5.1: Vectors in a Plane
Learning objectives
- Identify difference between vector and scalar quantities.
- Represent vector quantities graphically and using component notation.
- Compute the resultant of a set of vectors.
- Determine the unit vector to represent a given direction.
Introduction
In many cases, numbers alone are sufficient to model physical situations or to answer such questions as, “what temperature is it?” or “how much lumber will I need to fix the porch stairs?” or “what grade should I expect in this class?” Sometimes, however, a single number is not sufficient to convey the necessary information. For example, crime scene investigators know that both the speed and the direction of a car’s motion will affect the outcome of a crash. Quantities that can be represented by a single number are called scalar quantities. Those that require multiple numbers or directional representation are called vectors. In this section we will discuss both the geometric and algebraic properties of vectors in two dimensions.
Two Dimensional Vectors
Throughout this website we will be using bold-face type to indicate vector quantities and regular face type for scalar quantities, whether we are using the endpoints of the vector or another variable to represent the vector quantity: AB or r. Another way to symbolize vector quantities is to use letters with arrows over them. For example, physicists use \begin{align*}\overrightarrow {v}\end{align*}
Vectors can be represented graphically using an arrow which points from the vector’s initial point to its terminal point. The length of the arrow-shaft represents the magnitude of the vector. The arrowhead identifies the vector’s direction.
Vectors AB and CD have the same magnitude and the same direction, so we can say that AB = CD, even if they don’t have the same location in space. Vectors AB and EF have the same magnitude, but they have opposite directions, therefore EF = – AB. The magnitude of a vector is a scalar quantity which can be symbolized by | AB |, | r |, \begin{align*}|\overrightarrow {v}|\end{align*}
Vectors can also be represented algebraically, using component notation. The components tell us the extent of the vector along the coordinate directions. For example, the vector on the left in the image below has components in both the x-direction, AB_{x}, and the y-direction, AB_{y}. Here, AB_{x} = 4 and AB_{y} = 3 because point B is 3 units to the right and 3 units up from point A. Similarly, vector D has components D_{x} = 3 and D_{y} = –1.25. The negative sign in D_{y} indicates that the y-component of vector D is downward, in the negative y direction. As with other numbers, we usually only include the negative sign explicitly.
Sometimes bracket notation is used to identify the components of a vector. The components of the vector are written as an ordered set of values, similar to the coordinates of a point in space. For example, vector \begin{align*}AB = \left \langle 4,3\right \rangle\end{align*}
If the coordinate system is polar rather than rectangular, a vector is identified by its magnitude and its direction with respect to the x-axis (or r-axis). In the case of the vector below, A = 10 m @ 30^{o}.
The process of determining the components of a vector is also known as resolving the vector. For example, if we wish to convert the radial coordinate notation for vector A into rectangular coordinates, we can use right triangle trigonometry to resolve the vector. First, create a right triangle with the vector as the hypotenuse and with the two legs of the triangle parallel to the rectangular coordinate axes. Then, according to the definitions of sine and cosine, A_{x} = | A | cos 30 and A_{y} = | A | sin 30.
Example: What are the x and y components of the vectors shown in the diagram?
Solution:
AB | CD | EF | GH | IJ | KL | MN | |
---|---|---|---|---|---|---|---|
x - component | 2.25 | -1.5 | 0 | -1.5 | 2.0 | 1.5 | -2.25 |
y - component | 0.25 | -2.0 | -2.25 | -2.0 | -1.5 | 2.0 | 0 |
In the diagram, each division is 0.25 units. All vectors which point toward the left have negative x-components and those that point downward have negative y-components. Notice that for the horizontal vector, MN, the y-component is equal to 0. Likewise, for the vertical vector, EF, the x-component is equal to 0.
Example: (a) Which of the vectors in Example #1 is equal to vector CD? (b) Which vector is equal to –CD?
Solution: (a) Vector GH = CD. Both vectors have the same length and the same orientation. (b) Vector KL = –CD. Both vectors have the same length and the two vectors point in opposite directions.
