Now that students have (hopefully) a decent grasp of vectors from a geometric perspective, this section introduces vectors from an algebraic perspective. Topics include the component form of a vector, unit vectors, formulas for computing vector components and vector magnitude, normalizing a vector, and static equilibrium. In all, students might feel like this section is a repeat of the previous section; however, it will feel a bit "crunchier" because of the computations involved.
Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.
(click to expand)
Simplifying Expressions
Adding and Subtracting Polynomials: Adding vectors component-wise (\(\langle a_1 + a_2, b_1 + b_2 \rangle\)) mirrors combining like terms in polynomial addition.
The following is a list of learning objectives for this section.
Learning Objectives (click to expand)
Express a vector in coordinate form.
Convert a vector between geometric and coordinate forms.
Find the unit vector in the same direction as a given vector.
Scale a vector.
Perform arithmetic on vectors.
Use vectors in force and static equilibrium problems.
Components of a Vector
We have seen that we can add two vectors to get a third, or resultant, vector. We can also break down a vector into two or more component vectors. Breaking a vector into horizontal and vertical components is helpful for many applications.
For example, the beetle on the conveyor belt moved at 4.47 inches per second in the direction \(26.6^{\circ}\). If we set up coordinate axes, as shown in Figure \(\PageIndex{1}\), the vector \(\vec{v}\) representing its velocity is the sum of the beetle's motion in the \(x\)-direction, \(\vec{v}_x\), at 4 inches per second, and its motion in the \(y\)-direction, \(\vec{v}_y\), at 2 inches per second.
Figure \(\PageIndex{1}\)
We can break down any vector into its \(x\)- and \(y\)-components and the sum of those components is equal to the original vector. In other words, \[\vec{v} = \vec{v}_{x} + \vec{v}_{y}.\nonumber\]
Definition: Vector Components
A (two-dimensional) vector \( \vec{v} \) has two vector components - a horizontal vector component and a vertical vector component. These vector components are themselves vectors. The horizontal vector component is denoted \( \vec{v}_x \) and the vertical vector component is denoted \( \vec{v}_y \). These are the only two vectors (of horizontal and vertical direction) whose sum is the original vector \( \vec{v} \).
To find the component vectors, we use the following theorem.
Definition: Components
If \( \theta \) is the angle measured counter-clockwise from the positive \( x \)-axis to the vector \( \vec{v} \), where the tail of \( \vec{v} \) is at the origin, then the scalar quantities given by\[ \begin{array}{rcl}
v_x & = & \norm{\vec{v}} \cos\left( \theta \right) \\[6pt] & \text{and} & \\[6pt] v_y & = & \norm{\vec{v}} \sin\left( \theta \right) \\[6pt] \end{array} \nonumber \]are called simply the components of the vector \(\vec{v}\).
Note that the components \(v_x\) and \(v_y\) of a vector are scalars; they are not vectors. They can be either positive or negative (or zero), as shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\)
Again, for clarification, \(\vec{v}_x\) is the red horizontal vector in each of the images in Figure \(\PageIndex{2}\). At the same time, \(v_x\) is a scalar representing the signed length of \(\vec{v}_x\). A similar statement holds for the vertical vector \(\vec{v}_y\).
Example \(\PageIndex{1}\)
A plane flies 300 miles per hour at a heading of \(300^{\circ}\). Find the \(x\)- and \(y\)-components of its velocity.
Solution
We draw a triangle showing the plane's velocity, \(\vec{v}\), as the sum of components in the \(x\)- and \(y\)-directions.
Figure \(\PageIndex{3}\)
Since the heading is \(300^{\circ}\), \(\vec{v}\) makes an angle of \(30^{\circ}\) with the horizontal. The angle \(\theta\), on the other hand, is \(150^{\circ}\) because \(\theta\) is always measured from the positive \(x\)-axis. Therefore,\[\begin{array}{rcl}
v_x & = & 300 \cos\left(150^{\circ}\right) \approx -259.81 \\[6pt] & \text{and} & \\[6pt] v_y & = & 300 \sin \left(150^{\circ}\right) = 150 \\[6pt] \end{array} \nonumber\]
Checkpoint \(\PageIndex{1}\)
The wind blows 50 kilometers per hour at a heading of \(100^{\circ}\). Find the \(x\)- and \(y\)-components of its velocity.
