Chapter 7: Vectors and the Geometry of Space

Section 7.2 Space Coordinates and Vectors in

Space

Written by Karen Overman

Instructor of Mathematics

Tidewater Community College, Virginia Beach Campus

Virginia Beach, VA

With Assistance from a VCCS LearningWare Grant

In this lesson you will learn:

o 3 Space - The three-dimensional coordinate system o Points in space, ordered tripleso The distance between two points in spaceo The midpoint between two points in space o The standard form for the equation of a sphere

o Vectors in 3 Spaceo Different forms of vectorso Vector operationso Parallel vectorso Applications of vectors

Previously you studied vectors in the Cartesian plane or 2-dimensions, now we are going to expand our knowledge of vectors to 3-dimensions. Before we discuss vectors, let’s look at 3-dimensional space.

To construct a 3-dimensional system, start with a yz plane flat on the paper(or screen).

y

zNext, the x-axis is perpendicularthrough the origin. (Think of the x-axis as coming out of the screen towards you.)

For each axis drawn the arrow represents the positive end.

x

Three-Dimensional Space

y

z

x

This is considered a right-handed system.

To recognize a right-handed system, imagine your right thumb pointing up the positive z-axis, your fingers curl from the positive x-axis to the positive y-axis.

In a left-handed system, if your left thumb is pointing up the positive z-axis, your fingers will still curl from the positive x-axis to the positive y-axis. Below is an example of a left-handed system.

z

x

y

Throughout this lesson, we will use right-handed systems.

x

y

z

The 3-dimensional coordinate system is divided into eight octants. Three planes shown below separate 3 space into the eight octants.

The three planes are the yz plane which is perpendicular to the x-axis, the xy plane which is perpendicular to the z-axis and the xz plane which is perpendicular to the y-axis.

Think about 4 octants sitting on top of the xy plane and the other 4 octants sitting below the xy plane. yz plane

x

y

z

xy plane

x

y

z

xz plane

Every position or point in 3-dimensional space is identified by an ordered triple,(x, y, z).

Here is one example of plotting points in 3-dimensional space:

Plotting Points in Space

y

z

P (3, 4, 2)

The point is 3 units in front of the yz plane,4 points in front of the xz plane and 2 units up from the xy plane.

x

Here is another example of plotting points in space. In plotting the point Q (-3,4,-5) you will need to go back from the yz plane 3 units, out from the xz plane 4 units and down from the xy plane 5 units.

y

z

Q (-3, 4, -5)

As you can see it is more difficult to visualize points in 3 dimensions.

x

Distance Between Two Points in Space

The distance between two points

in space is given by the formula:

2122

122

12 zzyyxxd

222111 ,,and,, zyxQzyxP

Take a look at the next two slides to see how we come up with this formula.

Consider finding the distance between the two points, .

It is helpful to think of a rectangular solid with P in the bottom back corner and Q in the upper front corner with R below it at .

222111 ,,and,, zyxQzyxP

P

Q

R

Using two letters to represent the distance between the points, we knowfrom the Pythagorean Theorem that PQ² = PR² + RQ²

Using the Pythagorean Theorem againwe can show that

PR² =

122 ,, zyx

2122

12 yyxx

12 xx

12 yy

Note that RQ is . 12 zz

12 zz

P

Q

R

12 xx

12 yy

12 zz

Starting with PQ² = PR² + RQ²

Make the substitutions: PR² = and RQ = 2122

12 yyxx 12 zz

Thus, PQ² =

Or the distance from P to Q,

PQ =

2122

122

12 zzyyxx

2122

122

12 zzyyxx

That’s how we get the formula for the distance between any two points in space.

Find the distance between the points P(2, 3, 1) and Q(-3,4,2).

2.53327

1125

115

123423222

222

212

212

212

d

d

d

d

zzyyxxd

Example 1:

We will look at example problems related to the three-dimensional coordinate system as we look at the different topics.

Solution: Plugging into the distance formula:

Example 2:

Find the lengths of the sides of triangle with vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3). Then determine if the triangle is a right triangle, an isosceles triangle or neither.

Solution: First find the length of each side of the triangle by finding thedistance between each pair of vertices.

(0, 0, 0) and (5, 4, 1)

42

11625

010405 222

d

d

d

(0, 0, 0) and (4, -2, 3)

29

9416

030204 222

d

d

d

(5, 4, 1) and (4, -2, 3)

41

4361

134254 222

d

d

d

These are the lengths of the sides of the triangle. Since none of them are equal we know that it is not an isosceles triangle and since we know it is not a right triangle. Thus it is neither.

