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# 5.1: Two-Dimensional Vectors

Brian was telling Tanya about his trip across the river earlier that day. If he wanted to be accurate and describe how the current made it very difficult for him to actually drive straight across, how could he clearly show the different forces of the motor on his boat pushing one way and the current pushing another? Specifically, how could he show that the boat was crossing the stream at 35km/hr, and the current was pushing him downstream at 25km/hr?

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### Guidance

Throughout the study of vectors 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 $\overrightarrow {v}$ to indicate the velocity of an object.

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 |, $|\overrightarrow {v}|$, or v.

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, ABx, and the y-direction, ABy. Here, ABx = 4 and ABy = 3 because point B is 3 units to the right and 3 units up from point A. Similarly, vector D has components Dx = 3 and Dy = –1.25. The negative sign in Dy 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 $AB = \left \langle 4,3\right \rangle$ and $D = \left \langle 3, -1.25\right \rangle.$

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 @ 30o.

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, Ax = | A | cos 30 and Ay = | A | sin 30.

#### Example A

Express the statements in your own words:

a) $AB_x = CD_x : AB_y = -CD_y$
b) Vector $AB = \left \langle 4,6\right \rangle$

Solutions

a) Vector AB component x is equal to vector CD component x, AB component y is the same magnitude as CD component y, but in the opposite direction.
b) Vector AB has x component of 4 units, and y component of 6 units.

#### Example B

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 C

(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.

Concept question wrap-up If a boat is being motored perpendicularly at 35km/hr across a stream which is flowing at 25km/hr, how can the direction and speed that it travels be clearly shown using vectors?

There are a number of possible notations, one example is:

In vector notation, the x component is 30 units, and the y component is 25 units.

$\therefore AB = \left \langle 30,25\right \rangle$

Brian could also draw a diagram and represent the speed of the boat with an arrow 35units long, and the current with an arrow 25units long.

### Vocabulary

The magnitude of a value is a description of how "big" it is, in other words, the absolute value of a number.

A scalar is a value representing only a magnitude, or size. Scalars can expressed with a single number.

A vector is a value representing a magnitude and a direction. Vectors cannot be expressed with only a single value.

### Guided Practice

Questions

1) Two vectors, $\overrightarrow{A}$ and $\overrightarrow{B}$, are shown in the diagram. Identify the $\hat{x}$ and $\hat{y}$ components of both vectors.

2) For the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ shown in the diagram, determine the components, magnitude, and orientation angle for vector $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$.

3) $\overrightarrow{A} = \left \langle 12, 7, 9.5 \right \rangle$ and $\overrightarrow{B} = \left \langle 8, 8, 11 \right \rangle$ Determine the vector $\overrightarrow{C}$ that makes the following equation true: $\overrightarrow{A} + \frac{1} {2}\overrightarrow{C} = \frac{2} {3}\overrightarrow{B}$.

4)A vector can be written as $\overrightarrow{R} = 2.74 m$ @ $60^\circ$. Identify this same vector using component notation.

Solutions

1) Two vectors, $\overrightarrow{A}$ and $\overrightarrow{B}$, are shown in the diagram. Identify the $\hat{x}$ and $\hat{y}$ components of both vectors.

The $\hat{x}$ and $\hat{y}$ components of a vector are the extensions of the vector along the $\hat{x}$ and $\hat{y}$ 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 Ax = 2.75 - (-2) = 4.75 and its y-component is given by Ay = 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 Bx = 1.75 - 4 = -2.25 and its y-component is given by By = -1.5 - 2 = -3.5

2) For the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ shown in the diagram, determine the components, magnitude, and orientation angle for vector $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B}$.

