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# 5.2: Vectors in Space

Created by: CK-12

## Learning objectives

• Identify useful coordinate systems to describe a given problem.
• Identify points in 3-dimensional space using vector notation.
• Identify the displacement of an object from one point to another.
• Identify the midpoint between two points.

## Introduction

In the previous section, we confined our discussion to points and vectors located in a 2-dimensional space, but these tools are not limited to describing points in a plane. In fact, mathematicians and scientists use these same techniques to describe a variety of abstract spaces as well, such as imaginary numbers, the phase relationships between a set of sinusoidal functions, or the social relationships within a high-school class. In this section, we will see that vector notation is also useful to describe the position of points or objects in a world that has at least three dimensions. Vectors can be used for more dimensions as well (but four dimensional objects are hard to draw!)

## Coordinate Systems and Position Vectors in Two Dimensions

The Cartesian coordinate system used to describe a plane consists of an origin and four open axes that extend outwards from the origin towards infinity. These axes are divided into two, co-linear pairs: +x with –x and +y with –y. Usually, the x-axes are oriented horizontally or left-to-right across a page and the y-axes are oriented vertically or top-to-bottom on a printed page as shown below. The green square represents the x-y plane; remember, however, that the x-y plane is not confined to this square but rather extends outward toward infinity. On most North American maps, +y corresponds to North and +x corresponds to East. We used this coordinate system to describe Allen’s trip to his mother’s home across Blue Lake in one of the examples of the previous sections introducing vector addition.

The choice of coordinate axes is arbitrary. Nothing about the motion of an object is limited by our choice of how to describe it. Therefore, we can choose a coordinate system that is most convenient. If we go back to the example of Blue Lake, we can identify the location of the Lake Park Lodge (L) with respect to Allen’s home (P), to Aunt Frances’s home (F), or to the Blue Lake Post Office (P). The point we choose as the reference position becomes our origin of coordinates (O). The position vector which identifies the Lodge’s position within the reference frame is represented by an arrow starting at the origin and ending at the Lodge. This is like taking the rectangular coordinate system and putting the origin at a convenient location. The diagram below shows the position vector for the Lodge in three reference frames, with origins at F, A, and P respectively.

Example: The motion of an object along an inclined plane is a very common problem in introductory physics. The diagram below shows one such situation.

Stickman Beauford has taken his niece Brynna to the park and waves to her as she plays on the slide. Choose two coordinate systems that could be used to describe Brynna’s motion and identify the position vectors for points A and B in both coordinate systems.

Solution: If we want to describe Brynna’s motion as she moves from point A to point B along the slide, we could use a standard horizontal and vertical coordinate system with the origin at the base of the slide’s ladder, but then the vector describing her motion would have components in both the x and y directions. Our mathematical description of her motion can be greatly simplified if we choose point A to be the origin and if we rotate the coordinate system such that the x-axis is parallel to the slide and the y-axis is perpendicular to the slide. Now Brynna’s motion from point A to point B is only along the x-axis. Note, other choices of origin are possible.

Once we have identified an origin and coordinate axes for of our reference frames, we can use vector notation to identify the location of points A and B. The position vector for point A is the vector starting at the origin and ending at point A, $\overrightarrow{OA}$. For the standard coordinate system shown on the left above, the position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are shown on the left below. For the rotated coordinate system, shown on the right above, the position vector $\overrightarrow{OA} = 0$ and $\overrightarrow{OB} = \overrightarrow{AB}$.

Example: Identify the position vectors for the three points shown on the grid below.

Solution: The position vectors begin at the origin, (0, 0) and end at each point:

$\overrightarrow{OA} = \left \langle -3, 1 \right \rangle, \ \overrightarrow{OB} = \left \langle 1, 2 \right \rangle$ and $\overrightarrow{OC} = \left \langle 2.5, 0 \right \rangle$

## Coordinate Systems and Position Vectors in Three Dimensions

The rectangular (or Cartesian) coordinate system is used to describe a plane divided into four quadrants, as shown below left. (Note, the colored squares are used to help you visualize the space, remember that the coordinate planes actually extend outward toward infinity.)

The Cartesian coordinate system use to describe three-dimensional space consists of an origin and six open axes, +z and –z are perpendicular to the x-y plane. These axes define three planes which divide the space into eight quadrants as shown above right. Think of these planes as cutting space three ways: left to right, top to bottom, and front to back.

