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5.7: Component Vectors

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Learning Objectives

  • Perform scalar multiplication with vectors.
  • Find the resultant as a sum of two components.
  • Find the resultant as magnitude and direction.
  • Use component vectors to solve real-world and applied problems.

A car has traveled 216 miles in a direction of 46^\circ north of east. How far east of its initial point has it traveled? How far north has the car traveled?

The car traveled on a vector distance called a displacement. It moved in a line to a particular distance from the starting point. Having two components in their expression, vectors are confusing to some. A diagram helps sort out confusion. Looking at vectors by separating them into components allows us to deal with many real-world problems. The components often relate to very different elements of the problem, such as wind speed in one direction and speed supplied by a motor in another.

In order to find how far the car has traveled east and how far it has traveled north, we will need to find the horizontal and vertical components of the vector. To find \vec{x}, we use cosine and to find \vec{y} we use sine.

\cos 46 & = \frac{|\vec{x}|}{216} = \frac{x}{216} && \sin 46 = \frac{|\vec{y}|}{216} = \frac{y}{216} \\\cos 46 & = \frac{x}{216} && \sin 46 = \frac{y}{216} \\216 \cos 46 & = x && 216 \sin 46 = y \\x & = 150.0 && y = 155.4

In this section, we will learn about component vectors and how to find them. We will also explore other ways of finding the magnitude and direction of a resultant of two or more vectors. We will be using many of the tools we learned in the previous sections dealing with right and oblique triangles.

Vector Multiplied by a Scalar

In working with vectors there are two kinds of quantities employed. The first is the vector, a quantity that has both magnitude and direction. The second quantity is a scalar. Scalars are just numbers. The magnitude of a vector is a scalar quantity. A vector can be multiplied by a real number. This real number is called a scalar. The product of a vector \vec{a} and a scalar k is a vector, written \vec{ka}. It has the same direction as \vec{a} with a magnitude of k|\vec{a}| if k > 0. If k < 0, the vector has the opposite direction of \vec{a} and a magnitude of k |\vec{a}|.

Example 1: The speed of the wind before a hurricane arrived was 20 mph from the SSE (N 22.5^\circ W). It quadrupled when the hurricane arrived. What is the current vector for wind velocity?

Solution: The wind is coming now at 80 mph from the same direction.

Example 2: A sailboat was traveling at 15 knots due north. After realizing he had overshot his destination, the captain turned the boat around and began traveling twice as fast due south. What is the current velocity vector of the ship?

Solution: The ship is traveling at 30 knots in the opposite direction.

If the vector is expressed in coordinates with the starting end of the vector at the origin, this is called standard form. To perform a scalar multiplication, we multiply our scalar by both the coordinates of our vector. The word scalar comes from “scale.” Multiplying by a scalar just makes the vectors longer or shorter, but doesn't change their direction.

Example 3: Consider the vector from the origin to (4, 6). What would the representation of a vector that had three times the magnitude be?

Solution: Here k = 3 and \vec{v} is the directed segment from (0,0) to (4, 6).

Multiply each of the components in the vector by 3.

\vec{kv} & = (0,0)\ to\ (12,18)

The new coordinates of the directed segment are (0, 0), (12, 18).

Example 4: Consider the vector from the origin to (3, 5). What would the representation of a vector that had -2 times the magnitude be?

Solution: Here, k =-2 and \vec{v} is the directed segment from (0, 0) to (3, 5).

\vec{kv} = (-2(3), -2(5)) = (-6,-10)

Since k < 0, our result would be a directed segment that is twice and long but in the opposite direction of our original vector.

Translation of Vectors and Slope

What would happen if we performed scalar multiplication on a vector that didn’t start at the origin?

Example 5: Consider the vector from (4, 7) to (12, 11). What would the representation of a vector that had 2.5 times the magnitude be?

Solution: Here, k = 2.5 and \vec{v} = the directed segment from (4, 7) to (12, 11).

Mathematically, two vectors are equal if their direction and magnitude are the same. The positions of the vectors do not matter. This means that if we have a vector that is not in standard position, we can translate it to the origin. The initial point of \vec{v} is (4, 7). In order to translate this to the origin, we would need to add (-4, -7) to both the initial and terminal points of the vector.