Addition and Subtraction of Vector Quantities
Vectors, like scalars, can be combined arithmetically, but, in contrast to adding scalars, we need to take the direction of vectors into account. In the diagram below, vector C is the resultant or sum of vectors A and B. A + B = C. To add the vectors graphically, position the individual vectors head to tail without changing their orientations. By head-to-tail, we mean that the initial point of the second vector is coincident with the terminal point of the first vector and so on. As you can see in the figure below, the order of the head-to-tail addition of the vectors does not affect the magnitude or direction of the resultant. In other words, the addition of the vectors is commutative: A + B = B + A.
The head-to-tail addition process also works to combine groups of three or more vectors. In the diagram below, we can also see that the associative law also applies to vector addition:
A + B + C = (A + C) + B = A + (B + C).
Vectors can be added graphically, as in the diagrams above. They can also be added using their components. For any set of vectors, the x-component of the resultant vector is equal to the sum of the x-components of the individual vectors. Likewise, the y-component of the resultant is the sum of the individual y-components. For example, in the diagram below, \begin{align*}\overrightarrow {A} + \overrightarrow {B} = \overrightarrow {C}\end{align*}
Example: Demonstrate the associative law using vectors \begin{align*}\overrightarrow {AB}, \ \overrightarrow {CD},\end{align*}
Solution: The associative law states that A + B + C = (A + C) + B = A + (B + C). Here, that means that \begin{align*}(\overrightarrow {AB} + \overrightarrow {CD}) + \overrightarrow {EF} = \overrightarrow {AB} + (\overrightarrow {CD} + \overrightarrow {EF}) = \overrightarrow {AB} + \overrightarrow {CD} + \overrightarrow {EF}\end{align*}
Firstly, \begin{align*}(\overrightarrow {AB} + \overrightarrow {CD}) = \left \langle (2.25 - 1.5), (0.25 - 2)\right \rangle = \left \langle 0.75, -1.75\right \rangle\end{align*}
then, \begin{align*}(\overrightarrow {AB} + \overrightarrow {CD}) + \overrightarrow {EF} = \left \langle (0.75 + 0), (-1.75 - 2.25)\right \rangle = \left \langle 0.75, -4\right \rangle\end{align*}
Compare this to the other order for the addition.
Firstly, \begin{align*}(\overrightarrow {CD} + \overrightarrow {EF})= \left \langle (-1.5 + 0), (-2 - 2.25)\right \rangle = \left \langle -1.5, -4.25\right \rangle\end{align*}
then, \begin{align*}\overrightarrow{AB} + (\overrightarrow{CD} + \overrightarrow{EF}) = \left \langle (2.25 - 1.5),(0.25 - 4.25)\right \rangle = \left \langle 0.75, -4\right \rangle\end{align*}
Also, \begin{align*}\overrightarrow{AB} + \overrightarrow{CD} + \overrightarrow{EF} = \left \langle (2.25 - 1.5 + 0),(0.25 - 2 - 2.25)\right \rangle = \left \langle 0.75, -4\right \rangle\end{align*}
Example
My aunt Frances and her son Allen both have homes in the hamlet of Blue Lake. Usually, my cousin drives around the lake to reach his mother’s home, but when the weather is fine, he sometimes takes his canoe across the lake instead. The black path (A, B, C, E, D, F) marks his driving route and the red path marks his trip in the canoe. Use vector addition techniques to determine both (a) the distance he travels across the lake and (b) his heading, θ. A map of the area around Blue Lake and the distances for each leg of his journey are shown in the figure below.
