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Trigonometry Through Wayfinding and Navigation Across the Pacific

Kamuela E Yong

Section 4.3 Geometric Vectors

Hōkūleʻa’s inaugural voyage from Hawaiʻi to Tahiti in 1976 sparked a cultural resurgence in Hawaiʻi. By showcasing the effectiveness of traditional navigation techniques learned from Micronesia, it soon became clear that there was a need for constructing a waʻa kaulua (double-hulled voyaging canoe) using traditional methods and native materials. This canoe, named Hawaiʻiloa, aimed to recover knowledge and skills associated with traditional canoe building in Hawaiʻi.
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However, during the years 1989-1990, an exhaustive search across Hawaiʻi revealed a significant challenge: the lack of accessible koa trees of sufficient size for constructing a canoe of this size. The search then expanded beyond Hawaiʻi. This decision was guided by historical evidence indicating that storms had transported large trees from the Pacific Coast of North America to islands like Hawaiʻi, where they were traditionally used for canoe construction. As a result, the search extended to North America in pursuit of suitable materials.
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Two trees, each 200 feet high and over 400 years old, were discovered on Shelikof Island in Soda Bay, Prince of Wales Island, west of Ketchikan, Alaska. After conducting traditional tree-cutting ceremonies rooted in Hawaiian and Tlingit traditions to seek permission, both trees were felled to be used for constructing the canoe. The logs then embarked on a 2,235-nautical-mile journey to Hawaiʻi, following a heading of 256∘. This journey is an example of a vector, a quantity possessing both magnitude and direction.
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In physics, vectors are often represented with arrows, denoting their direction, while the length of the arrow signifies the vector’s magnitude. Vectors play a fundamental role in describing various things in the world around us. For example, movement of objects requires a magnitude and direction, the force exerted on an object requires the magnitude of the force used and the direction of the force. In this section, we will learn how to express vectors, both geometrically and analytically as well as the various properties of vectors.
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Subsection 4.3.1 Geometric Vectors

Definition 4.3.1.

A vector is a quantity that has both magnitude and direction. The vector PQ→ represents the directed segment from point P to point Q. The point P is called the initial point, and the point Q is called the terminal point. The direction of PQ→ is from P to Q.
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Definition 4.3.2.

The magnitude of a vector PQ→, denoted as ‖PQ→‖, is the distance between its initial point P and its terminal point Q.
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Definition 4.3.3.

The angle θ represents the direction of a vector PQ→. It is the smallest positive angle measured in standard position between the positive x-axis and the vector PQ→, where 0∘≤θ<360∘.
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Remark 4.3.4.

In physical contexts, PQ→ can also be interpreted as the displacement vector, representing the change in position from P to Q.
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Remark 4.3.5. Notation.

Another way to write a vector is with a lowercase boldface letter, with or without an arrow: v or v→. Typically, if we are writing a vector by hand, we will want to use the arrow, since it is difficult to tell if you are writing in bold or not.
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Remark 4.3.6. Calculating Magnitude and Direction.

Let PQ→ be a vector with initial point P=(px,py) and terminal point Q=(qx,qy). The change in the x-direction (run) is denoted by △x=qx−px, and the change in the y-direction (rise) is denoted by △y=qy−py.
The magnitude of the vector is the length of the segment from P to Q, which can be calculated using the distance formula:
‖PQ→‖=(qx−px)2+(qy−py)2=(△x)2+(△y)2.
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To determine the direction (θ) of the vector, we use the slope:
slope=qy−pyqx−px=△y△x.
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This is illustrated in the figure below.
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If rise = 0 or run = 0, use the figure to determine θ. By Right Triangle Trigonometry, the tangent of the direction angle is:
tan⁡θ=△y△x.
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Since −90∘<tan−1⁡(△y△x)<90∘, the inverse tangent function returns an angle in Quadrants I or IV. To determine the correct direction θ, consider the quadrant in which the vector lies and make the appropriate adjustments:
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  • If θ is in Quadrant I, then θ=tan−1⁡(△y△x).
  • If θ is in Quadrant II or III, then θ=tan−1⁡(△y△x)+180∘.
  • If θ is in Quadrant IV, then θ=tan−1⁡(△y△x)+360∘.
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Example 4.3.7.