As we saw in the previous problem, $\overrightarrow{A} = \left \langle 4.75, 0.5 \right \rangle$ and $\overrightarrow{B} = \left \langle -2.25, -3.5 \right \rangle$. To add the two vectors, we add the x-components together and we add the y-components together to give $\overrightarrow{C} = \left \langle (4.75 + (-2.25)),(-3.5 + 0.5)\right \rangle = \left \langle 2.5, -3 \right \rangle$
We can also add these two vectors graphically by positioning $\overrightarrow{A}$ and $\overrightarrow{B}$ head to tail. Vector $\overrightarrow{A}$ is the single vector that begins where $\overrightarrow{A}$ begins and ends where $\overrightarrow{B}$ ends. As you can see from the diagram, the components of vector $\overrightarrow{C}$ are $C_x = 2.5$ and $C_y = -3$.
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3) $\overrightarrow{A} = \left \langle 12, 7, 9.5 \right \rangle$ and $\overrightarrow{B} = \left \langle 8, 8, 11\right \rangle$ Determine the vector $\overrightarrow{C}$ that makes the following equation true: $\overrightarrow{A} + \frac{1} {2} \overrightarrow{C} = \frac{2} {5} \overrightarrow{B}$.

Use standard algebraic techniques to solve for $\overrightarrow{C}$:
$\frac{1} {2} \overrightarrow{C} = \frac{2} {5} \overrightarrow{B} - \overrightarrow{A}$
$\overrightarrow{C} = \frac{4} {5} \overrightarrow{B} - 2\overrightarrow{A}$
Remember that multiplying a vector by a scalar means multiplying each of the vector’s components by that vector. Therefore,
$\frac{4} {5} \overrightarrow{B} = \left \langle \left (\frac{4} {5}* 8\right ), \left (\frac{4} {5}* 8\right ), \left (\frac{4} {5}* 11\right )\right \rangle = \left \langle \left (\frac{32} {5}\right ), \left (\frac{32} {5}\right ), \left (\frac{44} {5}\right )\right \rangle = \left \langle 6.4, 6.4, 8.8 \right \rangle$
$2\overrightarrow{A} = \left \langle(2*12), (2*7), (2*9.5)\right \rangle = \left \langle 24, 14, 19\right \rangle$
Therefore
$\overrightarrow{C} = \left \langle(6.4 - 24), (6.4 - 14), (8.8 - 19)\right \rangle = \left \langle -17.6, -7.6, -10.2 \right \rangle$

4) A vector can be written as $\overrightarrow{R} = 2.74m$ @ $60^\circ$.

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 = 0o and the y-axis as pointing along θ = 90o, the x-component is given by Rx = R cos θ and Ry = R sin θ. In this case,
Rx = R cos θ = (2.74m) cos 60o = 1.37m
Ry = R sin θ = (2.74m) sin 60o = 2.37m

### Practice

1. Can the x or y component of a vector ever have a greater magnitude than the vector itself?
2. If two vectors have magnitudes that are not equal, can the sums of their magnitudes ever be zero?

Use the counter-clockwise (from East) convention to determine the direction. Use the indicated scale and a scale conversion to determine the magnitude.

1. Given the scale: $1cm = 10m/s$, determine the magnitude and direction of the vector in the diagram.
2. Given the scale: $1cm = 50 km/hr$, determine the magnitude and direction of this vector.

Using the Cartesian coordinate system, draw the vectors.

1. $A = 1x + 2y$
2. $B = -2x + 2y$
3. $C = -2x - 3y$
4. Identify and plot the three angles, (a,b,c) to describe the direction of each vector.

Calculate the magnitude and angles of each vector.

1. $A = 1x + 2y$ a)Magnitude: b)Angle:
2. $B = -2x + 2y$ a)Magnitude: b)Angle:
3. $C = -2x - 3y$ a)Magnitude: b)Angle:

Describe and sketch the set of all points (x; y) in $IR^2$ that satisfy

1. $x = y$
2. $x + y = 1$
3. $x^2 + y^2 = 4$
4. $x^2+ y^2 = 2y$

Identify the vectors using component notation

1. A vector can be written as $\overrightarrow{R} = 3.45m$ @ $170^\circ$.
2. A vector can be written as $\overrightarrow{R} = 11m$ @ $90^\circ$.
3. A vector can be written as $\overrightarrow{R} = 7.23m$ @ $45^\circ$.
4. A vector can be written as $\overrightarrow{R} = 2.7m$ @ $235^\circ$.
5. A vector can be written as $\overrightarrow{R} = 23.75m$ @ $255^\circ$.

Nov 01, 2012

Dec 12, 2013

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