Solution:
Both the canoe trip and the drive will take Allen from his house to his mother’s home. The vector equation
\begin{align*}\overrightarrow{AF} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} + \overrightarrow{EF}\end{align*}
AF_{x} = AB_{x} + BC_{x} + CD_{x} + DE_{x} + EF_{x} = 0 + 295 m + 0 – 225 m + 0 = 70 m
AF_{y} = AB_{y} + BC_{y} + CD_{y} + DE_{y} + EF_{y} = –50 m + 0 + 310 m + 0 – 60 m = 200 m
Now that we know the x and y components of vector AF we can determine the magnitude and direction of the vector. According to the Pythagorean Theorem,
\begin{align*}AF^2 = AF^2_x + AF^2_y\end{align*}
\begin{align*}AF = \sqrt{(AF_x)^2 + (AF_y)^2} = \sqrt{70^2 + 200^2} = \sqrt{4900 + 40,000} = 212\mbox{m}\end{align*}
Then, since we know two sides of a right triangle, we can use trigonometry to determine Allen’s heading.
\begin{align*}\tan\ \theta = \frac{AF_y} {AF_x} = \frac{200} {70} = 0.2857\end{align*}
To subtract one vector from another, we simply add its opposite. For example, \begin{align*}\overrightarrow{A} - \overrightarrow{B} = \overrightarrow{A} + (-\overrightarrow{B})\end{align*}
Multiplying By a Scalar
If we need to add several vectors that have the same magnitude and direction, we can utilize the fact that vectors can use a simplification commonly used when adding scalars. Specifically, multiply the scalar by the number of times it appears as a factor in the addition. For example,
a + b + c + b + b + a + b = 2a + 4b + c
When multiplying a vector by a positive scalar, one changes the magnitude of the vector without changing its direction. Let’s look at the set of vectors shown below. Three identical vectors, A, placed end to end will reach from the origin to point P. This same displacement can be represented by the single vector B. Therefore 3A = B. Here \begin{align*}\overrightarrow{A} = \left \langle 1.5, 1 \right \rangle\end{align*}
When a vector is multiplied by a scalar, each of the components of the vector are multiplied by that same scalar. In component notation this becomes,
\begin{align*}\overrightarrow{A} + \overrightarrow{A} + \overrightarrow{A} = \left \langle (1.5 + 1.5 + 1.5),(1 + 1 + 1) \right \rangle = \left \langle 4.5, 3 \right \rangle\end{align*}
\begin{align*}3(\overrightarrow{A}) = \left \langle(3*1.5),(3*1)\right \rangle = \left \langle 4.5, 3\right \rangle\end{align*}
Vectors can be divided by scalar quantities as well, since by definition \begin{align*}\frac{x} {3} = \left (\frac{1} {3}\right )x\end{align*}.
Example: A force is a vector quantity indicting the magnitude and direction of a push or pull on an object. Stickman Beauford and his friends Akiko and Carlos, are trying to push their car out of the mud. Beauford and Carlos push with a force of 150 N, but Akiko has a sore wrist and can only push with a force of 85 N. (Note, 1.0 N = 0.225 lbs). If all three of them push horizontally on back of the trunk, what is their total force on the car?
\begin{align*}\overrightarrow{F_{total}} = \overrightarrow{F_{Akiko}} + \overrightarrow{F_{Beauford}} + \overrightarrow{F_{Carlos}}\end{align*}
Since Beauford and Carlos push with equal force, \begin{align*}\overrightarrow{F_{Beauford}} + \overrightarrow{F_{Carlos}} = 2\overrightarrow{F_{Beauford}}\end{align*}, therefore \begin{align*}\overrightarrow{F_{total}} = \overrightarrow{F_{Akiko}} + 2\overrightarrow{F_{Beauford}} = [85N + (2*150N)]\hat{x} = (385 N)\hat{x}\end{align*}
Unit Vectors
Dividing any vector by its own magnitude produces a new vector with the same direction but with unit magnitude (i.e. magnitude equal to 1), called a unit vector.
\begin{align*}\hat{r} = \frac{\overrightarrow{r}} {r}\end{align*}
A unit vector is indicated by placing a carrot above the symbol rather than the usual vector arrow. Standard unit vectors are used to represent each of the rectangular coordinate directions: \begin{align*}\hat{x}, \hat{y},\end{align*} and \begin{align*}\hat{z}\end{align*} or \begin{align*}\hat{i}, \hat{j},\end{align*} and \begin{align*}\hat{k}\end{align*} respectively.
