Vectors and the Geometry of Space 2015

Section 10.1 The Three Dimensional

Coordinate System

In this lesson you will learn:

o 3 Space - The three-dimensional coordinate system

o Points in space, ordered triples

o The distance between two points in space

o The midpoint between two points in space

o The standard form for the equation of a sphere

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

z

y

xz plane

The 3-dimensional coordinate system is divided into eight octants as shown in the diagram.

Notice we draw the x- and y-axes in the opposite direction

X = directed distance from yz-plane to some point P

Y= directed distance from xz-plane to some point P

Z= directed distance from xy-plane to some point P

(x,y,z)

So, to plot points you go out or back, left or right, up or down

Plotting Points in Space

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

Find the lengths of the sides of triangle with vertices (1, -3, -2), (5, -1, 2) and (-1, 1, 2).

Then determine if the triangle is a right triangle, an isosceles triangle or neither.

You Try:

Isosceles Triangle

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.

2 4 3 4 0 2 2 7 2Solution : , , , ,

2 2 2

71, ,122 2 2

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

You Try:

Find the midpoint of the points P(5, -2, 3) and Q(0,4,4).

5 7,1,2 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. Points satisfying the

equation of a sphere are “surface points”, not “interior points.”

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: 2 2 21 4 4 19x y z

Find the equation of the sphere with endpoints of a diameter (2, -2, 2) and (-1, 4, 6).

You Try:

2

2 21 611 4

2 4x y z

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.

Traces

• The intersection of a sphere (or anything) with one of the three coordinate planes is called a trace. A trace occurs when the sphere is sliced by one of the coordinate planes. The trace of a sphere is a circle.

• To find the xy-trace, use the fact that every point in the xy-plane has a z-coordinate of 0. Substitute z=0 into the original equation and the resulting equation will represent the intersection of the sphere with the xy-plane.

Example 7:

Find the xy-trace of the sphere given by:

2 2 23 2 4 25x y z

2 23 2 16 25x y

2 23 2 9x y

Example 8: You Try:

Find the yz-trace of the sphere given by:

2 2 23 2 4 25x y z

2 22 4 16y z

Homework

• Day 1: Pg.711 1-9, 17-55 odds

• Day 2: Pg.711 2-10, 18-56 evens