Initial point: (4, 7) + (-4, -7) = (0,0)

Terminal point: (12, 11) + (-4, -7) = (8, 4)

Now, to calculate \vec{kv} :

\vec{kv} & = (2.5(8), 2.5(4)) \\\vec{kv} & = (20,10)

The new coordinates of the directed segment are (0, 0) and (20, 10). To translate this back to our original terminal point:

Initial point: (0, 0) + (4, 7) = (4, 7)

Terminal point: (20, 10) + (4, 7) = (24, 17)

The new coordinates of the directed segment are (4, 7) and (24, 17).

Vectors with the same magnitude and direction are equal. This means that the same ordered pair could represent many different vectors. For instance, the ordered pair (4, 8) can represent a vector in standard position where the initial point is at the origin and the terminal point is at (4, 8). This vector could be thought of as the resultant of a horizontal vector with a magnitude or 4 units and a vertical vector with a magnitude of 8 units. Therefore, any vector with a horizontal component of 4 and vertical component of 8 could also be represented by the ordered pair (4, 8).

If you think back to Algebra, you know that the slope of a line is the change in y over the change in x, or the vertical change over the horizontal change. Looking at our vectors above, since they all have the same horizontal and vertical components, they all have the same slope, even though they do not all start at the origin.

Unit Vectors and Components

A unit vector is a vector that has a magnitude of one unit and can have any direction. Traditionally \hat{i} (read “i hat”) is the unit vector in the x direction and \hat{j} (read “j hat”) is the unit vector in the y direction. |\hat{i}| = 1 and |\hat{j}| = 1. Unit vectors on perpendicular axes can be used to express all vectors in that plane. Vectors are used to express position and motion in three dimensions with \hat{k} (“k hat”) as the unit vector in the z direction. We are not studying 3D space in this course. The unit vector notation may seem burdensome but one must distinguish between a vector and the components of that vector in the direction of the x- or y-axis. The unit vectors carry the meaning for the direction of the vector in each of the coordinate directions. The number in front of the unit vector shows its magnitude or length. Unit vectors are convenient if one wishes to express a 2D or 3D vector as a sum of two or three orthogonal components, such as x- and y-axes, or the z-axis. (Orthogonal components are those that intersect at right angles.)

Component vectors of a given vector are two or more vectors whose sum is the given vector. The sum is viewed as equivalent to the original vector. Since component vectors can have any direction, it is useful to have them perpendicular to one another. Commonly one chooses the x and y axis as the basis for the unit vectors. Component vectors do not have to be orthogonal.

A vector from the origin (0, 0) to the point (8, 0) is written as 8 \hat{i}. A vector from the origin to the point (0, 6) is written as 6 \hat{j}.

The reason for having the component vectors perpendicular to one another is that this condition allows us to use the Pythagorean Theorem and trigonometric ratios to find the magnitude and direction of the components. One can solve vector problems without use of unit vectors if specific information about orientation or direction in space such as N, E, S or W is part of the problem.

Resultant as the Sum of Two Components

We can look at any vector as the resultant of two perpendicular components. If we generalize the figure above, |\vec{r}|\hat{i} is the horizontal component of a vector \vec{q} and |\vec{s}| \hat{j} is the vertical component of \vec{q}. Therefore \vec{r} is a magnitude, |\vec{r}|, times the unit vector in the x direction and \vec{s} is its magnitude, |\vec{s}|, times the unit vector in the y direction. The sum of \vec{r} plus \vec{s} is: \vec{r} + \vec{s} = \vec{q}. This addition can also be written as |\vec{r}|\hat{i} + |\vec{s}| \hat{j} = \vec{q}.

If we are given the vector \vec{q}, we can find the components of \vec{q},\vec{r}, and \vec{s} using trigonometric ratios if we know the magnitude and direction of \vec{q}.

Example 6: If |\vec{q}| = 19.6 and its direction is 73^\circ, find the horizontal and vertical components.

Solution: If we know an angle and a side of a right triangle, we can find the other remaining sides using trigonometric ratios. In this case, \vec{q} is the hypotenuse of our triangle, \vec{r} is the side adjacent to our 73^\circ angle, \vec{s} is the side opposite our 73^\circ angle, and \vec{r} is directed along the x-axis.