Using unit vector notation, the vector shown above can be written as \begin{align*}\overrightarrow{A} = (10\ \mbox{cos}\ 30^\circ)\hat{x} + (10\ \mbox{sin}\ 30^\circ)\hat{y}\end{align*} or \begin{align*}\overrightarrow{A} = (10\ \mbox{cos}\ 30^\circ)\hat{i} + (10\ \mbox{sin}\ 30^\circ)\hat{j}\end{align*}
Example: What are the components of the unit vector that represents the direction of the vector shown in the diagram below?
Solution: The unit vector is defined by \begin{align*}\hat{r} = \frac{\overrightarrow{r}} {r}\end{align*}, therefore, find the magnitude of this vector first.
\begin{align*}\overrightarrow{D} = \left \langle 3, -1.25 \right \rangle\end{align*}
\begin{align*}D = \sqrt{(D_x)^2 + (D_y)^2}\end{align*}
\begin{align*}D = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25}\end{align*}
\begin{align*}D = 5\end{align*}
Then,
\begin{align*}\overrightarrow{D} = \frac{\overrightarrow{D}} {| D |}\end{align*}
\begin{align*}\overrightarrow{D} = \frac{4\hat{x} + 3\hat{y}} {5} = \frac{4} {5} \hat{x} + \frac{3} {5} \hat{y}\end{align*}
\begin{align*}\overrightarrow{D} = \left \langle 0.8, 0.6 \right \rangle\end{align*}
Lesson Summary
Vectors are very powerful tools used by computer programmers, mathematicians, and scientists to describe quantities which have both a size and a direction. Vectors are represented visually using arrows whose length represents the magnitude of the vector and whose orientation represents the direction of the vector. We also have two ways to represent a vector symbolically. The first identifies the vector using its magnitude and direction, \begin{align*}\overrightarrow{V} = 2.5\end{align*} @ \begin{align*}60^\circ\end{align*}. The second method uses the vector’s components, which are the extension of the vector along each of the coordinate directions, \begin{align*}\overrightarrow{P} = 2\hat{x} + 3\hat{y}\end{align*}. The arrows over the V and P identify these quantities as vectors and the carrot over the x and y identify these as unit vectors. A unit vector is a vector which has unit magnitude. We can also use a shorthand version of the component notation by writing the components in order within a set of brackets, \begin{align*}\overrightarrow{P} = \left \langle 2, 3 \right \rangle\end{align*}.
Vectors can be combined mathematically using addition and subtraction, but adding vectors is different from adding scalars. When adding scalar numbers, you just add the numbers together, for example 2 + 3 = 5. To add vector quantities, add their x-components to one another, the y-components to one another, and so on. For example, if \begin{align*}\overrightarrow{P} = \left \langle 2, 3 \right \rangle\end{align*} and \begin{align*}\overrightarrow{R} = \left \langle 5,6 \right \rangle, \overrightarrow{P} + \overrightarrow{R} = \left \langle 7,9 \right \rangle\end{align*}.
Points to Consider
1. Can we use vectors to describe systems with more dimensions that just two?
2. Will the Pythagorean Theorem work to determine the length of a three-dimensional vector?
3. We saw in the Blue Lake problem that vectors can be used to identify paths between two points in 2D space? What are the general rules for identifying points and the distances between them?
Practice Problems
- 1. Two vectors, \begin{align*}\overrightarrow{A}\end{align*} and \begin{align*}\overrightarrow{B}\end{align*}, are shown in the diagram. Identify the \begin{align*}\hat{x}\end{align*} and \begin{align*}\hat{y}\end{align*} components of both vectors.
- For the two vectors \begin{align*}\overrightarrow{A}\end{align*} and \begin{align*}\overrightarrow{B}\end{align*} shown in the diagram, determine the components, magnitude, and orientation angle for vector \begin{align*}\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}\end{align*}.