Let v be the vector with initial point P=(−2,−1) and terminal point Q=(3,3).
(a)
Draw the vector v.
Solution.
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(b)
Calculate the magnitude of v, ‖v‖.
Solution.
The magnitude of a vector is the length of the vector. From the distance formula, we have
‖v‖=(3−(−2))2+(3−(−1))2.
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(c)
Calculate the direction of v, θ.
Solution.
To find the direction, we have:
θ=tan−1⁡(3−(−1)3−(−2))=tan−1⁡(45)≈38.7∘.
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Definition 4.3.8. Zero Vector.

The zero vector, written as 0 or 0→, is a vector with zero magnitude, meaning it has no length and it has no direction.
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Subsection 4.3.2 Equivalent Vectors

Definition 4.3.9. Equivalent Vectors.

It is important to note that a vector only requires a magnitude and direction, but it does not have a unique location. Thus, as long as the magnitude and direction aren’t changed, a vector may be translated or moved from one place to another. As a result, if two vectors u and v have the same direction and the same magnitude, then they are equivalent, or equal, written as:
u=v.
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Example 4.3.10.

Seven canoes, reconstructed versions of the traditional double-hulled Polynesian voyaging vessels, were constructed in Aotearoa. These vessels, collectively referred to as Vaka Moana (canoes of the ocean), were individually named Marumaru Atua (Cook Islands), Gaualofa (Samoa), Te Matau a Māui (Aotearoa), Faʻafaite (Tahiti), Hinemoana (Aotearoa), Haunui (Aotearoa), and Uto Ni Yalo (Fiji). In 2011-2012, these sister canoes embarked on a voyage collectively titled Te Mana o Te Moana (The Spirit of the Ocean), covering a combined distance of 200,000 nautical miles through open waters. On average, each vaka traveled 120 nautical miles per day. Suppose one day, the vaka sailed northeast in the house Manu. The resulting vectors from each vaka are equivalent, as they all traveled in the same direction (NE) with the same magnitude (120 nm). This is depicted in the figure below.
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Example 4.3.11.

Show that vectors PQ→ and RS→, shown in the figure below, are equivalent.

Solution.

To show that two vectors are equivalent, we need to show they have the same magnitude and same direction. We will first calculate the magnitude of each vector.
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‖PQ→‖=(−2−0)2+(3−0)2=13‖RS→‖=(1−3)2+(5−2)2=13
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To determine the direction of each vector, we will look at their slopes. The slope of PQ→ is
change in ychange in x=3−0−2−0=−32,
and the slope of RS→ is
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change in ychange in x=5−21−3=−32.
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Given that both vectors PQ→ and RS→ have the same slopes and are directed upwards and to the left, it follows that they have the same direction. Therefore, since they have the same magnitude and direction, we can conclude that PQ→ and RS→ are equivalent.
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Subsection 4.3.3 Vector Addition

If we stand in Tahiti and face Hawaiʻi, we would be looking in the house Haka Hoʻolua or north by west (NbW). When the Hōkūleʻa sails from Tahiti back home to Hawaiʻi, she doesn’t head directly towards Hawaiʻi in the house Haka Hoʻolua. Instead, Hōkūleʻa initially sails north. Once the navigator determines that Hōkūleʻa has reached the latitude of Hawaiʻi, approximately 20∘ N, the waʻa then changes course to head westward towards home. This navigation technique is known as latitude sailing, where knowing the latitude is crucial. The navigator can determine the latitude by measuring the altitude of Hōkūpaʻa (North Star), observing the altitude of stars as they reach their highest point in the sky, or watching for pairs of stars that rise or set together.
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We can represent the location of Tahiti as T, the location with the same latitude as Hawaiʻi and directly north of Tahiti as L, and the location of Hawaiʻi as H. Then, Hōkūleʻa’s northward path can be represented by the vector TL→, and its westward path by the vector LH→. The total displacement is equivalent to sailing directly from Tahiti to Hawaiʻi, represented by TH→. See Figure 4.3.12. We refer to the vector TH→ as the sum of the vectors TL→ and LH→, written as
TL→+LH→=TH→.
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Figure 4.3.12. Combining the northward and westward vectors, similar to the navigation technique of latitude sailing, results in the direct vector from Tahiti to Hawaiʻi.
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Definition 4.3.13. Vector Addition.

