This section introduces vectors from a geometric perspective. It is important to realize that all vectors in this section are represented either geometrically or through verbal description (in the case of applications). Components of a vector, \(\vec{i}\)-\(\vec{j}\) notation, and the like are held off for the next section. The topics include:
geometrically defining vectors, operations on vectors, vector magnitude, and resultant vectors;
interpreting true heading and velocity as associated with the resultant vector; and
computing the angle a vector makes with the positive \(x\)-axis and the magnitude of a vector (not given in component form).
The following is a list of learning objectives for this section.
Learning Objectives (click to expand)
Sketch vectors and sums of vectors.
Find the magnitude of the horizontal and vertical vector components for a vector.
Find a vector's magnitude and its angle with the positive \(x\)-axis.
Solve an applied problem using vectors.
Introduction
One way to specify a location is to give a direction and a distance from a fixed landmark. For example, the airport is 8 miles northeast of the town hall, or a ship has been sighted 20 miles from the lighthouse in the direction \(10^{\circ}\) west of north. It's not enough to give just the distance to the object or just the direction; we need both to describe the object's location. Thus, we have the following definition.
Definition: Vector
In Mathematics, a quantity defined by both a magnitude and a direction is called a vector.
Vectors are used to analyze motion and velocity in multiple dimensions, study forces such as gravity and electric fields, and design computer graphics and animation. They are also used throughout Engineering, Mathematics, and Physics.
The types of vectors we will concern ourselves with in this section are position vectors.
Definition: Position Vector
A position vector is a vector used to designate a location relative to a fixed landmark.
We often illustrate a vector with an arrow. The length of the arrow represents the magnitude (size) of the vector, and the head of the arrow indicates its direction.
Example \(\PageIndex{1}\)
Draw an arrow to represent each vector.
The campsite is 3 miles away, in a direction \(30^{\circ}\) north of east.
The wind is blowing due west at 50 kilometers per hour.
Solutions
We draw an arrow making an angle of \(30^{\circ}\) from east. The length of the arrow is 3 units, representing 3 miles. See Figure \(\PageIndex{1a}\).
We draw an arrow pointing due west. That is, making an angle of \(180^{\circ}\) from east. The arrow's length represents the wind's speed, 50 kilometers per hour. See Figure \(\PageIndex{1b}\).
Figure \(\PageIndex{1}\)
In both images in Figure \(\PageIndex{1}\), we intuitively understand that the direction of the vector is where the arrow is pointing; however, to be able to discuss vectors better, it's best to laboriously define the geometric parts of the graphical representation of a vector.
Definition: Head and Tail of a Vector
Given the graph of a vector, which is a finite-length ray, we define the head as the location of the terminal end of the ray (the arrowhead) and the tail as the initial point of the ray (the non-arrowhead end).
In Figure \(\PageIndex{1a}\), the head of the vector is the campsite, and the tail is the origin.
Checkpoint \(\PageIndex{1}\)
Draw an arrow representing the velocity of an airplane traveling southeast at 300 miles per hour.
Answer
Figure \(\PageIndex{2}\)
Notation for Vectors
Some texts use boldface characters such as \(\mathbf{u}\) and \(\mathbf{v}\) to represent vectors; however, this is difficult to replicate by hand. Therefore, we choose the alternative notation, \(\vec{u}\) and \(\vec{v}\), which uses arrows above a variable to indicate that it is a vector.
It is vital to distinguish vector quantities from ones with magnitude only, such as length or temperature.
Definition: Scalar
A quantity having only magnitude (and no direction) is called a scalar.
Therefore, scalars are the constants and variables you are used to dealing with, usually denoted by italic letters such as \(x\) or \(k\). They have a magnitude (size), which we typically consider their absolute value; however, we refine the definition of magnitude when speaking of vectors.
Definition: Magnitude of a Vector (geometric)
The length of a vector \(\vec{v}\) is called its magnitude and is denoted by \(\norm{\vec{v}}\).
Caution: Magnitude is a Scalar Quantity
Note that the magnitude of a vector is a scalar quantity. Thus, \(\vec{v}\) denotes a vector (a distance and direction), but \(\norm{\vec{v}}\) represents a scalar (the length of the vector).
