Vectors in Trigonometry Study Guide (page 3)
In this lesson, we explore the concept of a vector. We see how to break a vector down into component vectors—pieces that make for easy computation. We also explore a variety of word problems that are best solved by using vectors.
A vector is a length with a direction. Vectors are often depicted as arrows that point in the direction and have the correct length. The vector with a 2-inch length and northwest direction is shown in Figure 16.1.
In physics, vectors are used to represent forces. The arrow points in the direction of the push or pull, and the length represents the magnitude (strength) of the force. In navigation, vectors represent the components of journeys. The length represents the distance traveled in the direction of the arrow. Vectors can also represent the flow of wind and water, the stresses on objects, and anything else that has both a quantity and a direction.
There are several ways to represent a vector. One is to draw an arrow with the length and direction, as described above. Another is to give the length as a number and the direction as an angle. If we use the usual notation of 0° pointing straight to the right and positive angles measured counterclockwise, then the vector with length 40 and direction 200° would be as illustrated in Figure 16.2.
Usually, we consider that north is straight up (along the positive y-axis), south is straight down, east is to the right (along the positive x-axis), and west is to the left. Using these, the vector in Figure 16.2 is closest to pointing west. More specifically, it is 20° south of west. This is how the directions of navigation vectors are usually described (see Figure 16.3).
Unfortunately for us, navigators consider north to be 0° and then measure their angles clockwise, so northeast is 45°, east is 90°, south is 180°, and west is 270°. Because this conflicts with the way mathematicians measure angles (measured counterclockwise with east 0°), we shall stick with the "20° south of west" notation.
A last way to represent a vector is to put its start at the origin and give the coordinates of its endpoint. This is where trigonometry is used. The vector in Figure 16.3 can be used as the hypotenuse of a right triangle, as shown in Figure 16.4. The angle to the nearest axis, here 20°, is called the reference angle.
Thus, this vector can be described by the endpoint (–37.59,–13.68) or as a combination of the two component vectors shown in Figure 16.5. The vector can be broken into a component vector of 37.59 west, and another component of 13.68 south.
What is the length and direction of the vector that runs from the origin to the point (8,–15)?
We sketch the vector in Figure 16.6.
The length L of this vector can be found by the Pythagorean theorem.
L2 = 82 + 152, so L = √64 + 225 = √289 = 17
The reference angle θ is formed by the vector and the nearest axis, which in this case is the negative y-axis. The length opposite this angle is 8 and the adjacent length is 15, so
This vector has a length of 17 and points 28° east of south. We could also say that the angle of this vector is 298°. This is depicted in Figure 16.7.
In general, the length of the vector from the origin to (x,y) is L = √X2 + Y2. The reference angle can be found with an inverse tangent.
Find the endpoint of the vector that starts at the origin, has a length of 16, and makes an angle of 143° with the positive x-axis (measured counterclockwise).
If our vector had a length of 1, then the endpoint would be on the unit circle. The y-coordinate would be sin(143°) ≈ 0.602 and the x-coordinate would be cos(143°) ≈ –0.799. Because of similar triangles, all we need to find our endpoint is to multiply each of these coordinates by 16. Because 16(–0.799) = –12.784 and 16(0.602) = 9.632, the endpoint of this vector is (–12.784,9.632). This is illustrated in Figure 16.8.
In general, the endpoint of a vector with angle θ and length L is (L · cos(θ),L · sin(θ)).
The components of a vector can be very useful in solving problems. When an object is being pushed or pulled, only the component that points in the direction of movement actually contributes to the movement.
For example, suppose a heavy weight sits on the ground. A person ties a rope to the weight and pulls so that the rope makes an angle of 40° with the ground. Suppose that the person pulls with 150 pounds of force, enough force to lift a 150-pound object (see Figure 16.9).
The vector with magnitude 150 pounds and 40° can be broken into two component vectors. The one in the x-direction is 150 · cos(40°) ≈ 115 pounds. The one in the y-direction is 150 · sin(40°) ≈ 96 pounds. This means that only 115 pounds of force is pulling the weight toward the person. The rest of the force is acting to lift the weight off the ground. Whether or not this is enough to make the weight move depends on how heavy it is and how much friction the floor has.
If the person were to pull with the same 150 pounds of force, but at a 30° angle instead, then the x-component of the vector would be 150 · cos(30°) ≈ 130 pounds of force. This is illustrated in Figure 16.10. This explains why people bend down low to push cars and pull heavy objects.
Similarly, if an object is on a slope, only a component of its weight acts to pull it down the slope.
Suppose a 200-pound weight is put on a 15° ramp, as illustrated in Figure 16.11. How much force is pulling the weight down the slope?
Gravity is pulling the weight straight down. The direction down the slope is 15° from horizontal, thus 75° from straight down, as illustrated in Figure 16.12.
We want to know the length x of the component vector that points down the slope. This is adjacent to the 75° angle of a right triangle with hypotenuse 200. Thus,
Thus, about 51.8 pounds of the object's weight are pushing in the direction of the slope. If a person pushed up the slope with a force of more than 51.8 pounds, then the weight would slide uphill (after friction was overcome).
To add two vectors, convert them into components and add each component separately. For example, the sum of (8,3) and (–4,4) is (8 + (–4),3 + 4) = (4,7). If the two vectors represent two trips, the sum will be the result of traveling along first one path and then the second. This is illustrated in Figure 16.15.
If the two vectors represent forces, the sum represents the overall effect of applying both forces at the same time to a single object. This is illustrated in Figure 16.16.
Suppose a hiker travels 15° west of north for 5 miles, and then 10° north of west for 8 miles. How far and in what direction is the hiker from the starting point?
The first vector has magnitude 5 and direction 105°, so the components are (5 · cos(105°),5 · sin(105°)) (–1.3,4.8). The second vector has magnitude 8 and direction 170°, so its components are (8 · cos(170°), 8 · sin(170°)) = (–7.9,1.4). The sum is thus (–1.3 - 7.9,4.8 + 1.4) = (–9.2,6.2). The magnitude of this vector is √(–9.2)2 + (6.2)2 = √123.08 ≈ 11.1, so the hiker is now 11.1 miles away from the starting point. The angle θ this vector makes with the x-axis is . Looking at the sketch in Figure 16.17, this direction is 34° north of west.
Two people pull on a sack of potatoes. The angle between them is 40°. One person pulls with 100 pounds of force and the other with 70 pounds. In which direction will the sack move and with how much force?
To make things easier, we can suppose that the larger force is headed in the 0° direction. We thus have the situation in Figure 16.18.
The component form for the larger vector is (100,0). The component form of the second vector is (70 · cos(40°),70 · sin(40°)) = (53.6,45). The sum is thus (153.6,45). This has a magnitude of √(153.6)2 + 452 = √25,617.96 ≈ 160 pounds. The angle of this vector is .
Thus, the sack will move in a direction between the two people that is 16.3° from the person who is pulling with 100 pounds of force. It will move as if pulled in that direction by 160 pounds of weight.
Practice problems for this study guide can be found at:
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