Answer
\(v_x \approx 49.2\) kph and \(v_y \approx 8.7\) kph
We now know how to resolve a vector into horizontal and vertical vector components. Thus, every vector can be expressed as the sum of a horizontal vector and a vertical vector.
Example \(\PageIndex{2}\)
A vector \(\vec{w}\) has magnitude 4 and direction \(\theta=29^{\circ}\), where \(\theta\) is measured counterclockwise from the positive \(x\)-axis. Express \(\vec{w}\) as the sum of a horizontal vector, \(\vec{w}_x\), and a vertical vector, \(\vec{w}_y\).
Figure \(\PageIndex{4}\)
Then \(\vec{w}=\vec{w}_x+\vec{w}_y\), where \(\vec{w}_x\) is a horizontal vector of magnitude 3.50, and \(\vec{w}_y\) is a vertical vector of magnitude 1.94. See the figure above.
Checkpoint \(\PageIndex{2}\)
A vector \(\vec{u}\) has magnitude 2 and direction \(\theta=116^{\circ}\), where \(\theta\) is in standard position. Express \(\vec{u}\) as the sum of a horizontal vector, \(\vec{u}_x\), and a vertical vector, \(\vec{u}_y\).
Answer
\(\vec{u}=\vec{u}_x+\vec{u}_y\), where \(\norm{\vec{u}_x}=-1.943,\norm{\vec{u}_y}=0.473\)
Using Components
We can describe a vector completely using either magnitude and direction, or components. Many calculations with vectors are more straightforward when we use components. For example, to add two vectors using components, we don't need the Law of Sines and the Law of Cosines. We resolve each vector into its horizontal and vertical components, add the corresponding components, and then compute the magnitude and direction of the resultant vector.
Before we get there, however, let's introduce a couple of theorems that should make sense to you.
To calculate magnitude and direction from the components, we need only Right Triangle Trigonometry.
Example \(\PageIndex{3}\)
After flying for some time, the airplane in the previous example encounters a steady wind blowing at 40 miles per hour at a heading of \(100^{\circ}\). What are the actual speed and heading of the airplane relative to the ground?
Solution
We want to add the vectors \(\vec{v}\), representing the plane's intended velocity, and \(\vec{u}\), representing the velocity of the wind. We first resolve \(\vec{u}\) into its components.\[\begin{array}{rcl}
u_x & = & 40 \cos \left(-10^{\circ}\right) \approx 39.39 \\[6pt] & \text{and} & \\[6pt] u_y & = & 40 \sin \left(-10^{\circ}\right) \approx -6.95 \\[6pt] \end{array} \nonumber\]We find the components of the resultant, \(\vec{w}\), by adding the components of \(\vec{u}\) and \(\vec{v}\), as shown below.
Figure \(\PageIndex{5}\)
That is,\[\begin{array}{rcccl}
w_x & = & u_x + v_x & \approx & 39.39 - 259.81 = -220.42 \\[6pt] & & & \text{and} & \\[6pt] w_y & = & u_y + v_y & \approx & -6.95 + 150 = 143.05 \\[6pt] \end{array} \nonumber\]Finally, we compute the magnitude and direction of the resultant vector.\[\norm{\vec{w}}^2 \approx (-220.49)^2 + (143.05)^2 \approx 69{,}079.14\nonumber\]Therefore, \(\norm{\vec{w}} \approx 262.83\). So, the airplane's ground speed is approximately 262.83 miles per hour.
To find the plane's true heading we compute\[\tan\left( \theta \right) \approx \dfrac{143.05}{-220.42} \implies \widehat{\theta} \approx \tan^{-1}\left( \dfrac{143.05}{220.42} \right).\nonumber\]We don't need to compute the value of \(\theta\) because we are going to state the true heading of the plane. We only need to know that \(\theta\) is a second-quadrant angle. We will use the reference angle, \(\widehat{\theta}\), to get the heading.