Figure \(\PageIndex{3}\) shows two vectors, \(\vec{u}\) and \(\vec{v}\), that have the same magnitude but different directions. For this example, \(\norm{\vec{u}} = \norm{\vec{v}}\) but \(\vec{u} \neq \vec{v}\). It makes sense that two vectors may have the same length but point in different directions.
Figure \(\PageIndex{3}\)
This begs the question: what does it mean for two vectors to be equal? Believe it or not, we have to define what this means!
Definition: Vector Equality (geometric)
Two vectors are equal if and only if they have the same length and direction.
Thus, we could have two equal vectors (same magnitude and direction) but with their tails (and, therefore, heads) at different locations. The fact that we can move vectors from one location to another is valuable.
Example \(\PageIndex{2}\)
Which of the vectors are equal?
Figure \(\PageIndex{4}\)
Solution
Only vectors \(\vec{c}\) and \(\vec{g}\) are equal. Vectors \(\vec{b}\), \(\vec{c}\), and \(\vec{f}\) are all the same length but have different directions. Vectors \(\vec{a}\) and \(\vec{d}\) also have equal lengths but different directions. Vectors \(\vec{d}\) and \(\vec{e}\) have the same direction but different lengths.
Checkpoint \(\PageIndex{2}\)
Sketch a vector \(\vec{w}\) that is equal to the vector \(\vec{v}\), but with tail at the point \((-4, 2)\).
Figure \(\PageIndex{5}\)
Answer
Figure \(\PageIndex{6}\)
Scalar Multiplication of Vectors
We can define multiplication of a vector by a scalar.
Definition: Scalar Multiplication (geometric)
The notation \( k \vec{v} \), where \( k \) is a scalar, is called the scalar multiple (or scalar product) of the vector \( \vec{v} \) with the scalar \( k \) and it is defined to be the vector having same direction as \( \vec{v} \), but magnitude \( k \) times the magnitude of \( \vec{v} \).
Figure \(\PageIndex{7}\) shows the vector \(\vec{w}\) and the vector \(\vec{v} = 3\vec{w}\). Multiplying by a positive scalar changes a vector's length (magnitude) but not its direction.
Figure \(\PageIndex{7}\)
Thus, the vector \(\vec{v}\) is 3 times as long as the vector \(\vec{w}\). If we multiply a vector by a negative scalar, we alter its length and reverse its direction; that is, we change the direction by \(180^{\circ}\). For example, the vector \(\vec{u}=-\frac{1}{2} \vec{w}\). It is half the length of \(\vec{w}\) and points in the opposite direction.
In general, if \(k\) is a real number, then \(k \vec{v}\) represents the vector with magnitude \(k\) times the magnitude of \(\vec{v}\). It points in the same direction as \(\vec{v}\) when \(k \gt 0\) and in the opposite direction from \(\vec{v}\) when \(k \lt 0\). Real numbers are called scalars because they "scale" vectors in this way.
Example \(\PageIndex{3}\)
The figure below shows the vector \(\vec{v}\) and two scalar multiples of \(\vec{v}\).
Figure \(\PageIndex{8}\)
The vector \(\frac{2}{3} \vec{v}\) points in the same direction as \(\vec{v}\), but is only two-thirds as long as the vector \(\vec{v}\). The vector \(-\sqrt{3} \vec{v}\) points in the direction opposite to \(\vec{v}\) and is \(\sqrt{3}\) or approximately 1.7 times as long as \(\vec{v}\).
Note that because \(\vec{v}\) has a slope of \(\frac{-2}{5}\), so does any nonzero multiple of \(\vec{v}\).
Checkpoint \(\PageIndex{3}\)
For the vector \(\vec{w}\) shown below, draw the vectors \(-0.6 \vec{w}\) and \(\sqrt{2} \vec{w}\).
Figure \(\PageIndex{9}\)
Answer
Figure \(\PageIndex{10}\)
Addition of Vectors
Definition: Displacement Vector
A displacement vector represents the change in position from one point to another.
For example, you leave home and travel 6 miles east and 8 miles north. These two displacements are represented by the vectors \(\vec{u}\) and \(\vec{v}\), in Figure \(\PageIndex{11}\). When we follow one displacement vector with a second one, the net displacement is a new vector, \(\vec{w}\), starting at the tail of the first vector and ending at the head of the second vector.