Vector addition is the process of combining two or more vectors to produce a resultant vector. To find the sum of vectors u and v, we place the initial point of the second vector at the terminal point of the first vector. The vector formed from the initial point of the first vector to the terminal point of the second vector represents the sum of the two vectors, denoted by u+v. Since vectors can be translated, any two vectors can be added by shifting one vector’s starting point to the endpoint of the other vector. Notice that adding vector v to vector u produces the same result as adding u to v, as demonstrated below. Thus, we have
u+v=v+u.
Figure 4.3.14. The vectors u and v can be added to obtain u+v and v+u, resulting in the same resultant vector. The arrangement is a parallelogram, demonstrating the commutative property of vector addition.
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Example 4.3.15.

Let u be a vector whose initial point is at (3,5) and terminal point at (−2,4) and v be a vector whose initial point is at (0,−1) and terminal point at (6,−2). Draw the vector u+v.

Solution.

We will begin by drawing the vectors u and v:
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To find u+v, we will translate vector v so that its initial point coincides with the terminal point of u, which is at (−2,4). The translated vector will have the same magnitude and direction as v:
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The sum of the two vectors is the vector starting from the initial point of u and ending at the terminal point of v, in this case from (3,5) to (4,3). This resultant vector is u+v:
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Subsection 4.3.4 Multiplying Vectors by a Scalar

When a waʻa sails for two days at 5 knots, it covers a distance of 240 miles. This distance traveled is known as the magnitude of the vector and is represented by a real number, which we refer to as a scalar.
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The magnitude of the vector representing a waʻa’s path for two days will be twice that of the vector representing the path sailed for only one day. We can now define scalar multiplication of vectors.
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Definition 4.3.16. Scalar Multiplication of Vectors.

If c is a scalar (real number) and v is a vector, then the scalar multiple cv has magnitude equal to the absolute value of c times the magnitude of v, |c|‖v‖. Additionally:
  • If c>0, the scalar multiple cv has the same direction as v.
  • If c<0, the scalar multiple cv has the opposite direction as v.
If c=0 or v=0, then cv=0.
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Example 4.3.17. Scalar Multiplication.

To demonstrate scalar multiplication, consider a vector v representing the displacement of a canoe. Let’s say v is a vector with magnitude 100 miles, pointing northeast. We can visualize this vector as representing the journey of a canoe from one point to another.
Now, let’s explore scalar multiplication applied to this vector:
  • 1v: This represents the same direction and magnitude as v.
  • 12v: This represents half the magnitude of v, so it would have a magnitude of 50 miles, still pointing northeast.
  • −v: This represents the opposite direction of v, so it would have the same magnitude of 100 miles but pointing southwest instead.
  • 2v: This represents doubling the magnitude of v, so it would have a magnitude of 200 miles, still pointing northeast.
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The figure below illustrates the original vector v and the resulting vectors 1v, 12v, −v, and 2v.
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Subsection 4.3.5 Vector Subtraction

Definition 4.3.18. Vector Subtraction.

Vector subtraction is defined as the sum of one vector with the negative of another vector. In other words, u−v=u+(−v).
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Example 4.3.19.

Given vectors u and v, shown in the figure below, find u−v.

Solution.