- \begin{align*}\overrightarrow{A} = \left \langle 12, 7, 9.5 \right \rangle\end{align*} and \begin{align*}\overrightarrow{B} = \left \langle 8, 8, 11 \right \rangle\end{align*} Determine the vector \begin{align*}\overrightarrow{C}\end{align*} that makes the following equation true: \begin{align*}\overrightarrow{A} + \frac{1} {2}\overrightarrow{C} = \frac{2} {3}\overrightarrow{B}\end{align*}.
- A vector can be written as \begin{align*}\overrightarrow{R} = 2.74 m\end{align*} @ \begin{align*}60^\circ\end{align*}. Identify this same vector using component notation.
- Two vectors are identified as \begin{align*}\overrightarrow{R} = 4.5 m\end{align*} @ \begin{align*}20^\circ\end{align*} and \begin{align*}\overrightarrow{R} = 6.3m\end{align*} @ \begin{align*}135^\circ\end{align*}. Determine the magnitude and orientation angle for the sum of these two vectors.
- My neighbors recently adopted a retired racing greyhound named Flash. When Josef and Zara return home from work, they let Flash out into the back yard where she runs counter-clockwise around a circular path, similar to the one shown in the diagram. Joe recently used this circular motion to help his son Ahmed learn the difference between the vector quantity displacement and the scalar quantity distance. An object’s displacement is defined as the vector beginning at the object’s initial position and ending at the object’s final position. The exact path does not matter to the displacement. In contrast, the distance traveled by an object does depend on the path taken in getting from its initial position to its final position. If Flash’s circular path has a radius of 3.7 m, identify the displacement vector, \begin{align*}\overrightarrow{\triangle r}\end{align*}, and the distance traveled, d, as Flash runs from point 1 to point 2.
- Two displacement vectors have magnitudes of 2.5 cm and 6.0 cm. Is it possible to add these two vectors to obtain a vector having a magnitude of 7.5 cm? Explain why or why not using a diagram to illustrate your reasoning.
- Three vectors are identified by the following component equations: \begin{align*}\overrightarrow{P} = \left \langle 25, 17, 32 \right \rangle, \ \overrightarrow{L} = \left \langle 14, 23, 57 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{Q} = \left \langle 49, 11, 27 \right \rangle\end{align*}. Determine the components and magnitude of the vector \begin{align*}\overrightarrow{R} = 3\overrightarrow{P} + 2\overrightarrow{L} - \overrightarrow{Q}\end{align*}.
- Sasha’s three puppies frequently fight over their favorite toy, a bright green Frisbee. Smoky pulls on the Frisbee with a force \begin{align*}\overrightarrow{F_{SonF}} = (-5.5 N)\hat{x}\end{align*}, Ginger pulls with a force \begin{align*}\overrightarrow{F_{GonF}} = (-5.2 N)\hat{y}\end{align*}, and Bruno pulls with a force \begin{align*}\overrightarrow{F_{BonF}} = (+6.0 N)\hat{x}\end{align*}. What is the total force acting on the Frisbee?
- Aunt Francis frequently drives around Blue Lake to visit her friend, Rhoda, who lives at the corner marked C in the diagram below. Determine the components of the unit vector which identifies the direction from Aunt Francis’s home to Rhoda’s house?