Thus, the true heading of the plane is \(270^{\circ} + \widehat{\theta} \approx 302.98^{\circ}\).
Caution: Use Reference Angles and Quadrants
In Example \(\PageIndex{3}\), the airplane's heading is not \(\tan ^{-1}\left( -\frac{143.02}{220.42} \right) \approx -32.98^{\circ}\). Remember that there are always two angles with a given tangent. This is why we only consider reference angles when computing the values of inverse trigonometric functions. We can refer to a sketch of the vector or to the signs of its components to decide which of those two angles is appropriate.
Checkpoint \(\PageIndex{3}\)
A plane is flying in a wind blowing 50 kilometers per hour in a direction \(10^{\circ}\) south of due west. The plane has an airspeed of \(200 \mathrm{kph}\) and heading \(142^{\circ}\). What is the ground speed and actual direction of the plane?
Answer
groundspeed \(182 \mathrm{kph}, 66^{\circ} \mathrm{S}\) of \(\mathrm{E}\)
Unit Vectors
It is often helpful to describe a vector by giving its horizontal and vertical vector components instead of its magnitude and direction. To make the notation easier, we give names to the vectors of length 1 that point in the \(x\)- and \(y\)-directions.
Definition: Unit Vector
A vector of magnitude 1 is called a unit vector.
We can have unit vectors in any direction, but the unit vector in the \(x\)-direction is traditionally denoted by \(\vec{i}\), and the unit vector in the \(y\)-direction is represented by \(\vec{j}\), as shown in Figure \(\PageIndex{6}\).
Figure \(\PageIndex{6}\)
By taking scalar multiples of \(\vec{i}\) and \(\vec{j}\), we can describe any vector that lies in the directions of the coordinate axes. For example, \(4 \vec{i}\) represents the vector of magnitude 4 pointing in the \(x\)-direction, and \(3 \vec{j}\) represents the vector of magnitude 3 pointing in the \(y\)-direction. By adding multiples of \(\vec{i}\) and \(\vec{j}\), we can represent any vector we like. The vector \(\vec{v}=4 \vec{i}+3 \vec{j}\) is shown in Figure \(\PageIndex{6}\).
Because the components of the vector are chosen to align with the coordinate system, we call this the vector's coordinate form.
Definition: Coordinate Form (of a vector)
The vector \( \vec{v} \) is said to be wirtten in coordinate form when it is written as\[\vec{v}=v_x \vec{i}+v_y \vec{j},\nonumber \]where \( v_x \) and \( v_y \) are the horizontal and vertical components of \( \vec{v} \), respectively.
The coordinate form of the vector \(\vec{w}\) in the previous example is \(\vec{w}=3.50 \vec{i}+1.94 \vec{j}\).
Example \(\PageIndex{4}\)
State the coordinate form of the vector shown below.
Figure \(\PageIndex{7}\)
Solution
From its base to its tip, the vector extends 4 units in the negative \(\vec{i}\) direction and 6 units in the \(\vec{j}\) direction. Thus, \(\vec{v}=-4 \vec{i}+6 \vec{j}\).
Checkpoint \(\PageIndex{4}\)
State the coordinate form of the vector shown below.
Figure \(\PageIndex{8}\)
Answer
\(\vec{w}=3 \vec{i}-5 \vec{j}\)
Converting Between Geometric and Coordinate Form
Finding the magnitude and direction of a vector given in coordinate form is a simple matter. The vector \(\vec{v}=4 \vec{i}+3 \vec{j}\) has magnitude\[\norm{\vec{v}}=\sqrt{3^2+4^2}=\sqrt{25}=5\nonumber\]and direction \(\theta=\tan ^{-1}\left(\frac{3}{4}\right) \approx 36.9^{\circ}\). Thus, we can readily convert vectors from geometric form to coordinate form or vice versa.