Figure \(\PageIndex{11}\)
In this example, notice that \(\vec{w}\) forms the hypotenuse of a right triangle, so we can calculate its magnitude and direction.\[\norm{\vec{w}}^2 = \norm{\vec{u}}^2 + \norm{\vec{v}}^2 = 6^2 + 8^2 = 100.\nonumber\]Therefore,\[\norm{\vec{w}} = 10.\nonumber\]To find the direction of the vector \(\vec{w}\), we rely on the methods we have established up to this point in Trigonometry.\[\begin{array}{rrclcl}
& \tan\left( \theta \right) & = & \dfrac{8}{6} & & \\[6pt] \implies & \tan\left( \theta \right) & = & \dfrac{4}{3} & & \\[6pt] \implies & \widehat{\theta} & = & \tan ^{-1}\left(\dfrac{4}{3}\right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array}\nonumber\]Since \(0^{\circ} \lt \theta \lt 90^{\circ}\), \(\theta = \widehat{\theta} \approx 53.1^{\circ}\).
The net displacement gives your current position relative to home: 10 miles in the direction \(53.1^{\circ}\) north of east.
When we follow one vector by a second vector as described above, we are adding the two vectors. As Figure \(\PageIndex{11}\) illustrates, the sum of vectors results in a new vector. This new vector gets a special name.
Definition: Resultant Vector
The vector resulting from a sum of vectors is the resultant vector.
Example \(\PageIndex{4}\)
For each pair of vectors, draw the resultant vector \(\vec{w}=\vec{u}+\vec{v}\).
Figure \(\PageIndex{12}\)
Solutions
We draw a copy of the vector \(\vec{v}\) so that its tail (or starting point) is placed on the head (or ending point) of \(\vec{u}\), and then we draw vector \(\vec{w}\) from the tail of \(\vec{u}\) to the head of \(\vec{v}\).
Figure \(\PageIndex{13}\)
Checkpoint \(\PageIndex{4}\)
Draw the resultant vector \(\vec{w}=\vec{u}+\vec{v}\)
Figure \(\PageIndex{14}\)
Answer
Figure \(\PageIndex{15}\)
When the resultant vector forms the third side of a triangle, we can calculate its length and direction using the Law of Sines and the Law of Cosines.
Example \(\PageIndex{5}\)
You are camping in a state park. In the morning, you start from your campground at point \(C\) and hike 4 miles bearing \(\mathrm{S} \, 45^{\circ} \, \mathrm{W}\) to point \(B\). After taking a break, you continue hiking, and this time you cover 3 miles in the bearing \(\mathrm{N} \, 30^{\circ} \, \mathrm{E}\), and arrive at point \(A\), as shown below. What is the position of the campground relative to your position?
Figure \(\PageIndex{16}\)
Solution
The resultant vector, \(\vec{w}\), forms the side opposite a \(15^{\circ}\) angle.\[\begin{array}{rrclcl}
& \norm{\vec{w}}^2 & = & a^2 + c^2 - 2 a c \cos\left( \angle ABC \right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & \norm{\vec{w}} & = & \sqrt{a^2 + c^2 - 2 a c \cos\left( \angle ABC \right)} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \end{array}\nonumber\]In this case, \(a = 4\), \(c = 3\), and \(\angle ABC = 15^{\circ}\). Substituting these values into our result, we get \(\norm{\vec{w}} \approx 1.348\), so you are approximately 1.348 miles from the campground. To find the direction back to the campground, we must calculate the angle \(\angle ACB\). Following the advice in the section covering the Law of Cosines, we find the angle opposite the smallest remaining side. Luckily, that angle is \(\angle ACB\).\[\begin{array}{rrclcl}
& \dfrac{\sin \left(\angle ACB \right)}{c} & = & \dfrac{\sin\left( \angle ABC \right)}{b} & \quad & \left( \text{Law of Sines} \right) \\[6pt] \implies & \sin\left( \angle ACB \right) & = & \dfrac{c \, \sin\left( \angle ABC \right)}{b} & \quad & \left( \text{multiplying both sides by }c \right) \\[6pt] \implies & \widehat{\angle ACB} & = & \sin^{-1}\left( \dfrac{c \, \sin\left( \angle ABC \right)}{b} \right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array}\nonumber\]Since \(\angle ACB \lt 90^{\circ}\), \(\angle ACB = \widehat{\angle ACB} \approx 35.2^{\circ}\). It's best to get a visual of our angles, so we redraw the figure with the newly found angles.