To find \vec{r}, we will use cosine and to find \vec{s} we will use sine. Notice this is a scalar equation so all quantities are just numbers. It is written as the quotient of the magnitudes, not the vectors.

\cos 73 & = \frac{|\vec{r}|}{|\vec{q}|} = \frac{r}{q} && \sin 73 = \frac{|\vec{s}|}{|\vec{q}|} = \frac{s}{q} \\\cos 73 & = \frac{r}{19.6} && \sin 73 = \frac{s}{19.6} \\r & = 19.6 \cos 73 && \qquad \ \ s = 19.6 \sin 73 \\r & = 5.7 && \qquad \ \ s = 18.7

The horizontal component is 5.7 and the vertical component is 18.7. One can rewrite this in vector notation as 5.7\hat{i} + 18.7\hat{j} = \vec{q}. The components can also be written \vec{q} = \big \langle{5.7, 18.7}\big\rangle, with the horizontal component first, followed by the vertical component. Be careful not to confuse this with the notation for plotted points.

Resultant as Magnitude and Direction

If we don’t have two perpendicular vectors, we can still find the magnitude and direction of the resultant without a graphic estimate with a construction using a compass and ruler. This can be accomplished using both the Law of Sines and the Law of Cosines.

Example 7: \vec{A} makes a 54^\circ angle with \vec{B}. The magnitude of \vec{A} is 13.2. The magnitude of \vec{B} is 16.7. Find the magnitude and direction the resultant makes with the smaller vector.

There is no preferred orientation such as a compass direction or any necessary use of x and y coordinates. The problem can be solved without the use of unit vectors.

Solution: In order to solve this problem, we will need to use the parallelogram method. Since vectors only have magnitude and direction, one can move them on the plane to any position one wishes, as long as the magnitude and direction remain the same. First, we will complete the parallelogram: Label the vectors. Move \vec{b} so its tail is on the tip of \vec{a}. Move \vec{a} so its tail is on the tip of \vec{b}. This makes a parallelogram because the angles did not change during the translation. Put in labels for the vertices of the parallelogram.

Since opposite angles in a parallelogram are congruent, we can find angle A.

\angle{CBD} +  \angle{CAD} + \angle{ACB} + \angle{BDA} & = 360 \\2 \angle{CBD} + 2 \angle{ACB} & = 360 \\\angle{ACB} & = 54^\circ \\2 \angle{CBD} & = 360 - 2(54) \\\angle{CBD} & = \frac{360-2(54)}{2} = 126

Now, we know two sides and the included angle in an oblique triangle. This means we can use the Law of Cosines to find the magnitude of our resultant.

x^2 & = 13.2^2 + 16.7^2 - 2(13.2)(16.7) \cos 126 \\x^2 & = 712.272762  \\x & = 26.7

To find the direction, we can use the Law of Sines since we now know an angle and a side across from it. We choose the Law of Sines because it is a proportion and less computationally intense than the Law of Cosines.

\frac{\sin \theta}{16.7} & = \frac{\sin 126}{26.7} \\\sin \theta & = \frac{16.7 \sin 126}{26.7} \\\sin \theta & = 0.5060143748 \\\theta & = \sin ^{-1} \ 0.5060 = 30.4^\circ

The magnitude of the resultant is 26.7 and the direction it makes with the smaller vector is 30.4^\circ counterclockwise.

We can use a similar method to add three or more vectors.

Example 8: Vector A makes a 45^\circ angle with the horizontal and has a magnitude of 3. Vector B makes a 25^\circ angle with the horizontal and has a magnitude of 5. Vector C makes a 65^\circ angle with the horizontal and has a magnitude of 2. Find the magnitude and direction (with the horizontal) of the resultant of all three vectors.

Solution: To begin this problem, we will find the resultant using Vector A and Vector B. We will do this using the parallelogram method like we did above.

Since Vector A makes a 45^\circ angle with the horizontal and Vector B makes a 25^\circ angle with the horizontal, we know that the angle between the two (\angle{ADB}) is 20^\circ.