Theorem: Conversion Between Geometric and Coordinate Forms
Given the vector \(\vec{v}\) with direction \(\theta\) (in standard position) and coordinate form \(\vec{v}=v_x \, \vec{i}+v_y \, \vec{j}\), then\[ \begin{array}{rcl}
v_x & = & \norm{\vec{v}} \cos\left( \theta \right) \\[6pt] & \text{and} & \\[6pt] v_y & = & \norm{\vec{v}} \sin\left( \theta \right) \\[6pt] \end{array} \nonumber \]where \(\norm{\vec{v}} = \sqrt{v_x^2+v_y^2}\) and \( \tan\left( \theta \right) = \frac{v_y}{v_x} \).
Example \(\PageIndex{5}\)
Find the geometric form of the vector \(\vec{w}=-3 \vec{i}-2 \vec{j}\).
Find the coordinate form of the vector whose magnitude is 5 and whose direction is the angle \(\theta=135^{\circ}\).
Solutions
The magnitude of \(\vec{w}\) is \(\norm{\vec{w}}=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}\). The direction satisfies the equation \(\tan\left( \theta \right)=\frac{-2}{-3} = \frac{2}{3}\). Because the point \((-3,-2)\) lies in the third quadrant, we know that \(\theta\) must be a third quadrant angle. Thus, \(\theta=\arctan\left( \frac{2}{3}\right)+180^{\circ} \approx 213.69^{\circ}\).
The coordinates of the vector are given by\[a=5 \cos \left(135^{\circ}\right)=-\dfrac{5}{\sqrt{2}}, \quad b=5 \sin \left(135^{\circ}\right)=\dfrac{5}{\sqrt{2}},\nonumber\]so the coordinate form is \(\vec{w}=-\frac{5}{\sqrt{2}} \vec{i}+\frac{5}{\sqrt{2}} \vec{j}\).
Checkpoint \(\PageIndex{5}\)
Find the geometric form of the vector \(\vec{F}=-6 \vec{i}+8 \vec{j}\).
Answer
magnitude 10, direction \(127^{\circ}\)
Scalar Multiples of Vectors in Coordinate Form
Scalar multiplication is easy to compute in coordinate form. This is because the sides of similar triangles are proportional.
Figure \(\PageIndex{9}\) shows the vectors \(\vec{u}=3 \vec{i}+2 \vec{j}\) and \(\vec{v}=2 \vec{u}\). The vector \(\vec{v}\) is twice as long as \(\vec{u}\) and points in the same direction as \(\vec{u}\). You can see that each component of \(\vec{v}\) is twice the corresponding component of \(\vec{u}\), so that \(\vec{v}=6 \vec{i}+4 \vec{j}\).
Figure \(\PageIndex{9}\)
Theorem: Scalar Multiplication (coordinate form)
If \(\vec{v}=v_x \, \vec{i}+v_y \, \vec{j}\) and \(k\) is a scalar, then\[k \vec{v}=k v_x \, \vec{i}+k v_y \, \vec{j}.\nonumber \]
In other words, to find a scalar multiple of a vector in coordinate form, we multiply each component by the scalar.
Example \(\PageIndex{6}\)
The vector \(\vec{v}\) has coordinate form \(\vec{v}=4 \vec{i}-7 \vec{j}\). Find the coordinate form for the vector \(\vec{w}=-\frac{1}{2} \vec{v}\).
Solution
We multiply each component of \(\vec{v}\) by \(-\frac{1}{2}\) to get \(\vec{w}=-2 \vec{i}+\frac{7}{2} \vec{j}\).
Checkpoint \(\PageIndex{6}\)
Find the coordinate form for the vector \(\vec{u}=3 \vec{v}\), where \(\vec{v}=-12 \vec{i}+15 \vec{j}\).
Answer
\(\vec{u}=-36 \vec{i}+45 \vec{j}\)
If we divide a non-zero vector \(\vec{v}\) by its length, we create a unit vector that points in the same direction as \(\vec{v}\).
Definition: Normalizing (a vector)
The process of dividing a vector by its magnitude is called normalizing.
