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# 5.3: Three-Dimensional Positions

Difficulty Level: At Grade Created by: CK-12
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Practice Three-Dimensional Positions

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Brian just entered the museum. Looking at the "You are here" map, he sees that his favorite exhibit, "Dinosaurs in Motion!" has been moved to the third floor, and is in the fourth room on the right.

Brian knows the museum well, so he knows the entrance is the first room on the left on the first floor, and is already familiar with the museum layout of 8 rooms on each side of the main hall on each floor, with the stairs at the front of the building.

How could the location of the dinosaur exhibit be described in 3-dimensional coordinates? What is the displacement of the dinosaur room from his position at the front entrance?

Embedded Video:

### Guidance

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 parts knowns as octants 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:

|P|=P2x+P2y+P2z\begin{align*}|\overrightarrow{P}| = \sqrt{P_x^2 + P_y^2 + P_z^2}\end{align*}

#### Example A

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. Darnell later looked at a topographical map and identified his trip across the field. He traveled 300 yards south and 750 yards west from where he left his car. The map showed 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 P=Peast,Pnorth,Pup=2250,900,50\begin{align*}\overrightarrow{P} = \left \langle P_{east}, P_{north}, P_{up} \right \rangle = \left \langle -2250,-900, 50 \right \rangle\end{align*}.

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.

#### Example B

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.

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.

Solution

One possible origin of coordinates is located at Zeke’s starting position. In this case, the initial position vector is given by ri=0m,0m,0m\begin{align*}\overrightarrow{r_i} = \left \langle 0m, 0m, 0m \right \rangle\end{align*} and his final position is given by rf=6.1m,2.3m,0m\begin{align*}\overrightarrow{r_f} = \left \langle 6.1m, 2.3m, 0m \right \rangle\end{align*}. Zeke’s displacement is the difference between these two vectors, r=rfri=6.1m,2.3m,0m0m,0m,0m=6.1m,2.3m,0m\begin{align*}\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\end{align*}

Another possible origin of coordinates is at the point marked O in the diagram below. In this case his original position is given by ri=6.1m,0m,0m\begin{align*}\overrightarrow{r_i} = \left \langle -6.1m, 0m, 0m \right \rangle\end{align*} and his final position is given by rf=0m,2.3m,0m\begin{align*}\overrightarrow{r_f} = \left \langle 0m, 2.3m, 0m \right \rangle\end{align*}. Zeke’s displacement is the difference between these two vectors, r=rfri=0m,2.3m,0m6.1m,0m,0m=6.1m,2.3m,0m\begin{align*}\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\end{align*}

Note that the position vectors representing this motion depend on the choice of coordinate system, but the displacement vector is independent of the coordinate system.

#### Example C

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.

Solution

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 rA=R,0,0=3.5m,0,0\begin{align*}\overrightarrow{r_A} = \left \langle -R, 0, 0 \right \rangle = \left \langle -3.5m, 0, 0 \right \rangle\end{align*}.

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: rB=R cos θ, R sin θ, H=(3.5m) cos 45,(3.5m) sin 45,11m\begin{align*}\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\end{align*} =2.475m,2.475m,11m\begin{align*} = \left \langle -2.475m, 2.475m, 11m \right \rangle\end{align*}

The displacement vector between these two points is the vector which obeys the equation: r=rBrA\begin{align*}\overrightarrow{\triangle r} = \overrightarrow{r_B} - \overrightarrow{r_A}\end{align*}

r=2.475m,2.475m,11m3.5m,0m,0m\begin{align*}\overrightarrow{\triangle r} = \left \langle -2.475m, 2.475m, 11m \right \rangle - \left \langle -3.5m, 0m, 0m \right \rangle\end{align*}

r=(2.475m(3.5m)),(2.475m0m),(11m0m)=1.025m,2.475m,11m\begin{align*}\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\end{align*}

Concept question wrap-up "Brian knows the museum well, so he knows the entrance is the first room on the left on the first floor, and is already familiar with the museum layout of 8 rooms on each side of the main hall on each floor, with the stairs at the front of the building.

How could the location of the dinosaur exhibit be described in 3-dimensional coordinates? What is the displacement of the dinosaur room from his position at the front entrance?"

