y_A)\text{,}\) which are the components of the vector we are finding the magnitude of.
Because the distance formula in \(\R^n\) uses the sum of the squares of the change in each coordinate, we can see that given a vector \(\vv=\langle v_1, v_2, \ldots, v_n \rangle \in \R^n\) will have magnitude \(\vecmag{\vv} = \sqrt{v_1^2+v_2^2+\ldots+v_n^2}\text{.}\) Vectors of length one will be useful for describing directions without a focus on the magnitude. A unit vector is a vector whose magnitude is 1. Scalar multiplication gives us a very useful tool for describing a unit vector in a particular direction because we will need to stretch/shrink a given vector to length 1. Our vector given by \(\overrightarrow{AB}=\langle 2,4\rangle\) will have magnitude \(\sqrt{20}\text{,}\) so the vector \(\frac{1}{\sqrt{20}}\langle2,4\rangle\) will have length 1 and will be in the same direction as \(\overrightarrow{AB}\text{.}\)
Let \(\vu = \langle 2,3\rangle\) and \(\vv = \langle -1,2\rangle\text{.}\) Find \(\vecmag{\vu}\text{,}\)\(\vecmag{\vv}\text{,}\) and \(\vecmag{\vu+\vv}\text{.}\) Is it true that \(\vecmag{\vu+\vv} = \vecmag{\vu}+\vecmag{\vv}\text{?}\)
With the vector \(\vu = \langle 2,3\rangle\text{,}\) find the lengths of \(2\vu\text{,}\)\(3\vu\text{,}\) and \(-2\vu\text{,}\) respectively, and use proper notation to label your results.
A vector is an object that possesses the attributes of magnitude and direction. Examples of vector quantities are position, velocity, acceleration, and force.
Two vectors are equal if they have the same direction and magnitude. Notice that position is not considered, so a vector is independent of its location.
If \(\vu = \langle u_1, u_2, \ldots, u_n \rangle\) and \(\vv = \langle v_1, v_2, \ldots, v_n \rangle\) are two vectors in \(\R^n\text{,}\) then their vector sum is the vector
A vector \(\vu\) is a unit vector provided that \(\vecmag{\vu} = 1\text{.}\) If \(\vv\) is a nonzero vector, then the vector \(\frac{\vv}{\vecmag{\vv}}\) is a unit vector with the same direction as \(\vv\text{.}\)
(b) The vector \(\vec w\) in 3-space of length 3 lying in the \(xz\)-plane pointing upward at an angle of \(\pi/4\) measured from the positive \(x\)-axis.
Find all vectors \(\vec v\) in 2 dimensions having \(||\vec v|| = 17\) where the \(\,\mathit{\vec j}\)-component of \(\vec v\) is \(8 \,\mathit{\vec j}\text{.}\)
Which is traveling faster, a car whose velocity vector is \(28\vec i + 33\vec j\text{,}\) or a car whose velocity vector is \(40\vec i\text{,}\) assuming that the units are the same for both directions?
A truck is traveling due north at \(50\) km/hr approaching a crossroad. On a perpendicular road a police car is traveling west toward the intersection at \(55\) km/hr. Both vehicles will reach the crossroad in exactly one hour. Find the vector currently representing the displacement of the truck with respect to the police car.
Let \(\vu = \langle 2, 1 \rangle\) and \(\vv = \langle 1, 2 \rangle\text{.}\)
Determine the components and draw geometric representations of the vectors \(2\vu\text{,}\)\(\frac{1}{2}\vu\text{,}\)\((-1)\vu\text{,}\) and \((-3)\vu\) on the same set of axes.
Determine the components and draw geometric representations of the vectors \(\vu + \vv\text{,}\)\(\vu + 2\vv\text{,}\) and \(\vu + 3\vv\) on the same set of axes.
Determine the components and draw geometric representations of the vectors \(\vu - \vv\text{,}\)\(\vu - 2\vv\text{,}\) and \(\vu - 3\vv\) on the same set of axes.
Recall that \(\vu - \vv = \vu + (-1)\vv\text{.}\) Sketch the vectors \(\vu\text{,}\)\(\vv\text{,}\)\(\vu+\vv\text{,}\) and \(\vu-\vv\) on the same set of axes. Use the “tip to tail” perspective for vector addition to explain the geometric relationship between \(\vu\text{,}\)\(\vv\text{,}\) and \(\vu - \vv\text{.}\)
Recall that given any vector \(\vv\text{,}\) we can calculate its length, \(|\vv|\text{.}\) Also, we say that two vectors that are (non-zero) scalar multiples of one another are parallel.
Let \(\vv = \langle 3,4 \rangle\) in \(\R^2\text{.}\) Compute \(|\vv|\text{,}\) and determine the components of the vector \(\vu = \frac{1}{|\vv|} \vv\text{.}\) What is the magnitude of the vector \(\vu\text{?}\) How does its direction compare to \(\vv\text{?}\)
Let \(\vv = \langle 2, 3, 5 \rangle\) in \(\R^3\text{.}\) Compute \(|\vv|\text{,}\) and determine the components of the vector \(\vu = \frac{1}{|\vv|} \vv\text{.}\) What is the magnitude of the vector \(\vu\text{?}\) How does its direction compare to \(\vv\text{?}\)
A force (like gravity) has both a magnitude and a direction. If two forces \(\vu\) and \(\vv\) are applied to an object at the same point, the resultant force on the object is the vector sum of the two forces. When a force is applied by a rope or a cable, we call that force tension. Vectors can be used to determine tension.
As an example, suppose a painting weighing 50 pounds is to be hung from wires attached to the frame as illustrated in Figure 9.2.11. We need to know how much tension will be on the wires to know what kind of wire to use to hang the picture. Assume the wires are attached to the frame at point \(O\text{.}\) Let \(\vu\) be the vector emanating from point \(O\) to the left and \(\vv\) the vector emanating from point \(O\) to the right. Assume \(\vu\) makes a \(60^{\circ}\) angle with the horizontal at point \(O\) and \(\vv\) makes a \(45^{\circ}\) angle with the horizontal at point \(O\text{.}\) Our goal is to determine the vectors \(\vu\) and \(\vv\) in order to calculate their magnitudes.
Treat point \(O\) as the origin. Use trigonometry to find the components \(u_1\) and \(u_2\) so that \(\vu = u_1 \vi + u_2 \vj\text{.}\) Since we don’t know the magnitude of \(\vu\text{,}\) your components will be in terms of \(|\vu|\) and the cosine and sine of some angle. Then find the components \(v_1\) and \(v_2\) so that \(\vv = v_1 \vi + v_2 \vj\text{.}\) Again, your components will be in terms of \(|\vv|\) and the cosine and sine of some angle.
The total force holding the picture up is given by \(\vu+\vv\text{.}\) The force acting to pull the picture down is given by the weight of the picture. Find the force vector \(\vw\) acting to pull the picture down.
The picture will hang in equilibrium when the force acting to hold it up is equal in magnitude and opposite in direction to the force acting to pull it down. Equate these forces to find the components of the vectors \(\vu\) and \(\vv\text{.}\)