Normalizing vectors becomes incredibly important in Linear Algebra. Dividing a vector by a scalar \(k\) is the same as multiplying the vector by \(\frac{1}{k}\).
Example \(\PageIndex{7}\)
Find a unit vector pointing in the same direction as \(\vec{v}=2 \vec{i}-4 \vec{j}\).
Solution
We first compute the magnitude of \(\vec{v}\).\[\norm{\vec{v}}=\sqrt{2^2+(-4)^2}=\sqrt{20}=2 \sqrt{5}.\nonumber\]Next we multiply \(\vec{v}\) by the reciprocal of its length. The unit vector in the same direction as \(\vec{v}\) is then\[\begin{array}{rcl}
\vec{u} & = & \dfrac{1}{\norm{\vec{v}}} \vec{v} \\[6pt] & = & \dfrac{1}{2 \sqrt{5}}(2 \vec{i}-4 \vec{j}) \\[6pt] & = & \dfrac{1}{2 \sqrt{5}} \cdot 2 \vec{i}+\dfrac{1}{2 \sqrt{5}} \cdot(-4 \vec{j}) \\[6pt] & = & \dfrac{1}{\sqrt{5}} \vec{i}-\dfrac{2}{\sqrt{5}} \vec{j} \\[6pt] \end{array}\nonumber\]
By computing its length, you can check that the vector \(\vec{u}\) found in Example \(\PageIndex{7}\) is a unit vector.\[\norm{\vec{u}}=\sqrt{\left(\dfrac{1}{\sqrt{5}}\right)^2+\left(-\dfrac{2}{\sqrt{5}}\right)^2}=\sqrt{\dfrac{1}{5}+\dfrac{4}{5}}=\sqrt{1}=1.\nonumber\]Once we have a unit vector, \(\vec{u}\), that points in a given direction, we can create a vector of any length in that direction, simply by scaling \(\vec{u}\) by the length we desire. For example, the vector of length 10 pointing in the same direction as \(\vec{v}\) in the previous example is\[\begin{array}{rcl}
10 \vec{u} & = & 10 \cdot\left(\dfrac{1}{\sqrt{5}} \vec{i}+\dfrac{-2}{\sqrt{5}} \vec{j}\right) \\[6pt] & = & 10 \cdot \dfrac{1}{\sqrt{5}} \vec{i}+10 \cdot \dfrac{-2}{\sqrt{5}} \vec{j} \\[6pt] & = & 2 \sqrt{5} \vec{i}-4 \sqrt{5} \vec{j} \\[6pt] \end{array} \nonumber\]
Theorem: Unit Vector Scaling
The unit vector \(\vec{u}\) in the direction of \(\vec{v}\) is given by \(\vec{u}=\frac{1}{\norm{\vec{v}}} \vec{v}\).
A vector \(\vec{w}\) of length \(k\) in the direction of \(\vec{v}\) is given by \( \vec{w}=\frac{k}{\norm{\vec{v}}} \vec{v}\).
Checkpoint \(\PageIndex{7}\)
Find a unit vector \(\vec{u}\) and a vector \(\vec{w}\) of length 10 pointing in the same direction as \(\vec{v}=-\vec{i}+3 \vec{j}\).
It is also easy to add vectors in coordinate form. Figure \(\PageIndex{10}\) shows the sum of \(\vec{u}=\vec{i}+2 \vec{j}\) and \(\vec{v}=5 \vec{i}+3 \vec{j}\). Remember that we add two vectors by following the first vector by the second.
Figure \(\PageIndex{10}\)
The vector sum describes the following path: 1 unit in the \(x\)-direction, then 2 units in the \(y\)-direction, then 5 units in the \(x\)-direction, then 3 units in the \(y\)-direction.
However, we arrive at the same endpoint by traveling 1 + 5 = 6 units in the \(x\)-direction, then 2 + 3 = 5 units in the \(y\)-direction. The resultant vector has components that are just the sums of the components of the vectors \(\vec{u}\) and \(\vec{v}\). To add two vectors in coordinate form, we add the corresponding components.