A coordinate system can be assumed to use any units you wish, in this case the location information is given in units of "rooms" and "floors". Let the x axis represent the horizontal left and right-hand sides of the main halls, y can be the vertical axis, and z can be the front-back axis of rooms down each hall.

Let Brian's starting position at the entrance represent (0,0,0)\begin{align*}(0, 0, 0)\end{align*} in three dimensional rectangular coordinates. "Dinosaurs in Motion!" is on the third floor, and is in the fourth room on the right.

a) (2,3,4)\begin{align*}(2, 3, 4)\end{align*} Would represent the dinosaur exhibit room
b) Since we were clever, and oriented our coordinate system on the entrance, the vector displacement is the same as the room's location: 2rooms,3floors,4rooms\begin{align*}\left \langle 2rooms, 3floors, 4rooms \right \rangle\end{align*}

### Vocabulary

A quadrant is any one of the four "corners" of the standard 2-dimensional rectangular Cartesian Coordinate system.

An octant is any one of the eight "corners" of the 3-dimensional rectangular coordinate system.

### Guided Practice

Questions

1) Identify the midpoint between points P = (3.7, 8.4, -2.1) and Q = (5.5, -1.9, -8.6).

2) Identify the midpoint between points A = (4.2, -13.6, 4.2) and B = (1.2, 3.9, -2.7).

3) Draw the three dimensional vector A=3i^+2j^+5k^\begin{align*}\overrightarrow{A} = 3\hat{i} + 2\hat{j} +5\hat{k}\end{align*}

4) Draw the three dimensional vector A=3i^+4j^5k^\begin{align*}\overrightarrow{A} = -3\hat{i} + -4\hat{j} -5\hat{k}\end{align*}

Solutions

1) To find the midpoint between two points, determine the average of the two positions.

M=(12(3.7+5.5),12(8.4+(1.9)),12(2.1+(8.6)))=(12(9.2),12(6.5)),12(10.7)))\begin{align*}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 )\end{align*}
M=(4.6,3.25,5.35)\begin{align*}M = (4.6, 3.25, -5.35)\end{align*}

2) To find the midpoint between points A = (4.2, -13.6, 4.2) and B = (1.2, 3.9, -2.7), determine the average of the two positions.

M=(12(4.2+1.2),12(13.6+3.9),12(4.2+(2.7)))=(12(5.4),12(9.7)),12(1.5)))\begin{align*}M = \left (\frac{1}{2}(4.2 + 1.2), \frac{1}{2}(-13.6 + 3.9), \frac{1}{2}(4.2 + (-2.7)) \right ) = \left ( \frac{1}{2}(5.4), \frac{1}{2} (-9.7)), \frac{1}{2} (1.5))\right )\end{align*}
M=(2.7,4.85,.75)\begin{align*}M = (2.7, -4.85, .75)\end{align*}

3) I, j and k are unit vectors respectively in the positive direction of the x, y, and z axes.

4) I, j and k are unit vectors respectively in the negative direction of the x, y, and z axes.

### Practice

Given: initial and final position vectors the coordinate system. Identify the displacement vector or midpoint from the initial to the final position.

1. What is the displacement or difference between these two vectors? The initial position vector ri=0m,0m,0m\begin{align*}\overrightarrow{r_i} = \left \langle 0m, 0m, 0m \right \rangle\end{align*} and the final position vector rf=16.1m,7.5m,3m\begin{align*}\overrightarrow{r_f} = \left \langle 16.1m, 7.5m, 3m \right \rangle\end{align*}.
2. What is the displacement or difference between these two vectors? The initial position vector ri=0mi,0mi,0mi\begin{align*}\overrightarrow{r_i} = \left \langle 0mi, 0mi, 0mi \right \rangle\end{align*} and the final position vector rf=9.4mi,12.5mi,6.6mi\begin{align*}\overrightarrow{r_f} = \left \langle 9.4mi, 12.5mi, 6.6mi \right \rangle\end{align*}.
3. What is the displacement or difference between these two vectors? The initial position vector ri=5km,3km,8km\begin{align*}\overrightarrow{r_i} = \left \langle 5km, 3km, 8km \right \rangle\end{align*} and the final position vector rf=10km,20km,19km\begin{align*}\overrightarrow{r_f} = \left \langle 10km, 20km, 19km \right \rangle\end{align*}.
4. What is the displacement or difference between these two vectors? The initial position vector ri=1cm,3cm,1cm\begin{align*}\overrightarrow{r_i} = \left \langle 1cm, 3cm, 1cm \right \rangle\end{align*} and the final position vector rf=5.1cm,2cm,5cm\begin{align*}\overrightarrow{r_f} = \left \langle 5.1cm, 2cm, 5cm \right \rangle\end{align*}.
5. What is the displacement or difference between these two vectors? The initial position vector ri=5.6mm,10.2mm,2.2mm\begin{align*}\overrightarrow{r_i} = \left \langle 5.6mm, 10.2mm, 2.2mm \right \rangle\end{align*} and the final position vector rf=20.4mm,31.1mm,1.1mm\begin{align*}\overrightarrow{r_f} = \left \langle 20.4mm, 31.1mm, 1.1mm \right \rangle\end{align*}.
6. What is the displacement or difference between these two vectors? The initial position vector ri=1in,2in,3in\begin{align*}\overrightarrow{r_i} = \left \langle 1in, 2in, 3in \right \rangle\end{align*} and the final position vector rf=4in,5in,3in\begin{align*}\overrightarrow{r_f} = \left \langle 4in, 5in, 3in \right \rangle\end{align*}.