By convention, we number the four quadrants of the x-y plane in this way: points in quadrant 1 have +x and +y coordinates, those in quadrant 2 have –x and +y, those in quadrant 3 have –x and –y, and those in quadrant 4 have +x and –y. There is currently no standardized numbering system for the octants in three-dimensional space, although most people identify the region with +x, +y, and +z as the first octant. The method used to identify the octants is to indicate verbally the portion of space they occupy. For example, the first octant could also be identified as (top, front, right).

Position vectors in 3D space are still represented by arrows that begin at the origin and end at the point in question. The diagram above shows a point, P, located in the front, lower, right octant. The three components of the position vector (Px, Py, and Pz) are shown in the diagram. According to the Pythagorean Theorem, the magnitude of the position vector is given by:

$|\overrightarrow{P}| = \sqrt{P_x^2 + P_y^2 + P_z^2}$

Example: Darnell was driving home from a football game in a nearby town when he swerved to avoid a deer which had run onto the road. Fortunately for Darnell, he was able to avoid hitting the deer. Unfortunately, his car ended up in the ditch beside the road. When he was unable to remove the car from the ditch by himself, he walked across a nearby field to the Tucker family farm to ask for help. The topographical map below shows Darnell’s trip across the field. He traveled 300 yards south and 750 yards west from where he left his car. The map shows that he also walked uphill from an altitude of 800 feet to an altitude of 850 feet above sea level. If we treat the location of Darnell’s car as the origin of coordinates, what the position vector of the Tucker farm?

Solution: Define a coordinate system where x = E, y = N, and z = up. Since Darnell walked south and west from the car, the x and y coordinates of the farm are both negative. If we measure all of the distances in feet (1 yard = 3 feet), the farm’s position vector can be written as $\overrightarrow{P} = \left \langle P_{east}, P_{north}, P_{up} \right \rangle = \left \langle -2250,-900, 50 \right \rangle$.

Note in this example that Darnell’s walking distance was given in yards, while the elevation change was given in feet. You need to watch out for these small changes when you are solving real-life problems.

## Vectors Between Two Points

The displacement vector represents the motion beginning at one point and ending at another. In the diagram below, vector C is the vector between points A and B. It starts at point A and ends at point B. As we saw ((in section 5.1.2)), this means that $\overrightarrow{A} + \overrightarrow{C} = \overrightarrow{B}$ and $\overrightarrow{C} = \overrightarrow{B} - \overrightarrow{A}$. In this case, $\overrightarrow{A} = \left \langle 1, 3 \right \rangle$ and $\overrightarrow{B} = \left \langle 3, 2 \right \rangle$, therefore $\overrightarrow{C} = \left \langle (3 - 1), (2 - 3) \right \rangle = \left \langle 2, - 1 \right \rangle$ matching what we can see in the diagram.

Example: Determine the coordinates and magnitude of the vector, D, beginning at the point $\overrightarrow{P_1} = \left \langle 12, 7 \right \rangle$ and ending at $\overrightarrow{P_2} = \left \langle 8, 10 \right \rangle$.

Solution: The displacement vector D is the difference between the two position vectors:.

$D = \left \langle P_{2x} - P_{1x}, P_{2y} - P_{1y} \right \rangle = \left \langle 8 - 12, 10 - 7 \right \rangle = \left \langle -4, 3 \right \rangle$.

The magnitude of the vector, D, can be found using the Pythagorean Theorem:

$|\overrightarrow{D}| = \sqrt{(-4)^2 + (3)^2} = \sqrt{25} = 5$.

## Vector to a Point Between Two Points

Computer graphics artists frequently need to know the location of a point which lies midway between two other points. Once we know the position vectors for two discrete points, we can determine the midpoint between them using their coordinates. Specifically, the midpoint between points A and B is the “average” of the two positions, therefore the coordinates of the midpoint are given by $x_{mp} = \frac{1}{2} (x_A + x_B), \ y_{mp} = \frac{1}{2} (y_A + y_B)$, and $z_{mp} = \frac{1}{2} (z_A + z_B)$ and the position vector for the midpoint can be written as $P_{mp} = \left \langle x_{mp}, y_{mp}, z_{mp} \right \rangle$. The vector from any other point to this midpoint can then be calculated using the method described in our discussion of displacement vectors.