Solutions
- Two vectors, \begin{align*}\overrightarrow{A}\end{align*} and \begin{align*}\overrightarrow{B}\end{align*}, are shown in the diagram. Identify the \begin{align*}\hat{x}\end{align*} and \begin{align*}\hat{y}\end{align*} components of both vectors. The \begin{align*}\hat{x}\end{align*} and \begin{align*}\hat{y}\end{align*} components of a vector are the extensions of the vector along the \begin{align*}\hat{x}\end{align*} and \begin{align*}\hat{y}\end{align*} directions. Here we can obtain that information off of the grid in the diagram. Vector A begins at (-2, 1) and ends at (2.75, 1.5), therefore its x-component is given by A_{x} = 2.75 - (-2) = 4.75 and its y-component is given by A_{y} = 1.5 - 1 = 0.5. Vector B begins at (4, 2) and ends at (1.75, -1.5), therefore its x-component is given by B_{x} = 1.75 - 4 = -2.25 and its y-component is given by B_{y} = -1.5 - 2 = -3.5
- For the two vectors \begin{align*}\overrightarrow{A}\end{align*} and \begin{align*}\overrightarrow{B}\end{align*} shown in the diagram, determine the components, magnitude, and orientation angle for vector \begin{align*}\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}\end{align*}. As we saw in the previous problem, \begin{align*}\overrightarrow{A} = \left \langle 4.75, 0.5 \right \rangle\end{align*} and \begin{align*}\overrightarrow{B} = \left \langle -2.25, -3.5 \right \rangle\end{align*}. To add the two vectors, we add the x-components together and we add the y-components together to give \begin{align*}\overrightarrow{C} = \left \langle (4.75 + (-2.25)),(-3.5 + 0.5)\right \rangle = \left \langle 2.5, -3 \right \rangle\end{align*} We can also add these two vectors graphically by positioning \begin{align*}\overrightarrow{A}\end{align*} and \begin{align*}\overrightarrow{B}\end{align*} head to tail. Vector \begin{align*}\overrightarrow{A}\end{align*} is the single vector that begins where \begin{align*}\overrightarrow{A}\end{align*} begins and ends where \begin{align*}\overrightarrow{B}\end{align*} ends. As you can see from the diagram, the components of vector \begin{align*}\overrightarrow{C}\end{align*} are \begin{align*}C_x = 2.5\end{align*} and \begin{align*}C_y = -3\end{align*}.
- \begin{align*}\overrightarrow{E} = \left \langle 12, 7, 9.5 \right \rangle\end{align*} and \begin{align*}\overrightarrow{F} = \left \langle 8, 8, 11\right \rangle\end{align*} Determine the vector \begin{align*}\overrightarrow{G}\end{align*} that makes the following equation true: \begin{align*}\overrightarrow{E} + \frac{1} {2} \overrightarrow{G} = \frac{2} {5} \overrightarrow{F}\end{align*}. Use standard algebraic techniques to solve for \begin{align*}\overrightarrow{G}\end{align*}: \begin{align*}\frac{1} {2} \overrightarrow{G} = \frac{2} {5} \overrightarrow{F} - \overrightarrow{E}\end{align*} \begin{align*}\overrightarrow{G} = \frac{4} {5} \overrightarrow{F} - 2\overrightarrow{E}\end{align*} Remember that multiplying a vector by a scalar means multiplying each of the vector’s components by that vector. Therefore, \begin{align*}\frac{4} {5} \overrightarrow{F} = \left \langle \left (\frac{4} {5}* 8\right ), \left (\frac{4} {5}* 8\right ), \left (\frac{4} {5}* 11\right )\right \rangle = \left \langle \left (\frac{32} {5}\right ), \left (\frac{32} {5}\right ), \left (\frac{44} {5}\right )\right \rangle = \left \langle 6.4, 6.4, 8.8 \right \rangle\end{align*} \begin{align*}2\overrightarrow{E} = \left \langle(2*12), (2*7), (2*9.5)\right \rangle = \left \langle 24, 14, 19\right \rangle\end{align*} Therefore \begin{align*}\overrightarrow{G} = \left \langle(6.4 - 24), (6.4 - 14), (8.8 - 19)\right \rangle = \left \langle -17.6, -7.6, -10.2 \right \rangle\end{align*}
- A vector can be written as \begin{align*}\overrightarrow{R} = 2.74m\end{align*} @ \begin{align*}60^\circ\end{align*}. Identify this same vector using component notation. When resolving a two-dimensional vector into components, remember that the vector itself is always the hypotenuse of a right triangle. If we define the x-axis as pointing from the origin along θ = 0^{o} and the y-axis as pointing along θ = 90^{o}, the x-component is given by R_{x} = R cos θ and R_{y} = R sin θ. In this case, R_{x} = R cos θ = (2.74m) cos 60^{o} = 1.37m R_{y} = R sin θ = (2.74m) sin 60^{o} = 2.37m
