Post on 14-Apr-2017
Section 9.1Coordinates and Distance
Math 21a
February 4, 2008
Announcements
I Grab a bingo card and start playing!I Homework for Wednesday 2/6:
I 9.1: 5, 6, 7, 8, 10, 14, 18, 30, 32, 34;I 9.2.1*, 9.2.3*, 9.3.1*
Outline
Bingo
Axes and Coordinates in spaceAxesOrientationCoordinate lines and planes
DistanceThe Pythagorean TheoremSimple curves and surfaces
Outline
Bingo
Axes and Coordinates in spaceAxesOrientationCoordinate lines and planes
DistanceThe Pythagorean TheoremSimple curves and surfaces
Placing points—spaceland
Example
Place the point P(3, 4, 5) in space.
Solution
x
y
z
3
|
|
|
4| | | |
5
−
−
−
−
−
Placing points—spaceland
Example
Place the point P(3, 4, 5) in space.
Solution
x
y
z
3
|
|
|
4| | | |
5
−
−
−
−
−
Placing points—spaceland
Example
Place the point P(3, 4, 5) in space.
Solution
x
y
z
3
|
|
|
4| | | |
5
−
−
−
−
−
Placing points—spaceland
Example
Place the point P(3, 4, 5) in space.
Solution
x
y
z
3
|
|
|
4| | | |
5
−
−
−
−
−
Placing points—spaceland
Example
Place the point P(3, 4, 5) in space.
Solution
x
y
z
3
|
|
|
4| | | |
5
−
−
−
−
−
Meet the Mathematician: Rene Descartes
I French, 1596–1650
I Philosopher andmathematician
I Cogito ergo sum
I Cartesian coordinatesystem
Outline
Bingo
Axes and Coordinates in spaceAxesOrientationCoordinate lines and planes
DistanceThe Pythagorean TheoremSimple curves and surfaces
The Pythagorean Theorem
If a, b, and c are sides of aright triangle and c is thehypotenuse, then
a2 + b2 = c2
Meet the mathematician: Pythagoras
I Greek, c. 580 – c. 490BCE (pre-Socratic)
I Philosopher who believedall order is in number
I until one of his orderdiscovered irrationalnumbers
Meet the mathematician: Pythagoras
I Greek, c. 580 – c. 490BCE (pre-Socratic)
I Philosopher who believedall order is in number
I until one of his orderdiscovered irrationalnumbers
Distance in flatland
Given two points P1(x1, y1) and P2(x2, y2), we can use Pythagorasto find the distance between them:
x
y
P1
P2
x2 − x1
y2 − y1
|P1P2| =√
(x2 − x1)2 + (y2 − y1)2
Distance in spaceland
Example
Find the distance between the points P1(3, 2, 1) and P2(4, 4, 4).
Solution
x
y
z
P1
P2
1 2√5
3d
Distance in spaceland
Example
Find the distance between the points P1(3, 2, 1) and P2(4, 4, 4).
Solution
x
y
z
P1
P2
1 2√5
3d
Distance in spaceland
Example
Find the distance between the points P1(3, 2, 1) and P2(4, 4, 4).
Solution
x
y
z
P1
P2
1 2
√5
3d
Distance in spaceland
Example
Find the distance between the points P1(3, 2, 1) and P2(4, 4, 4).
Solution
x
y
z
P1
P2
1 2√5
3d
Distance in spaceland
Example
Find the distance between the points P1(3, 2, 1) and P2(4, 4, 4).
Solution
x
y
z
P1
P2
1 2√5
3d
Distance in spaceland—General
TheoremThe distance between (x1, y1, z1) and (x2, y2, z2) is√
(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2
A curve
In flatland, the set (or locus) of all points which are a fixeddistance from a fixed point is a
circle.
y
x
A curve
In flatland, the set (or locus) of all points which are a fixeddistance from a fixed point is a circle.
y
x
A surface
In spaceland, the locus of all points which are a fixed distance froma fixed point is a
sphere.
A surface
In spaceland, the locus of all points which are a fixed distance froma fixed point is a sphere.
A surface
In spaceland, the locus of all points which are a fixed distance froma fixed point is a sphere.
Munging an equation to see its surface
Example
Find the surface is represented by the equation
x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0
SolutionWe can complete the square:
0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9
So
(x − 2)2 + (y + 4) + (z − 5)2 = 9
=⇒ |(x , y , z)(2,−4, 5)| = 3
This is a sphere of radius 3, centered at (2,−4, 5).
Munging an equation to see its surface
Example
Find the surface is represented by the equation
x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0
SolutionWe can complete the square:
0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9
So
(x − 2)2 + (y + 4) + (z − 5)2 = 9
=⇒ |(x , y , z)(2,−4, 5)| = 3
This is a sphere of radius 3, centered at (2,−4, 5).
Munging an equation to see its surface
Example
Find the surface is represented by the equation
x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0
SolutionWe can complete the square:
0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9
So
(x − 2)2 + (y + 4) + (z − 5)2 = 9
=⇒ |(x , y , z)(2,−4, 5)| = 3
This is a sphere of radius 3, centered at (2,−4, 5).
Munging an equation to see its surface
Example
Find the surface is represented by the equation
x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0
SolutionWe can complete the square:
0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9
So
(x − 2)2 + (y + 4) + (z − 5)2 = 9
=⇒ |(x , y , z)(2,−4, 5)| = 3
This is a sphere of radius 3, centered at (2,−4, 5).