Example: Determine the position vector identifying the midpoint between points $\overrightarrow{P_1} = \left \langle 12, 7 \right \rangle$ and $\overrightarrow{P_2} = \left \langle 8, 10 \right \rangle$.

Solution: Since these two points are located in the x-y plane, the z-coordinate of both P1 and P2 is 0. The x-coordinate of the midpoint is given by

$x_{mp} = \frac{1}{2} (x_A + x_B) = \frac{1}{2}(12 + 8) = 10$

and the y-coordinate of the midpoint is given by

$y_{mp} = \frac{1}{2} (y_A + y_B) = \frac{1}{2}(7 + 10) = 8.5$.

Therefore the position vector for this midpoint can be written as

$P_{mp} = \left \langle 10,8.5,0 \right \rangle$, or if the problem context allows, you can omit the z-coordinates.

$P_{mp} = \left \langle 10,8.5 \right \rangle$

## Lesson Summary

Vectors are very useful in identifying the locations of objects or points of interest. Once we define an origin of coordinates for a problem, we can identify the position of an object using the vector from that origin to the location of the object. We can also use the average of two position vectors to identify the midpoint between two points.

## Practice Problems

1. The diagram shows two positions of a bicycle as it moves along a long straight road. Two possible coordinate systems for the motion are shown below. Determine the position vectors in each of the two coordinate systems for the bicycle at points A and B. Then determine the displacement vector from A to B in each case. (Not drawn to scale.)
2. Identify the position vectors for the three points shown in the diagram below.
3. Identify the midpoints between points A and C and between B and C.
4. Determine the length of the displacement vectors from points A and C to the midpoint determined in the previous practice problem.
5. Zeke is enjoying an afternoon at the local skate-park. The diagram below shows his starting position and his ending position at the highest point on the new hill. Following the model of practice problem #1, choose two different coordinate systems which could describe this system. Find Zeke’s initial and final position vectors in each of the two coordinate systems. Then identify the displacement vector from his starting position to his final position at the top of the hill.
6. An architecture student designs a spiral staircase, a model of which is shown below. The staircase winds its way around a cylinder of radius 3.5 m and height 11 m. The staircase makes 1 ⅞ turns progressing counter-clockwise from its beginning at point A to its end point at B. Using an origin of coordinates at the bottom center of the staircase, determine the position vectors of points A and B. Then find the displacement vector between the two points.
7. Identify the midpoint between points P = (3.7, 8.4, -2.1) and Q = (5.5, -1.9, -8.6).
8. Wilhelm and Armond head out into the marsh near their home to hunt goose for Michealmass dinner, but they disagree on the best place to set their blinds. They park their truck along the Bluffton road at an altitude of 840 m above sea level. Wilhelm heads along the riverbed, ending up 350 m west and 87 m north of the truck at an altitude of 780 m above sea level. Armond, on the other hand, heads toward another marshy area 738 m west and 92 m south of the truck at an altitude of 800 m. If the truck serves as the origin of coordinates, determine the position vector for each hunter and determine the displacement vector from Armond’s position to Wilhelm’s.