To find \angle{DBE}:

2 \angle{ADB} + 2 \angle{DBE} & = 360 \\\angle{ADB} & = 20^\circ \\2 \angle{DBE} & = 360 - 2(20) \\ \angle{DBE} & = \frac{360-2(20)}{2} = 160

Now, we will use the Law of Cosines to find the magnitude of DE.

DE^2 & = 3^2 + 5^2 - 2(3)(5) \cos 160 \\DE^2 & = 62 \\ DE & = 7.9

Next, we will use the Law of Sines to find the measure of angle EDB.

\frac{\sin 160}{7.9} & = \frac{\sin \angle{EDB}}{3} \\\sin \angle{EDB} & = \frac{3 \sin 160}{7.9} \\\sin \angle{EDB} & = .1299 \\\angle{EDB} & = \sin^{-1} \ 0.1299 = 7.46^\circ

We know that Vector B forms a 25^\circ angle with the horizontal so we add that value to the measure of \angle{EDB} to find the angle DE makes with the horizontal. Therefore, DE makes a 32.46^\circ angle with the horizontal.

Next, we will take DE, and we will find the resultant vector of DE and Vector C from above. We will repeat the same process we used above.

Vector C makes a 65^\circ angle with the horizontal and DE makes a 32^\circ angle with the horizontal. This means that the angle between the two (\angle{CDE}) is 33^\circ. We will use this information to find the measure of \angle{DEF}.

2\angle{CDE} + 2\angle{DEF} & = 360 \\ \angle{CDE} & = 33^\circ \\2 \angle{DEF} & = 360 - 2(33) \\\angle{DEF} & = \frac{360-2(33)}{2} = 147

Now we will use the Law of Cosines to find the magnitude of DF.

DF^2 & = 7.9^2 + 2^2 - 2(7.9)(2) \cos 147 \\DF^2 & = 92.9 \\DF & = 9.6

Next, we will use the Law of Sines to find \angle{FDE}.

\frac{\sin 147}{9.6} & = \frac{\sin \angle{FDE}}{2} \\\sin \angle{FDE} & = \frac{2 \sin 147}{9.6} \\\sin \angle{FDE} & = .1135 \\\angle{FDE} & = \sin^{-1} 0.1135 = 6.5^\circ = 7^\circ

Finally, we will take the measure of \angle{FDE} and add it to the 32^\circ angle that DE forms with the horizontal. Therefore, DF forms a 39^\circ angle with the horizontal.

Example 9: Two forces of 310 lbs and 460 lbs are acting on an object. The angle between the two forces is 61.3^\circ. What is the magnitude of the resultant? What angle does the resultant make with the smaller force?

Solution: We do not need unit vectors here as there is no preferred direction like a compass direction or a specific axis. First, to find the magnitude we will need to figure out the other angle in our parallelogram.

2 \angle{ACB} + 2 \angle{CAD} & = 360 \\\angle{ACB} & = 61.3^\circ \\2 \angle{CAD} & = 360 - 2(61.3) \\\angle{CAD} & = \frac{360 - 2 (61.3)}{2} = 118.7

Now that we know the other angle, we can find the magnitude using the Law of Cosines.

x^2 & = 460^2 + 310^2 - 2(460)(310) \cos 118.7^\circ \\x^2 & = 444659.7415  \\x & = 667

To find the angle the resultant makes with the smaller force, we will use the Law of Sines.

\frac{\sin \theta}{460} & = \frac{\sin 118.7}{666.8} \\\sin \theta & = \frac{460 \sin 118.7}{666.8} \\\sin \theta & = .6049283888 \\\theta & = \sin^{-1}\ 0.6049 = 37.2^\circ

Example 10: Two trucks are pulling a large chunk of stone. Truck 1 is pulling with a force of 635 lbs at a 53^\circ angle from the horizontal while Truck 2 is pulling with a force of 592 lbs at a 41^\circ angle from the horizontal. What is the magnitude and direction of the resultant force?

Solution: Since Truck 1 has a direction of 53^\circ and Truck 2 has a direction of 41^\circ, we can see that the angle between the two forces is 12^\circ. We need this angle measurement in order to figure out the other angles in our parallelogram.