- Two vectors are identified as \begin{align*}\overrightarrow{L} = 4.5m\end{align*} @ \begin{align*}20^\circ\end{align*} and \begin{align*}\overrightarrow{N} = 6.3m\end{align*} @ \begin{align*}155^\circ\end{align*}. Determine the magnitude and orientation angle for the sum of these two vectors. First we need to find the components of these two vectors. Again, let us define the x-axis as pointing from the origin along θ = 0^{o} and the y-axis as pointing along θ = 90^{o}. L_{x} = L cos θ = (4.5m) cos 20^{o} = 4.2m L_{y} = L sin θ = (4.5m) sin 20^{o} = 1.5m N_{x} = N cos θ = (6.3m) cos 155^{o} = -5.7m N_{y} = N sin θ = (6.3m) sin 155^{o} = 2.7m Now we can add the components of the two vectors to give the components of their sum \begin{align*}\overrightarrow{J} = \overrightarrow{L} + \overrightarrow{N} = \left \langle (4.2 - 5.7), (1.5 + 2.7)\right \rangle = \left \langle -1.5, 4.2 \right \rangle\end{align*} To find the magnitude of the sum, we use the Pythagorean Theorem: \begin{align*}c = \sqrt{a^2 + b^2}\end{align*} \begin{align*}J = \sqrt{J_x^2 + J_y^2} = \sqrt{(-1.5)^2 + (4.2)^2} = \sqrt{2.25 + 17.64} = \sqrt{19.89} = 4.46\mbox{m}\end{align*} To find the direction of the vector use right triangle trigonometry, specifically \begin{align*}\mbox{tan}\ \theta = \frac{J_y} {J_x} = \frac{4.2} {-1.5} = -2.8\end{align*} \begin{align*}\theta = -109.7^\circ\end{align*} Note, your calculator may give you -70.3^{o} as the angle. Many calculators only give angles from the 1^{st} and 4^{th} quadrant, but we know from the components that vector \begin{align*}\overrightarrow{J}\end{align*} is in the 2^{nd} quadrant, but -70.3^{o} lies within the 4^{th} quadrant. Add 180^{o} to the -70.3^{o} to obtain the actual orientation angle for this vector.
- My neighbors recently adopted a retired racing greyhound named Flash. When Josef and Zara return home from work, they let Flash out into the back yard where she runs counter-clockwise around a circular path, similar to the one shown in the diagram. Joe recently used this circular motion to help his son Ahmed learn the difference between the vector quantity scalar quantity distance. An object’s displacement is defined as the vector beginning at the object’s initial position and ending at the object’s final position. The exact path does not matter to the displacement. In contrast, the distance traveled by an object does depend on the path taken in getting from its initial position to its final position. If Flash’s circular path has a radius of 3.7 m, identify the displacement vector, \begin{align*}\overrightarrow{\triangle r}\end{align*}, and the distance traveled, d, as Flash runs from point 1 to point 2. The vector beginning at point 1 and ending at point 2 is the hypotenuse of a right isosceles triangle. The length of this vector is given by the Pythagorean Theorem \begin{align*}\triangle r = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R \sqrt{2}\end{align*} The distance traveled is the portion of the circumference traveled by Flash going counter-clockwise around the circle from point 1 to point 2. \begin{align*}d = \frac{3} {4} (2 \pi R) = \frac{3} {2} \pi R\end{align*}
- Two displacement vectors have magnitudes of 25 cm and 64 cm. Is it possible to add these two vectors to obtain a vector having a magnitude of 77 cm? Explain why or why not using a diagram to illustrate your reasoning. If these two numbers were scalars, then it would be impossible to add them and obtain a sum of 7.5. They are, however, vector quantities even though the problem statement did not define their directions. The largest possible sum for these two magnitudes would occur when the two vectors point in the same direction. For example if both vectors were aligned with the positive x-axis, as shown in (A) below, their sum would be 8.5 @ 0^{o}. The smallest possible sum for these two magnitudes would occur when the two vectors point in opposite directions. For example, if one vector was parallel to the positive y-axis and the other was parallel with the negative y-axis their sum will either be +3.5 @ 90^{o} or -3.5 @ 90^{o}, as shown in (B) below. If the two vectors are perpendicular to one another, as shown in (C) below, the magnitude of their sum is given by the Pythagorean Theorem, \begin{align*}\sqrt{a^2 + b^2} = \sqrt{2.5^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5\end{align*}. To see this in a more concrete way, take a pencil and a crayon and hold them together end to end to represent the head-to-tail addition of two vectors. The sum of the two vectors will always be the straight line that starts at the open end of the pencil and ends at the open end of the crayon.