Figure \(\PageIndex{17}\)
Since \(\angle ABC = 15^{\circ}\) and \(\angle ACB \approx 35.2^{\circ}\), \(\angle BAC \approx 129.8^{\circ}\). However, this means\[30^{\circ} + 90^{\circ} + \left( 90^{\circ} - \theta \right) \approx 129.8^{\circ} \implies \theta \approx 80.2^{\circ}.\nonumber\]Thus, the campground is 1.348 miles away at a bearing of \(\mathrm{N} \, 80.2^{\circ} \, \mathrm{E}\).
Checkpoint \(\PageIndex{5}\)
Ron has gone sailing with friends. After leaving the marina, they sail 5 miles on a heading of \(160^{\circ}\) and stop at Gull Island for lunch. After lunch, they have sailed for 3 miles on a heading of \(25^{\circ}\) when they get a phone call to return home. What heading is the most direct route back to the marina, and how far is it?
Answer
Heading of \(303.6^{\circ}\) for 3.58 miles
It doesn't matter in which order we choose to add two vectors. As with the ordinary addition of scalars, the addition of vectors is commutative, so that \(\vec{u}+\vec{v}=\vec{v}+\vec{u}\). To see this, first draw \(\vec{u}\) and \(\vec{v}\) starting at the same point.
Figure \(\PageIndex{18}\)
To represent \(\vec{u}+\vec{v}\), we place the tail of \(\vec{v}\) at the head of \(\vec{u}\), and draw the resultant vector \(\vec{u}+\vec{v}\) as shown in Figure \(\PageIndex{18}\).
To represent \(\vec{v}+\vec{u}\), we place the tail of \(\vec{u}\) at the head of \(\vec{v}\), and the resultant vector \(\vec{v}+\vec{u}\) is the same as the vector \(\vec{u}+\vec{v}\).
Because the vector sum forms the diagonal of a parallelogram in this picture, the rule for adding vectors is sometimes called the parallelogram rule.
We summarize the operations on vectors as follows.
Summary: Operations on Vectors
We can multiply a vector, \(\vec{v}\), by a scalar, \(k\).
If \(k \gt 0\), the magnitude of \(k \vec{v}\) is \(k\) times the magnitude of \(\vec{v}\). The direction of \(k \vec{v}\) is the same as the direction of \(\vec{v}\).
If \(k \lt 0\), the direction of \(k \vec{v}\) is opposite the direction of \(\vec{v}\).
We can add two vectors \(\vec{v}\) and \(\vec{w}\) with the parallelogram rule.
Caution: Lengths of vectors don't add (in general)
Unless \(\vec{u}\) and \(\vec{v}\) are parallel vectors, it is not true that the length of \(\vec{u}+\vec{v}\) is just the sum of the lengths of \(\vec{u}\) and \(\vec{v}\). Vector addition is a geometric operation; the length of \(\vec{u}+\vec{v}\) depends on the lengths of \(\vec{u}\) and \(\vec{v}\) and on the angle between them. Be careful to distinguish between regular addition of scalars, such as \(\norm{\vec{u}}+\norm{\vec{v}}\), and vector addition, \(\vec{u}+\vec{v}\), which requires the parallelogram rule.
Velocity
Physicists and mathematicians use the word velocity to mean not simply speed but the combination of speed and direction of motion. The interesting thing about velocities is that they add like vectors; if an object's motion consists of two simultaneous components, the resulting displacement is the same as if the motions had occurred one after the other.
