vectors and the geometry of space 2015 section 10.1 the three dimensional coordinate system
TRANSCRIPT
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