Theorem: Vector Addition (coordinate form)
If \(\vec{u}=a \vec{i}+b \vec{j}\) and \(\vec{v}=c \vec{i}+d \vec{j}\), then\[\vec{u}+\vec{v}=(a+c) \vec{i}+(b+d) \vec{j}.\nonumber \]
Example \(\PageIndex{8}\)
Mike flew 15 miles southwest, followed by 20 miles on a heading of \(260^{\circ}\). What is his current position relative to his starting point?
Solution
A sketch of Mike's journey is shown below. We begin by converting the vector for each leg of his journey into coordinate form.\[\begin{array}{rcl}
v_x & = & 15 \cos \left(225^{\circ}\right) \approx -10.607 \\[6pt] v_y & = & 15 \sin \left(225^{\circ}\right) \approx -10.607 \\[6pt] w_x & = & 20 \cos \left(190^{\circ}\right) \approx -19.696 \\[6pt] w_y & = & 20 \sin \left(190^{\circ}\right) \approx -3.473 \\[6pt] \end{array} \nonumber\]
Figure \(\PageIndex{11}\)
Thus,\[\vec{v}=-10.607 \vec{i}-10.607 \vec{j} \quad \text{ and } \quad \vec{w}=-19.696 \vec{i}-3.473 \vec{j}.\nonumber\]We find the resultant vector, \(\vec{r}\), by adding \(\vec{v}\) and \(\vec{w}\).\[\begin{array}{rcl}
\vec{r} & \approx & (-10.607-19.696) \vec{i}+(-10.607-3.473) \vec{j} \\[6pt] & = & -30.303 \vec{i}-14.08 \vec{j} \\[6pt] \end{array} \nonumber\]Mike's position vector is \(\vec{r}=30.303 \vec{i}-14.08 \vec{j}\), or about 30.3 miles west and 14.1 miles south of his starting position.
Checkpoint \(\PageIndex{8}\)
Min's yacht moves at a heading of \(350^{\circ}\) at 20 kilometers per hour relative to the water. However, the water moves at a heading of \(70^{\circ}\) at 4 kilometers per hour. What is Min's velocity relative to land?
Answer
22.7 kph due north
The coordinate form is especially efficient if we want to add more than two vectors.
Example \(\PageIndex{9}\)
While exploring an abandoned cottage in the woods, you discover an old map showing the location of a buried treasure. According to the map, from the cottage, go \(10\) km bearing \(\mathrm{N} \, 20^{\circ} \, \mathrm{E}\). From that point, go \(12.8\) km bearing \(\mathrm{S} \, 78^{\circ} \, \mathrm{E}\). Finally, go \(9.5\) km bearing \(\mathrm{N} \, 39^{\circ} \, \mathrm{W}\).
You wonder if the directions will lead you safely to the treasure. You draw vectors to illustrate the path described by the directions; however, instead of following the map directions, you plan to fly by helicopter from the abandoned cottage directly to the treasure. You will need to know the correct bearing and distance to fly. Draw the vector for the net displacement from the cottage to the treasure and give its length and direction.