## Solutions

1. The diagram shows two positions of a bicycle as it moves along a long straight road. Two possible coordinate systems for the motion are shown below. Determine the position vectors in each of the two coordinate systems for the bicycle at points A and B. Then determine the displacement vector from A to B in each case. For the upper coordinate system, the position vector of the bicycle at point A is given by $\overrightarrow{r_A} = \left \langle -300m, 0, 0 \right \rangle$ and that at point B is given by $\overrightarrow{r_B} = \left \langle 100m, 0, 0 \right \rangle$. This gives a displacement of $\overrightarrow{\triangle r_{A-B}} = \left \langle (100m - (-300m)), (0 - 0), (0 - 0)\right \rangle = \left \langle 400m, 0, 0 \right \rangle$. For the upper coordinate system, the position vector of the bicycle at point A is given by $\overrightarrow{r_A} = \left \langle 100m, 0, 0 \right \rangle$ and that at point B is given by $\overrightarrow{r_B} = \left \langle 500m, 0, 0 \right \rangle$. This gives a displacement of $\overrightarrow{\triangle r_{A-B}} = \left \langle (500m - 100m), (0 - 0), (0 - 0)\right \rangle = \left \langle 400m, 0, 0 \right \rangle$. The position vectors for the bicycle at point A are shown in red and the position vectors for point B are shown in blue. The displacement vector between points A and B is shown in gold. As you can see, the position vectors representing this motion depend on the choice of coordinate system, but the displacement vector is independent of the coordinate system. No matter how we define the origin, the bike moves 400 m in the +x direction and does not move in the y or z direction.
2. Identify the position vectors for the three points shown in the diagram below. $\overrightarrow{r_A} = \left \langle -2.63, 2.63, 0 \right \rangle, \ \overrightarrow{r_B} = \left \langle 3, 1.75, 0 \right \rangle, \ \overrightarrow{r_C} = \left \langle 0.25,1,0 \right \rangle$
3. Identify the midpoints between points A and C and between B and C. To find the midpoint between two points, determine the average of the two positions. $\overrightarrow{r_{AC,mid}} = \left \langle \frac{1}{2}(-2.63 + 0.25), \frac{1}{2}(2.63 + 1), \frac{1}{2}(0 + 0) \right \rangle = \left \langle \frac{1}{2}(-2.38), \frac{1}{2}(3.63), \frac{1}{2}(0) \right \rangle$ $= \left \langle -1.19, 1.815, 0 \right \rangle$ $\overrightarrow{r_{BC,mid}} = \left \langle \frac{1}{2}(3 + 0.25), \frac{1}{2}(1.75 + 1), \frac{1}{2}(0 + 0) \right \rangle = \left \langle \frac{1}{2}(3.25), \frac{1}{2}(2.75), \frac{1}{2}(0) \right \rangle$ $= \left \langle 1.625, 1.375, 0 \right \rangle$
4. Determine the length of the displacement vectors from points A and C to the midpoint determined in the previous practice problem. The distance between A and the midpoint is the magnitude of the displacement vector between these two points: $\overrightarrow{r_{AtoMid}} = \left \langle (-1.19 - (-2.63)), (1.815 - 2.63), (0 - 0) \right \rangle = \left \langle 1.44, -0.815, 0 \right \rangle$ We can use the Pythagorean Theorem to calculate the length of this vector: $| \overrightarrow{r_{AtoMid}}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{(1.44)^2 + (-0.815)^2 + 0^2} = \sqrt{2.0736 + 0.664225}$ $= 1.655$ The distance between C and the midpoint is the magnitude of the displacement vector between these two points: $\overrightarrow{r_{CtoMid}} = \left \langle (-1.19 - 0.25), (1.815 - 1), (0 - 0) \right \rangle = \left \langle -1.44, 0.815, 0 \right \rangle$ We can use the Pythagorean Theorem to calculate the length of this vector: $| \overrightarrow{r_{CtoMid}}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{(-1.44)^2 + (0.815)^2 + 0^2} = \sqrt{2.0736 + 0.664225}$ $= 1.655$ As you can see, the midpoint is the same distance from each of the endpoints.
5. Zeke is enjoying an afternoon at the local skate-park. The diagram below shows his starting position and his ending position at the highest point on the new hill. Following the model of practice problem #1, choose two different coordinate systems which could describe this system. Find Zeke’s initial and final position vectors in each of the two coordinate systems. Then identify the displacement vector from his starting position to his final position at the top of the hill. One possible origin of coordinates is located at Zeke’s starting position. In this case, the initial position vector is given by $\overrightarrow{r_i} = \left \langle 0m, 0m, 0m \right \rangle$ and his final position is given by $\overrightarrow{r_f} = \left \langle 6.1m, 2.3m, 0m \right \rangle$. Zeke’s displacement is the difference