222412942

The Midpoint Between Two Points in Space

The midpoint between two points, is given by: 222111 ,,and,, zyxQzyxP

2

,2

,2

Midpoint 212121 zzyyxx

Each coordinate in the midpoint is simply the average of the coordinatesin P and Q.

1,27

,122

,27

,22

220

,2

43,

242

:Solution

Example 3: Find the midpoint of the points P(2, 3, 0) and Q(-4,4,2).

Equation of a Sphere

A sphere is the collection of all points equal distance from a center point.

To come up with the equation of a sphere, keep in mind that the distance

from any point (x, y, z) on the sphere to the center of the sphere,

is the constant r which is the radius of the sphere.

Using the two points (x, y, z), and r, the radius in the distance

formula, we get:

ooo zyx ,,

222r ooo zzyyxx

If we square both sides of this equation we get:

The standard equation of a sphere is

where r is the radius and is the center.

2222r ooo zzyyxx

ooo zyx ,,

ooo zyx ,,

Example 4:

Find the equation of the sphere with radius, r = 5 and center, (2, -3, 1).

Solution: Just plugging into the standard equation of a sphere we get:

25132 222 zyx

Example 5:

Find the equation of the sphere with endpoints of a diameter (4, 3, 1) and (-2, 5, 7).

Solution: Using the midpoint formula we can find the center and using the distance formula we can find the radius.

4,4,1

271

,2

53,

224

Center

19

919

414314Radius 222

Thus the equation is: 19441 222 zyx

Example 6:

Find the center and radius of the sphere, .07864222 zyxzyx

Solution: To find the center and the radius we simply need to write the equation of the sphere in standard form, . Then we can easily identify the center, and the radius, r. To do this we will need to complete the square on each variable.

ooo zyx ,,

2222r ooo zzyyxx

36432

169471689644

7864

07864

222

222

222

222

zyx

zzyyxx

zzyyxx

zyxzyx

Thus the center is (2, -3, -4) and the radius is 6.

Vectors in Three-Dimensional Space

Now that we have an understanding of the three-dimensional system, we are ready to discuss vectors in the three-dimensional system. All the information you learned about vectors in the previous lesson will apply, only now we will add in the third component. Vectors in component form in three dimensions are written as ordered triples, in other words, now a vector in component form is .

321 ,, vvvv

In three dimensions the zero vector is O = < 0, 0, 0> and the standard unit vectors are . 1,0,0and0,1,0,0,0,1 kji

z

y

x

k

i

j

Each of the unit vectors represents one unit of change in the direction of each of their respective positive axes.

Given the initial point, and the terminal point, , the component form of the vector can be found the same way it was on the Cartesian Plane.

321 ,, pppP 321 ,, qqqQ

321332211 ,,,, vvvpqpqpqvPQ

Be sure to subtract the initial point’s coordinates from the terminal point’s coordinates.

Component form of a vector

The same vector can be written as a combination of the unit vectors.

kvjvivv

321

Standard Unit Vector Notation

We will look at examples using both forms.

More on Vectors in Three-Dimensions

Let and let c be a scalar. 321321 ,,and,, vvvvuuuu

o Vector Equality:

o Magnitude or Length of a Vector:

o Vector Addition:

o Scalar Multiplication:

o Unit Vector in the Direction of :

.and,ifonlyandif 332211 vuvuvuvu

232

22

1 uuuu

332211 ,, vuvuvuvu

321 ,, cucucuuc

u

321 ,,1

uuuuu

u

Note: This is simply the vector multiplied by the reciprocal of its magnitude.

Let’s look at some example problems involving vectors.

Example 1:Sketch the vector with initial point P(2, 1, 0) and terminal point Q(3, 5, 4). Then find the component form of the vector, the standard unit vector form and a unit vector in the same direction.

Solution: First draw a 3D system and plot P and Q. The vector connects P to Q.

P

Q

33334

,33334

,3333

33

4,

33

4,

33

14,4,1

33

1VectorUnit

3316161441 222PQ

Second, find the component form of the vector. Do this by subtracting the initial point’s coordinates from the terminal point’s coordinates.

Example 1 Continued:

kjiPQ

PQ

44

and

4,4,104,15,23

Last, find a unit vector in the same direction. Do this by multiplying the vector by the reciprocal of the magnitude.