For example, imagine a beetle walking across a moving conveyor belt, as shown in the leftmost image in Figure \(\PageIndex{19}\). Suppose the beetle starts at the lower left corner. The conveyor belt is moving at a speed of 4 inches per second, and the beetle walks at right angles to the motion of the belt at 2 inches per second. After 1 second, the beetle has traveled upwards from its starting point, a total distance of 2 inches. Simultaneously, the conveyor belt has moved the beetle to the right a distance of 4 inches. Thus, the beetle has moved a distance of \(\sqrt{2^2+4^2}=\sqrt{20}\), or about 4.47 inches. His actual velocity relative to the ground is approximately 4.47 inches per second at an angle of \(\theta=\tan ^{-1}\left(\frac{2}{4}\right)=26.6^{\circ}\) from the direction of the conveyor belt.
Figure \(\PageIndex{19}\)
Notice that the resulting displacement would be the same if the two motions were performed in succession instead of simultaneously, as shown in the rightmost image in Figure \(\PageIndex{19}\). In other words, if the beetle had walked across the belt for 1 second before the belt started moving and then ridden the belt for 1 second without walking, he would still end up at point \(Q\).
Thus, velocity is a vector quantity, and we can calculate the result of two simultaneous motions using the parallelogram rule. We treat the two motions as if they had occurred one after the other by starting one vector at the endpoint of the other. (Remember, we can move a vector from one location to another if preserve its length and direction.)
Example \(\PageIndex{6}\)
A ship travels 15 miles per hour relative to the water on a heading of \(280^{\circ}\). The water current flows 6 miles per hour on a heading of \(160^{\circ}\). What is the actual speed and direction of the ship?
Solution
We represent the ship's velocity by a vector \(\vec{v}\) and the velocity of the water current by vector \(\vec{w}\), as shown in Figure \(\PageIndex{20a}\). Note that the angle \(\vec{v}\) makes with the horizontal is \(10^{\circ}\), and angle \(\vec{w}\) makes with the vertical is \(20^{\circ}\) — these will be important in a few moments.
Figure \(\PageIndex{20}\)
The actual motion of the ship is the sum of these two vectors. We can calculate the sum as if the two motions had occurred separately, one after the other, as shown in Figure \(\PageIndex{20b}\). So we will add \(\vec{w}\) to \(\vec{v}\) by placing the tail of \(\vec{w}\) at the head of \(\vec{v}\). The resultant vector, \(\vec{u}\), represents the actual motion of the ship. Let's zoom in to Figure \(\PageIndex{20b}\).
Figure \(\PageIndex{21}\)
The angle \(\alpha\) in Figure \(\PageIndex{21}\) is alternate interior to the \(10^{\circ}\) angle \(\vec{v}\) makes with the horizontal, so \(\alpha = 10^{\circ}\). Moreover, we already mentioned that \(\vec{w}\) makes a \(20^{\circ}\) with the vertical. Therefore, if we let \(\beta\) be the angle between the head of \(\vec{v}\) and the tail of \(\vec{w}\) (as seen in Figure \(\PageIndex{21}\)), we have \(20^{\circ} + \beta + 10^{\circ} =90^{\circ}\), so \(\beta=60^{\circ}\). Hence,\[\begin{array}{rrclcl}
& \norm{\vec{u}}^2 & = & \norm{\vec{v}}^2 + \norm{\vec{w}}^2 - 2 \norm{\vec{v}} \cdot \norm{\vec{w}} \cos \left(\beta\right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & \norm{\vec{u}} & = & \sqrt{\norm{\vec{v}}^2 + \norm{\vec{w}}^2 - 2 \norm{\vec{v}} \cdot \norm{\vec{w}} \cos \left(\beta\right)} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \end{array}\nonumber\]Substitution in \(\norm{\vec{v}} = 15\), \(\norm{\vec{w}} = 6\), and \(\beta = 60^{\circ}\), we find \(\norm{\vec{u}} = \sqrt{171} \approx 13.1\). Thus, the ship's speed is approximately 13.1 miles per hour.