between these two vectors, $\overrightarrow{\triangle r} = \overrightarrow{r_f} - \overrightarrow{r_i} = \left \langle 6.1m, 2.3m, 0m \right \rangle - \left \langle 0m, 0m, 0m \right \rangle = \left \langle 6.1m, 2.3m, 0m \right \rangle$ Another possible origin of coordinates is at the point marked O in the diagram below. In this case his original position is given by $\overrightarrow{r_i} = \left \langle -6.1m, 0m, 0m \right \rangle$ and his final position is given by $\overrightarrow{r_f} = \left \langle 0m, 2.3m, 0m \right \rangle$. Zeke’s displacement is the difference between these two vectors, $\overrightarrow{\triangle r} = \overrightarrow{r_f} - \overrightarrow{r_i} = \left \langle 0m, 2.3m, 0m \right \rangle - \left \langle -6.1m, 0m, 0m \right \rangle = \left \langle 6.1m, 2.3m, 0m \right \rangle$ As we saw in Practice Problem #1, the position vectors representing this motion depend on the choice of coordinate system, but the displacement vector is independent of the coordinate system.
6. An architecture student designs a spiral staircase, a model of which is shown below. The staircase winds its way around a cylinder of radius 3.5 m and height 11 m. The staircase makes 1 ⅞ turns progressing counter-clockwise from its beginning at point A to its end point at B. Using an origin of coordinates at the bottom center of the staircase, determine the position vectors of points A and B. Then find the displacement vector between the two points. As you can see in the top view of the diagram, point A is directly to the left of the origin, therefore the position vector for point A is given by $\overrightarrow{r_A} = \left \langle -R, 0, 0 \right \rangle = \left \langle -3.5m, 0, 0 \right \rangle$. Point B is located 11m above point A and ⅞ of one turn counter-clockwise is equal to ⅛ of one turn clockwise, so θ = 45o. We can also use the geometry of the system to determine the x and z coordinates of point B: $\overrightarrow{r_B} = \left \langle -R \ \mbox{cos} \ \theta, \ R \ \mbox{sin} \ \theta, \ H \right \rangle = \left \langle (-3.5m)\ \mbox{cos} \ 45^\circ, (3.5m)\ \mbox{sin} \ 45^\circ, 11m \right \rangle$ $= \left \langle -2.475m, 2.475m, 11m \right \rangle$ The displacement vector between these two points is the vector which obeys the equation: $\overrightarrow{\triangle r} = \overrightarrow{r_B} - \overrightarrow{r_A}$ $\overrightarrow{\triangle r} = \left \langle -2.475m, 2.475m, 11m \right \rangle - \left \langle -3.5m, 0m, 0m \right \rangle$ $\overrightarrow{\triangle r} = \left \langle (-2.475m -(-3.5m)), (2.475m - 0m), (11m - 0m) \right \rangle = \left \langle 1.025m, 2.475m, 11m \right \rangle$
7. Identify the midpoint between points P = (3.7, 8.4, -2.1) and Q = (5.5, -1.9, -8.6). To find the midpoint between two points, determine the average of the two positions. $M = \left (\frac{1}{2}(3.7 + 5.5), \frac{1}{2}(8.4 + (-1.9)), \frac{1}{2}(-2.1 + (-8.6)) \right ) = \left ( \frac{1}{2}(9.2), \frac{1}{2} (6.5)), \frac{1}{2} (-10.7))\right )$ $M = (4.6, 3.25, -5.35)$
8. Wilhelm and Armond head out into the marsh near their home to hunt goose for Michealmass dinner, but they disagree on the best place to set their blinds. They park their truck along the Bluffton road at an altitude of 840 m above sea level. Wilhelm heads along the riverbed, ending up 350 m west and 87 m north of the truck at an altitude of 780 m above sea level. Armond, on the other hand, heads toward another marshy area 738 m west and 92 m south of the truck at an altitude of 800 m. If the truck serves as the origin of coordinates, determine the position vector for each hunter and determine the displacement vector from Armond’s position to Wilhelm’s. Define the truck as the origin of coordinates, north as the Define the truck as the origin of coordinates, north as the +y direction and east as the +x direction. Upward is the +z direction. Wilhelm’s location along the river is therefore given by $W = \left \langle -350m, 87m, (780m - 840m) \right \rangle = \left \langle -350m, 87m, -60m \right \rangle$ Armond’s position at the other marsh is given by $A = \left \langle -738m, -92m, (800m - 840m) \right \rangle = \left \langle -738m, -92m, -40m \right \rangle$ The displacement vector from Armond’s position to Wilhelm’s position is given by subtracting Armond’s position vector from Wilhelm’s position vector: $\overrightarrow{\triangle r} = \overrightarrow{W} - \overrightarrow{A} = \left \langle (-350m - (-738m)), (87m - (-92m), (-60m - (-40m))) \right \rangle$ $\overrightarrow{\triangle r} = \left \langle 388m, 179m, -20m \right \rangle$

Feb 23, 2012

Dec 29, 2014