- Three vectors are identified by the following component equations: \begin{align*}\overrightarrow{P} = \left \langle 25, 17, 32\right \rangle, \ \overrightarrow{L} = \left \langle 14, 23, 57\right \rangle\end{align*}, and \begin{align*}\overrightarrow{Q} = \left \langle 49, 11, 27\right \rangle\end{align*}. Determine the components and magnitude of vector \begin{align*}\overrightarrow{R} = 3\overrightarrow{P} + 2\overrightarrow{L} - \overrightarrow{Q}\end{align*}. The components of vector \begin{align*}\overrightarrow{R}\end{align*} are given by R_{x} = (3*25) + (2*14) - 49 = 75 + 28 - 49 = 54 R_{y} = (3*17) + (2*23) - 11 = 51 + 46 - 11 = 86 R_{z} = (3*32) + (2*57) - 27 = 64 + 114 - 27 = 151 The magnitude of vector \begin{align*}\overrightarrow{R}\end{align*} is given by using the Pythagorean Theorem. \begin{align*}R = \sqrt{a^2 + b^2 + c^2} = \sqrt{54^2 + 86^2 + 151^2} = \sqrt{2916 + 7396 + 22801}\end{align*} \begin{align*}= \sqrt{33133} = 182\end{align*}
- Sasha’s three puppies frequently fight over their favorite toy, a bright green Frisbee. Smoky pulls on the Frisbee with a force \begin{align*}\overrightarrow{F_{SonF}} = (-4.5N)\hat{x}\end{align*} , Ginger pulls with a force \begin{align*}\overrightarrow{F_{GonF}} = (-5.2N)\hat{y}\end{align*}, and Bruno pulls with a force \begin{align*}\overrightarrow{F_{BonF}} = (+6.0N)\hat{x}\end{align*}. What is the total force acting on the Frisbee? Find both the magnitude and the direction. To find the total force on the Frisbee, add the three force vectors. \begin{align*}\overrightarrow{F_{total}} = \left \langle(6.0 - 4.5), (-5.2)\right \rangle = \left \langle 1.5, -5.2\right \rangle N\end{align*} \begin{align*}\overrightarrow{F_{total}} = \sqrt{1.5^2 + (-5.2)^2} = \sqrt{29.29} = 5.4 N\end{align*}
- Aunt Francis frequently drives around Blue Lake to visit her friend, Rhoda, who lives at the corner marked C in the diagram below. Determine the components of the unit vector which identifies the direction from Aunt Francis’s home to Rhoda’s house? To find the unit vector that identifies the direction from Aunt Francis’s home to Rhoda’s house, we first add the three vectors \begin{align*}\overrightarrow{FE}, \ \overrightarrow{ED},\end{align*} and \begin{align*}\overrightarrow{DC}\end{align*}. \begin{align*}\overrightarrow{FC} = \left \langle(0 + 225 + 0), (60 + 0 - 310)\right \rangle = \left \langle 225, -250\right \rangle\end{align*} Then we determine the magnitude of \begin{align*}\overrightarrow{FC}\end{align*}. \begin{align*}FC = \sqrt{225^2 + (-250)^2} = \sqrt{113125} = 336m\end{align*} The unit vector is defined as the ratio of the vector to its magnitude, in this case \begin{align*}\overrightarrow{FC} = \frac{\overrightarrow{FC}} {FC} = \left \langle \frac{225} {336}, \frac{250} {336}\right \rangle = \left \langle 0.670, 0.774\right \rangle\end{align*}
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