Next, we calculate the ship's heading (also known as its true heading). This means we need to calculate another angle within the triangle shown in Figure \(\PageIndex{21}\). We already know \(\beta\). Of the remaining angles, we want to find the measure of \(\gamma\) (mostly because it is opposite the smallest remaining side, but also because this is the angle that will help us calculate the ship's true heading).\[\begin{array}{rrclcl}
& \dfrac{\sin\left( \gamma \right)}{\norm{\vec{w}}} & = & \dfrac{\sin\left( \beta \right)}{\norm{\vec{u}}} & \quad & \left( \text{Law of Sines} \right) \\[6pt] \implies & \sin\left( \gamma \right) & = & \dfrac{\norm{\vec{w}} \sin\left( \beta \right)}{\norm{\vec{u}}} & \quad & \left( \text{multiplying both sides by }\norm{\vec{w}} \right) \\[6pt] \implies & \widehat{\gamma} & = & \sin^{-1}\left(\dfrac{\norm{\vec{w}} \sin\left( \beta \right)}{\norm{\vec{u}}}\right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array}\nonumber\]Since \(\gamma\) is opposite the smallest remaining side, we can guarantee \(0^{\circ} \lt \gamma \lt 90^{\circ}\). Hence, \(\gamma = \widehat{\gamma} \approx 23.5^{\circ}\).
Looking back at Figure \(\PageIndex{21}\), we see the angle that \(\vec{u}\) makes with the horizontal is \(\gamma-10^{\circ}=13.4^{\circ}\). Hence, the ship is heading \(270^{\circ} - 13.4^{\circ} = 256.6^{\circ}\) at 13.1 miles per hour.
Checkpoint \(\PageIndex{6}\)
A plane heads due north at an airspeed of 120 miles per hour. A 45 mph wind blows at a heading of \(95^{\circ}\). What are the plane's actual speed and direction relative to the ground?
Answer
Heading of \(21.1^{\circ}\) moving at 124.4 mph.
In some situations, we would like to find a vector to produce a particular sum instead of calculating a vector. That is, we know vectors \(\vec{u}\) and \(\vec{w}\), and we want to find a vector \(\vec{v}\) so that \(\vec{u}+\vec{v}=\vec{w}\).
Example \(\PageIndex{7}\)
Phuong intends to travel west to an island at 15 miles per hour. However, she must compensate for a current running at a bearing of \(\mathrm{N} \, 45^{\circ} \, \mathrm{E}\) at a speed of 3 miles per hour. In what direction and at what speed should Phuong move her boat?
Solution
We draw a triangle using vectors to represent the desired velocity of Phuong's boat, \(\vec{w}\), and the velocity of the current, \(\vec{u}\), as shown in Figure \(\PageIndex{22}\). We want to find a vector \(\vec{v}\) that represents the speed and bearing Phuong should take to compensate for the current.
Figure \(\PageIndex{22}\)
We first use the Law of Cosines to calculate the length or magnitude of the vector \(\vec{v}\). The angle at \(C\) is \(135^{\circ}\). (Do you see why?). Thus,\[\begin{array}{rrcl}
& \norm{\vec{v}}^2 & = & 3^2+15^2-2(3)(15) \cos \left(135^{\circ}\right) \\[6pt] & & \approx & 297.64 \\[6pt] \implies & \norm{\vec{v}} & \approx & 17.25 \\[6pt] \end{array}\nonumber\]
So, Phuong should travel at about 17.25 miles per hour. To find her bearing, we use the Law of Sines to calculate the angle \(\theta\).\[\begin{array}{rrcl}
& \dfrac{\sin\left( \theta \right)}{3} & = & \dfrac{\sin \left(45^{\circ}\right)}{15} \\[6pt] \implies & \sin\left( \theta \right) & = & \dfrac{3/\sqrt{2}}{15} \\[6pt] \implies & \theta & = & \sin^{-1}\left( \dfrac{1}{5\sqrt{2}} \right) \\[6pt] & & \approx & 8.1^{\circ} \\[6pt] \end{array}\nonumber\]This means that the angle \(\vec{v}\) makes with the vertical is approximately \(81.9^{\circ}\). Therefore, Phuong should move her boat in the bearing of \(\mathrm{S} \, 81.9^{\circ} \, \mathrm{W}\).
Checkpoint \(\PageIndex{7}\)
Matt would like to sail at 20 kilometers per hour due west towards a whale reported at that position. However, a steady ocean current is moving \(52^{\circ}\) east of north at 8 kilometers per hour. At what speed and heading should Matt sail?
Answer
\(26.76 \mathrm{kph}\) and \(10.6^{\circ} \mathrm{S}\) of \(\mathrm{W}\)