Identify the midpoint between points A and B

1. A = 0m,0m,0m\begin{align*}\left \langle 0m, 0m, 0m\right \rangle\end{align*} and B =16.1m,7.5m,3m\begin{align*} \left \langle16.1m, 7.5m, 3m\right \rangle\end{align*}
2. A = 0mi,0mi,0mi\begin{align*}\left \langle 0mi, 0mi, 0mi\right \rangle\end{align*} and B = 9.4mi,12.5mi,6.6mi\begin{align*}\left \langle 9.4mi, 12.5mi, 6.6mi\right \rangle\end{align*}
3. A = 5km,3km,8km\begin{align*}\left \langle 5km, 3km, 8km\right \rangle\end{align*} and B = 10km,20km,19km\begin{align*}\left \langle 10km, 20km, 19km\right \rangle\end{align*}
4. A = 1cm,3cm,1cm\begin{align*}\left \langle 1cm, 3cm, 1cm\right \rangle\end{align*} and B = 5.1cm,2cm,5cm\begin{align*}\left \langle 5.1cm, 2cm, 5cm\right \rangle\end{align*}
5. A = 5.6mm,10.2mm,2.2mm\begin{align*}\left \langle 5.6mm, 10.2mm, 2.2mm\right \rangle\end{align*} and B = 20.4m,31.1mm,1.1mm\begin{align*}\left \langle 20.4m, 31.1mm, 1.1mm\right \rangle\end{align*}
6. A = 1in,2in,3in\begin{align*}\left \langle 1in, 2in, 3in\right \rangle\end{align*} and B = 4in,5in,3in\begin{align*}\left \langle 4in, 5in, 3in\right \rangle\end{align*}

Draw the three dimensional vector

1. A=3i^+2j^+5k^\begin{align*}\overrightarrow{A} = 3\hat{i} + 2\hat{j} +5\hat{k}\end{align*}
2. A=5i^3j^+1k^\begin{align*}\overrightarrow{A} = 5\hat{i} - 3\hat{j} +1\hat{k}\end{align*}
3. A=7i^+1j^4k^\begin{align*}\overrightarrow{A} = 7\hat{i} + 1\hat{j} - 4\hat{k}\end{align*}
4. A=7i^4j^+8k^\begin{align*}\overrightarrow{A} = -7\hat{i} - 4\hat{j} +8\hat{k}\end{align*}
5. A=4i^+7j^3k^\begin{align*}\overrightarrow{A} = -4\hat{i} + 7\hat{j} - 3\hat{k}\end{align*}
6. A=3i^2j^5k^\begin{align*}\overrightarrow{A} = -3\hat{i} - 2\hat{j} - 5\hat{k}\end{align*}

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### Vocabulary Language: English

midpoint

The midpoint of two vectors is the location in the center of their endpoints.

octant

An octant is any one of the eight "corners" of the three-dimensional rectangular coordinate system.

position vector

A position vector describes the straight-line travel between a starting point (usually the origin) and the location of a second point on a coordinate plane.

A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.

A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.

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Date Created:
Nov 01, 2012
May 26, 2016
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