Solution
The helicopter's flight plan is the vector sum of the three vectors \(\vec{v}_1\) (for the first part of the journey), \(\vec{v}_2\) (for the second part), and \(\vec{v}_3\) (for the third part). We resolve each of the three vectors into its components to find their sum. We start by finding the angles each vector makes with the positive \(x\)-axis. From the beginning to the end of the journey, these angles are \(\theta_1 = 70^{\circ}\), \(\theta_2 = -12^{\circ}\), and \(\theta_3 = 129^{\circ}\) (You should verify this.). In coordinate form, the three displacement vectors are\[\begin{array}{rclcl}
\vec{v}_1 & = & \left(10 \cos \left(70^{\circ}\right)\right) \vec{i}+\left(10 \sin \left(70^{\circ}\right)\right) \vec{j} & \approx & 3.42 \vec{i}+9.40 \vec{j} \\[6pt] \vec{v}_2 & = & \left(12.8 \cos\left( -12^{\circ}\right)\right) \vec{i}+\left(12.8 \sin \left(-12^{\circ}\right)\right) \vec{j} & \approx & 12.52 \vec{i}-2.66 \vec{j} \\[6pt] \vec{v}_3 & = & \left(9.5 \cos\left( 129^{\circ}\right)\right) \vec{i}+\left(9.5 \sin \left(129^{\circ}\right)\right) \vec{j} & \approx & -5.98 \vec{i}+7.38 \vec{j} \\[6pt] \end{array} \nonumber\]The net displacement is given by the resultant vector, \(\vec{r}\). We add the corresponding components of \(\vec{v}_1\), \(\vec{v}_2\), and \(\vec{v}_3\).\[\begin{array}{rcl}
\vec{r} & \approx & (3.42 \vec{i}+9.40 \vec{j})+(12.52 \vec{i}-2.66 \vec{j})+(-5.98 \vec{i}+7.38 \vec{j}) \\[6pt] & = & (3.42+12.52-5.98) \vec{i}+(9.40-2.66+7.38) \vec{j} \\[6pt] & = & 10 \vec{i}+14.12 \vec{j} \\[6pt] \end{array} \nonumber\]The treasure is at the point \(10\) km east and \(14.12\) km north of the cottage. To find the flight plan for the helicopter, we compute the magnitude and direction of the vector \(\vec{r}\).\[\begin{array}{rrcl}
& \norm{\vec{r}} & = & \sqrt{10^2+14.12^2}=17.3 \\[6pt] & & \text{and} & \\[6pt] & \tan\left( \theta \right) & = & \dfrac{14.12}{10} \\[6pt] \implies & \widehat{\theta} & = & \tan ^{-1}\left( \dfrac{14.12}{10} \right) \\[6pt] & & \approx & 54.69^{\circ} \\[6pt] \end{array}\nonumber\]However, since we know the treasure is to the east and north, we know \(0^{\circ} \lt \theta \lt 90^{\circ}\). Hence, \(\theta \approx 54.7^{\circ}\). This means the treasure is \(17.3\) km from the cottage at a bearing of \(\mathrm{N} \, 35.3^{\circ} \, \mathrm{E}\).
Checkpoint \(\PageIndex{9}\)
From your campsite you hike \(1.6\) km heading \(175^{\circ}\), then \(0.8\) km at the heading of \(65^{\circ}\), and finally \(1.2\) km in the heading \(350^{\circ}\).
Draw a diagram for your hike, using vectors to represent each of the three segments. Use a grid in which 1 square represents 0.1 km.
Resolve each vector into components, and determine your location after your hike.
When you push or pull on something, you exert a force on the object. For example, an object's weight is actually a force, the result of gravity pulling the object towards the earth. A force has magnitude (measured in pounds) and direction, so it is a vector quantity. A force applied to an object causes the object to accelerate in the direction of the force.
When two or more forces act simultaneously on an object, the object moves as if the sum of the individual force vectors acted on it. The sum of all the forces acting on an object is the resultant force.
Example \(\PageIndex{10}\)
As shown below, Ivan and Wyatt are trying to move a refrigerator by pulling on it with forces of 150 pounds and 120 pounds, respectively. \(\vec{A}\) represents the force with which Ivan is pulling, and \(\vec{B}\) represents the force with which Wyatt is pulling. What is the resulting force on the refrigerator, and in what direction will it move?