2\angle{ACB} + 2\angle{CAD} & = 360 \\ \angle{ACB} & = 12^\circ \\2 \angle{CAD} & = 360 - 2(12) \\\angle{CAD} & = \frac{360-2(12)}{2} = 168

Now, use the Law of Cosines to find the magnitude of the resultant.

x^2 & = 635^2 + 592^2 - 2(635)(592) \cos 168^\circ \\ x^2 &= 1489099 \\x & = 1220.3\ lbs

Now to find the direction we will use the Law of Sines.

\frac{\sin \theta}{635} & = \frac{\sin 168}{1220.3} \\\sin \theta & = \frac{635 \sin 168}{1220.3} \\\sin \theta & = 0.1082 \\\theta & = \sin^{-1}\ 0.1082 = 6^\circ

Since we want the direction we need to add the 6^\circ to the 41^\circ from the smaller force. The magnitude is 1220 lbs and 47^\circ counterclockwise from the horizontal.

Example 11: Two tractors are being used to pull down the framework of an old building. One tractor is pulling on the frame with a force of 1675 pounds and is headed directly north. The second tractor is pulling on the frame with a force of 1880 pounds and is headed 33^\circ north of east. What is the magnitude of the resultant force on the building? What is the direction of the result force?

Soultion: We are asked to find the resultant force and direction, which means we are dealing with vectors. In order to complete our diagram, we will need to connect our two vectors and draw in our resultant. We will refer to the magnitude of our resultant as x and the direction of our resultant as \theta.

When finding the resultant of two vectors, we can choose from either the triangle method or the parallelogram method. We will solve this problem using the parallelogram method. Looking at the diagram, we can see that the two vectors form an angle of 57, (90 - 33). This means that the angle opposite the angle formed by our two vectors is also 57. To find the other two angles in our parallelogram, we know that the sum of all the angles must add up to 360 and that opposite angles must be congruent, \frac{360 - (57+57)}{2} = 123.

Now, we can use two sides of our parallelogram and our resultant to form a triangle in which we know two sides and the included angle (SAS).

This means that we can use the Law of Cosines to find the magnitude (x) of the resultant.

x^2 & = 1675^2 + 1880^2 - 2(1675)(1880) \cos 123 \\x^2 & = 9770161.643 \\x & = 3125.7

To find the direction (\theta), we can use the Law of Sines since we now know an angle and the side opposite it.

\frac{\sin 123}{3125.7} & = \frac{\sin \theta}{1675} \\\frac{1675 \sin 123}{3125.7} & = \sin \theta \\0.449427 & = \sin \theta \\26.71 & = \theta

Now that we know \theta, in order to find the angle of the resultant, we must add the 33^\circ from the x-axis to \theta, 33^\circ + 26.71^\circ = 59.71^\circ.

Points to Consider

  • How you can verify if your answers to problems involving vectors that are not perpendicular are correct?
  • In what ways are solving problems with oblique triangles and solving problems involving vectors similar?
  • In what ways are they different?
  • When is it appropriate to use vectors instead of oblique triangles to solve problems?
  • When is it helpful to use unit vectors? When can one solve a problem without explicitly using them?

Review Questions

  1. Find the resulting ordered pair that represents \vec{a} in each equation if you are given \vec{b} = (0,0)\ to \ (5,4) and \vec{c} = (0,0)\ to\ (-3,7).
    1. \vec{a} = 2 \vec{b}
    2. \vec{a} = - \frac{1}{2} \vec{c}
    3. \vec{a} = 0.6\vec{b}
    4. \vec{a} = - 3 \vec{b}
  2. Find the magnitude of the horizontal and vertical components of the following vectors given that the coordinates of their initial and terminal points.
    1. \text{initial} = (-3,8) \qquad \qquad \quad \text{terminal} = (2, -1)
    2. \text{initial} = (7, 13)  \qquad \qquad \quad  \ \text{terminal} = (11, 19)
    3. \text{initial} = (4.2, -6.8)  \qquad \quad  \ \text{terminal} = (-1.3, -9.4)
  3. Find the magnitude of the horizontal and vertical components if the resultant vector’s magnitude and direction are given.
    1. \text{magnitude} = 75  \qquad \qquad \quad  \text{direction} = 35^\circ
    2. \text{magnitude} = 3.4  \qquad \qquad \quad \text{direction} = 162^\circ
    3. \text{magnitude} = 15.9  \qquad \qquad \ \text{direction} = 12^\circ
  4. Two forces of 8.50 Newtons and 32.1 Newtons act on an object at right angles. Find the magnitude of the resultant and the angle that it makes with the smaller force.
  5. Forces of 140 Newtons and 186 Newtons act on an object. The angle between the forces is 43^\circ. Find the magnitude of the resultant and the angle it makes with the larger force.
  6. An incline ramp is 12 feet long and forms an angle of 28.2^\circ with the ground. Find the horizontal and vertical components of the ramp.
  7. An airplane is traveling at a speed of 155 km/h. It's heading is set at 83^\circ while there is a 42.0 km/h wind from 305^\circ. What is the airplane's actual heading?
  8. A speedboat is capable of traveling at 10.0 mph, but is in a river that has a current of 2.00 mph. In order to cross the river at right angle, in what direction should the boat be heading?
  9. If \overrightarrow{AB} is any vector, what is \overrightarrow{AB} + \overrightarrow{BA}?