By Definition 4.3.18, finding u−v is equivalent to finding u+(−v) so we first need to find −v. We do this by flipping vector v so that it points in the opposite direction (from (0,−2) to (5,5)):
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To add u and −v, we will place the initial point of −v at the terminal point of vector u. This is at the point (−4,1).
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The resultant vector from the initial point of u to the terminal point of −v is the vector u−v.
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Remark 4.3.20. Vector Subtraction in a Parallelogram.

Given two vectors u and v, we can visualize their sum and difference using a parallelogram. Translating one of the vectors so that their initial points coincide, we construct a parallelogram with u and v as adjacent sides.
The sum of the two vectors, u+v, is represented by the diagonal of the parallelogram extending from the common initial point to the opposite corner. This diagonal shows the resultant vector when moving along u and then v.
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The difference of the two vectors, u−v, corresponds to the other diagonal of the parallelogram. This diagonal extends from the initial point of u to the terminal point of −v, meaning it represents the vector obtained by adding u and the negation of v.
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Subsection 4.3.6 Velocity

Vectors can represent movement of various physical objects. For instance, the velocity of a moving waʻa (canoe) is depicted by a vector, which indicates the direction of motion and the speed of the waʻa. It is called the velocity vector and can also describe other physical phenomena such as wind and currents.
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Example 4.3.21. Apparent Wind.

Standing on the deck of a moving canoe, we’ll feel the wind against our face. However, this sensation is not solely due to the wind moving over the water. It is influenced by two factors: the actual wind blowing across the ocean, known as the true wind, and the wind created by the canoe’s own motion, called the headwind. The combination of these two is what we perceive as the apparent wind.
Headwind is similar to the resistance felt when extending our hand out of a moving car window on a still day—it results purely from our motion through the air. Mathematically, headwind is the negative of the canoe’s velocity vector. The apparent wind is the vector sum of the true wind and the headwind.
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The velocity of the apparent wind is determined by adding the velocities of headwind and true wind. If the true wind comes from the bow (front) of the canoe, the apparent wind increases. Conversely, if the true wind comes from the stern (back), the apparent wind decreases. For instance, sailing at 5 knots into a true wind of 5 knots would result in an apparent wind of 10 knots, as shown in the figure below.
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However, sailing at 5 knots with the true wind also blowing at 5 knots from behind would result in a zero apparent wind, as we would be moving at the same velocity as the wind, with no wind moving past us. This is shown in the following figure.
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Understanding the apparent wind is crucial for sail trimming and optimizing energy from the wind. Navigators must also consider apparent wind’s impact on the canoe’s speed and course.
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Example 4.3.22. Apparent Wind.

The waʻa Mānaiakalani is sailing off the shore of Lāhainā at a speed of 5 knots. According to the weather report, the true wind is moving at 15 knots. By turning our face into the wind so that we can feel the wind blowing evening past both sides of our face, we can determine the direction of the apparent wind, which we measure to 50∘ to the left of our course. What is the magnitude of the apparent wind?

Solution.

We have the following triangle:
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We can analyze this scenario using the Law of Sines. Given that the known angle is 50∘ and the side opposite to it is 15 knots (the true wind), and the side adjacent to it is 5 knots (the canoe’s speed), we have a Side-Side-Angle (SSA) triangle situation, specifically Case 4, which results in only one possible triangle.
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First, we find angle B using the Law of Sines:
sin⁡50∘15=sin⁡B5.
From this, we get:
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B=sin−1⁡(5sin⁡50∘15))≈14.8∘.
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Now, we can now calculate angle A:
A=180∘−50∘−14.8∘=115.2∘.
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Next, to determine the magnitude of the apparent wind, we find side a using the Law of Sines:
asin⁡115.2∘=15sin⁡50∘.
Therefore,
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a=sin⁡115.2∘⋅15sin⁡50∘≈17.7.
Thus, the magnitude of the apparent wind is approximately 17.7 knots.
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Exercises 4.3.7 Exercises

Exercise Group.