Figure \(\PageIndex{13}\)
Solution
We write each force in its coordinate form.\[\begin{array}{rclcl}
\vec{A} & = & 150 \cos \left(20^{\circ}\right) \vec{i} + 150 \sin \left(20^{\circ}\right) \vec{j} & \approx & 140.95 \vec{i}+51.30 \vec{j} \\[6pt] \vec{B} & = & 120 \cos \left(80^{\circ}\right) \vec{i} + 120 \sin \left(80^{\circ}\right) \vec{j} & \approx & 20.84 \vec{i}+118.18 \vec{j} \\[6pt] \end{array} \nonumber\]Next, we add the forces applied by Ivan and Wyatt to find the resultant force, \(\vec{r}=\vec{A}+\vec{B}\).\[\begin{array}{rcl}
\vec{r} & \approx & (140.95 \vec{i}+51.30 \vec{j})+(20.84 \vec{i}+118.18 \vec{j}) \\[6pt] & = & 161.75 \vec{i}+169.48 \vec{j} \\[6pt] \end{array} \nonumber\]The magnitude of the resultant force is \(\norm{\vec{r}} \approx \sqrt{161.75^2+169.48^2} \approx 234.28\) and the direction of the force is given by\[\tan^{-1} \left( \dfrac{169.48}{161.75} \right) = 46.33^{\circ}.\nonumber\]
Figure \(\PageIndex{14}\)
Therefore, the force exerted on the refrigerator is about 234.3 pounds, and it will move at an angle of approximately \(46.33^{\circ}\) from the horizontal.
Checkpoint \(\PageIndex{10}\)
Two tugboats pull on a barge in the river with forces \(\vec{u}=20 \vec{i}+8 \vec{j}\) and \(\vec{v}=28 \vec{i}-6 \vec{j}\), measured in thousands of pounds. In what direction will the barge move, and what is the magnitude of the force propelling it?
Answer
\(2.4^{\circ},\) 48.04 thousand lbs
If vectors \(\vec{u}\) and \(\vec{v}\) have the same magnitude but opposite directions, then \(\vec{u}+\vec{v}\) has zero magnitude and is called the zero vector.
Definition: Zero Vector
A zero vector, also known as a null vector, is a vector with a magnitude of zero and an undefined direction. It is denoted \( \vec{0} \).
A zero displacement vector means that the object started and ended in the same place; a zero velocity vector means that the object is not moving.
The vector that has the same magnitude as \(\vec{v}\) but the opposite direction is called the opposite of \(\vec{v}\) and denoted by \(\vec{- v}\). Then\[\vec{v}+\left(-\vec{v}\right)=\vec{0}.\nonumber\]We now have enough language to tackle some of the most interesting (and important) applications of vectors—those involving static equilibrium.
Definition: Static Equilibrium
When the sum of the forces (as vectors) acting on a particle or object is \( \vec{0} \), we say the object is in static equilibrium (or simply equilibrium).
Example \(\PageIndex{11}\)
Recall the refrigerator in the previous example. Jorge does not want Ivan and Wyatt to move it. How hard must Jorge pull so that it remains motionless?
Solution
Let \(\vec{C}\) be the force that Jorge applies. It must be equal in magnitude to the sum of the forces applied by Ivan and Wyatt but point in the opposite direction. That is,\[\vec{C}=-\vec{r}=-(\vec{A}+\vec{B}).\nonumber\]So, Jorge must pull with 234.3 pounds in the direction \(180^{\circ}+46.33^{\circ}=226.33^{\circ}\).
Checkpoint \(\PageIndex{11}\)
Suppose that \(\vec{u}+\vec{v}+\vec{w}=\vec{0}, \quad\) and that \(\vec{u}=-8 \vec{i}+32 \vec{j}\) and \(\vec{v}=14 \vec{i}-26 \vec{j}\). What is the coordinate form of vector \(\vec{w}\)?
Answer
\(\quad \vec{w}=-6 \vec{i}-6 \vec{j}\)
Success in College: Groupwork Learning Preferences
Most students are very intimidated by working in groups. It usually takes quite a while for students to realize that studying together can be very rewarding.
Think about the following questions regarding learning or practicing complicated math.
Do you prefer to do homework by yourself, or do you like to do homework with others?
Do you like to do homework at school, or do you like to do homework at home?
If you need help, do you prefer to get it from friends or a spouse, or do you prefer to get it from your professor or a tutor?
Do you concentrate on learning what you don't know by yourself, or do you prefer to help others because it helps you learn it better?
Do you work in conditions that match these preferences when you do your math homework? Why or why not?
Spending time with old and new friends is an integral part of college. However, studying with friends is not always practical. Do you ever feel peer pressure to study in ways that are not the best for you?