Review Answers

    1. 2 \vec{b} = 2 \big \langle{5, 4}\big\rangle = \big \langle{10, 8}\big\rangle = 10 \hat{i} + 8 \hat{j}
    2. - \frac{1}{2} \vec{c} = - \frac{1}{2} \big \langle{-3, 7}\big\rangle = \big \langle{1.5, -3.5}\big\rangle = 1.5 \hat{i} - 3.5 \hat{j}
    3. 0.6\vec{b} = 0.6 \big \langle{5, 4}\big\rangle = \big \langle{3,2.4}\big\rangle = 3 \hat{i} + 2.4 \hat{j}
    4. -3 \vec{b} = -3 \big \langle{5,4}\big\rangle = \big \langle{-15, -12}\big\rangle = -15 \hat{i} - 12 \hat{j}
  1. All of these need to be translated to (0,0). Also, recall that magnitudes are always positive. (a) (-3, 8) + (3, -8) = (0, 0) \qquad \qquad (2, -1) + (3, -8) = (5, -9) horizontal = 5, vertical = 9 (b) (7, 13) + (-7, -13) = (0, 0)\qquad \qquad  (11, 19) + (-7, -13) = (4, 6) horizontal = 4, vertical = 6 (c) (4.2, -6.8) + (-4.2, 6.8) = (0, 0) \qquad \qquad (-1.3, -9.4) + (-4.2, 6.8) = (-5.5, -2.6) horizontal = 5.5, vertical = 2.6
    1. \cos 35^\circ = \frac{x}{75}, \sin 35^\circ = \frac{y}{75}, x = 61.4, y = 43
    2. \cos 162^\circ = \frac{x}{3.4}, \sin 162^\circ = \frac{y}{3.4}, x = 3.2, y = 1.1
    3. \cos 12^\circ = \frac{x}{15.9}, \sin 12^\circ  = \frac{y}{15.9}, x = 15.6, y = 3.3
  2. magnitude = 33.2, direction = 75.2^\circ from the smaller force
  3. magnitude = 304, 18.3^\circ between resultant and larger force
  4. y = 12 \sin 28.2^\circ = 5.7, x = 12 \cos 28.2^\circ = 10.6
  5. Recall that headings and angles in triangles are complementary. So, an 83^\circ heading translates to 7^\circ from the horizontal. Adding that to 35^\circ (270^\circ from 305^\circ) we get 42^\circ for two of the angles in the parallelogram. So, the other angles in the parallelogram measure 138^\circ each, \frac{360-2(42)}{2}. Using 138^\circ in the Law of Cosines, we can find the diagonal or resultant, x^2 = 42^2 + 155^2 - 2(42)(155) \cos 138, so x = 188.3. We then need to find the angle between the resultant and the speed using the Law of Sines. \frac{\sin a}{42} = \frac{\sin 138}{188.3}, so a = 8.6^\circ. To find the actual heading, this number needs to be added to 83^\circ, getting 91.6^\circ.
  6. The heading is just \tan \theta = \frac{2}{10}, or 11.3^\circ against the current.
  7. BA is the same vector as AB, but because it starts with B it is in the opposite direction. Therefore, when you add the two together, you will get (0,0).

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