Note: You can verify it’s a unit vector by finding its magnitude.

113333

3316

3316

331

33

4

33

4

33

1222

Component form

Standard Unit Vector Form

Example 2:Given the vectors find the following:

a. b. c.

6,5,0and4,7,1,3,5,2 zvu

vu

zu

3 zvu

2

Solution:

a.

1,2,1

43,75,12

4,7,13,5,2

vu

b.

21,10,2

183,155,02

6,5,033,5,2

3zu

c.

4,22,5

646,5710,014

6,5,04,7,13,5,22

2 zvu

Parallel Vectors

You may recall from the previous section that a nonzero scalar multiple of a vector has the same direction as the vector (positive scalar) or the opposite direction as the vector (negative scalar). Since this is the case, any nonzero scalar multiple of a vector is considered a parallel vector.

In other words, if two vectors, , are parallel, then there exists some scalar, c such that . The zero vector does not have direction so it cannot be parallel.

vu

anducv

To get the idea, look at these vectors on the Cartesian Plane.

x

y

u

uv

2

uz

2

Example 3:Determine if the vector with initial point, P(3,2,-2) and terminal point, Q(7,5,1-3) is parallel to the vector . 3,9,12v

Solution: First find the component form of the vector from P to Q.

1,3,4

23,25,37

PQ

PQ

Second, if the two vectors are parallel, then there exists some scalar, c, such that .,3,43,9,12or cccPQcv

Then –12 = 4c c = -3And -9 = 3c c = -3And 3 = -c c = -3

For the two vectors to be parallel, c would have to be the samefor each coordinate. Since it is, the two vectors are parallel.

Example 4:Determine whether the points A(2,3,-1), B(0,1,3) and C(-3,-2,8) are collinear.

Solution: We need to find two vector between the three points and determine if they are parallel. If the two vectors are parallel and pass through a common point then the three points must be in the same line.

The vector from A to B is kjikji

422133120

Now we need to find the vector from A to C or B to C.

The vector from A to C is kjikji

955183223

To be parallel: -2 = -5c c = 2/5 -2 = -5c c = 2/5 4 = 9c c = 4/9

Since c is not the same in each case, the vectors are not parallel and the points are not collinear.

Example 5: Find a vector parallel to the vector with magnitude 5.kjiv

23

Solution: Be careful. We might quickly assume that all we need to do is to multiply the vector by 5. This would be fine if we were dealing with a unit vector. Since we are not, we need to multiply by the reciprocal of the magnitude first to get a unit vector and then multiply by 5.

14149123 222 v

kjiv

2314

1ofDirectioninVectorUnit

kjikjikji

v

14

5

14

10

14

1523

14

523

14

15

ofDirectionin5magnitudewithVector

Solution to Example 5 Continued:

You can verify the new vector is parallel if you look at the form:

Obviously the scalar multiple is .

You can verify the magnitude is 5 by finding the magnitude of the form:

kji

2314

5

14

5

52514

3501425

14100

14225

14

5

14

10

14

15Magnitude

14

5

14

10

14

15

222

kji

Example 6:The weight of an 80lb. chandelier hanging 2.5 feet from the ceiling is distributed over 3 chains. If the chains are located as shown below, represent the force exerted on each chain with a vector.

1 ft 1 ft1 ft

1 ft (0,1,0)

(-1,-1,0)

(1,-1,0)

(0,0,-2.5)

Solution to Example 6:First find the vectors from the chandelier to the three points on the ceiling.Each force is a multiple of the vector since we can find the direction, but we don’t know the magnitude.

1 ft 1 ft1 ft

1 ft (0,1,0)

(-1,-1,0)

(1,-1,0)

(0,0,-2.5)

5.2,1,01 aF

5.2,1,12 bF

5.2,1,13 cF

Solution to Example 6 Continued:

The sum of the three forces must negate the downward force of the chandelier from its weight.

So,

cbacbacb

FFF

5.25.25.2,,080,0,0

80,0,0 321

This gives us a system of three equations in three unknowns, a,b and c.

805.25.25.2

0

0

cba

cba

cb

Solving the system, you get a = 16, b = 8 and c = 8.

Thus the three forces are .20,8,8and20,8,8,40,16,0 321 FFF

You can find your practice problems for this lesson in Blackboard under the Assignments button under Lesson 7.2.