Draw the vector with initial point P and terminal point Q.
1.
P=(2,−3); Q=(1,4)
Answer.
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2.
P=(0,−2); Q=(3,3)
Answer.
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3.
P=(3,−1); Q=(4,−3)
Answer.
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4.
P=(2,1); Q=(−3,4)
Answer.
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5.
P=(−4,0); Q=(3,−2)
Answer.
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6.
P=(4,3); Q=(−1,0)
Answer.
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Exercise Group.

Given the vector PQ→, where P is the initial point and Q is the terminal point, compute the magnitude ‖PQ→‖ and the direction θ. Express the direction in degrees, rounded to one decimal place.
7.
Answer.
41, 38.7∘
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8.
Answer.
61, 309.8∘
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9.
P=(2,−4); Q=(3,2)
Answer.
37, 80.5∘
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10.
P=(−2,2); Q=(0,1)
Answer.
5, 333.4∘
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11.
P=(−3,5); Q=(4,1)
Answer.
65, 330.3∘
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12.
P=(−1,−5); Q=(−4,0)
Answer.
34, 121.0∘
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13.
P=(5,4); Q=(1,3)
Answer.
17, 194.0∘
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14.
P=(−5,−1); Q=(0,−3)
Answer.
29, 338.2∘
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Exercise Group.

Determine if the given vectors u and v are equivalent.
15.
Solution.
Not Equivalent
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16.
Solution.
Equivalent
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17.
Solution.
Not Equivalent
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18.
Solution.
Equivalent
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19.
u has initial point at (−1,1) and terminal point at (3,−4); v has initial point at (0,2) and terminal point at (4,−3).
Answer.
Equivalent
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20.
u has initial point at (2,1) and terminal point at (−3,2); v has initial point at (1,−4) and terminal point at (6,−3);.
Answer.
Not Equivalent
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21.
u has initial point at (2,0) and terminal point at (4,4); v has initial point at (3,−2) and terminal point at (−4,1).
Answer.
Not Equivalent
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22.
u has initial point at (2,4) and terminal point at (0,−3); v has initial point at (1,2) and terminal point at (−1,−5).
Answer.
Equivalent
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Exercise Group.

Use the vectors u and v, shown below, to graph the following vectors:
26.
−v
Answer.
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28.
u−v
Answer.
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29.
v−u
Answer.
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30.
3u−v
Answer.
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Exercise Group.

Changes in the speed of the true wind can alter the apparent wind. In each scenario below, a canoe sails at 5 knots, with the true wind blowing perpendicular to the canoe, as depicted in the figure. Calculate both the direction, θ, and magnitude of the apparent wind for the given true wind speed, rounded to one decimal. Since this forms a right triangle, we use the formula:
θ=tan−1⁡true windheadwind.
31.
True wind: 5 knots
Answer.
45∘; 7.1 knots
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32.
True wind: 10 knots
Answer.
63.4∘; 11.2 knots
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33.
True wind: 15 knots
Answer.
71.6∘; 15.8 knots
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34.
How does the apparent wind change as the true wind increases?
Answer.
As the true wind increases, the angle and magnitude of the apparent wind increase.
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Exercise Group.

In addition to changes in the speed of the true wind, the apparent wind can be affected by changes in the speed of the canoe. In each scenario below, true wind blows perpendicular to the canoe at 10 knots, as depicted in the figure. Calculate both the direction, θ, and magnitude of the apparent wind for the given speed of the canoe, rounded to one decimal.
35.
Canoe speed: 3 knots
Answer.
73.3∘; 10.4 knots
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36.
Canoe speed: 6 knots
Answer.
59.0∘; 11.7 knots
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37.
Canoe speed: 9 knots
Answer.
48.0∘; 13.5 knots
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38.
How does the apparent wind change as the true wind increases?
Answer.
As the speed of the canoe increases, the angle decreases and the magnitude of the apparent wind increases.
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