1. Section 9.1
Coordinates and Distance
Math 21a
February 4, 2008
Announcements
Grab a bingo card and start playing!
Homework for Wednesday 2/6:
9.1: 5, 6, 7, 8, 10, 14, 18, 30, 32, 34;
9.2.1*, 9.2.3*, 9.3.1*
2. Outline
Bingo
Axes and Coordinates in space
Axes
Orientation
Coordinate lines and planes
Distance
The Pythagorean Theorem
Simple curves and surfaces
3. Outline
Bingo
Axes and Coordinates in space
Axes
Orientation
Coordinate lines and planes
Distance
The Pythagorean Theorem
Simple curves and surfaces
31. Coordinate planes in spaceland
Example
Draw the plane x = 3.
Solution
z
y
x
(3, 0, 0)
32. Outline
Bingo
Axes and Coordinates in space
Axes
Orientation
Coordinate lines and planes
Distance
The Pythagorean Theorem
Simple curves and surfaces
33. The Pythagorean Theorem
If a, b, and c are sides of a
right triangle and c is the
hypotenuse, then
a2 + b 2 = c 2
34. Meet the mathematician: Pythagoras
Greek, c. 580 – c. 490
BCE (pre-Socratic)
Philosopher who believed
all order is in number
until one of his order
discovered irrational
numbers
35. Meet the mathematician: Pythagoras
Greek, c. 580 – c. 490
BCE (pre-Socratic)
Philosopher who believed
all order is in number
until one of his order
discovered irrational
numbers
36. Distance in flatland
Given two points P1 (x1 , y1 ) and P2 (x2 , y2 ), we can use Pythagoras
to find the distance between them:
P2
y
y2 − y1
P1 x2 − x1
x
|P1 P2 | = (x2 − x1 )2 + (y2 − y1 )2
37. Distance in spaceland
Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
38. Distance in spaceland
Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
Solution
z
P2
y
P1
x
39. Distance in spaceland
Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
Solution
z
P2
y
2
1
P1
x
40. Distance in spaceland
Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
Solution
z
P2
y
√2
1 5
P1
x
41. Distance in spaceland
Example
Find the distance between the points P1 (3, 2, 1) and P2 (4, 4, 4).
Solution
z
P2
d 3
y
√2
1 5
P1
x
42. Distance in spaceland—General
Theorem
The distance between (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
43. A curve
In flatland, the set (or locus) of all points which are a fixed
distance from a fixed point is a
44. A curve
In flatland, the set (or locus) of all points which are a fixed
distance from a fixed point is a circle.
y
x
45. A surface
In spaceland, the locus of all points which are a fixed distance from
a fixed point is a
46. A surface
In spaceland, the locus of all points which are a fixed distance from
a fixed point is a sphere.
47. A surface
In spaceland, the locus of all points which are a fixed distance from
a fixed point is a sphere.
48. Munging an equation to see its surface
Example
Find the surface is represented by the equation
x 2 + y 2 + z 2 − 4x + 8y − 10z + 36 = 0
49. Munging an equation to see its surface
Example
Find the surface is represented by the equation
x 2 + y 2 + z 2 − 4x + 8y − 10z + 36 = 0
Solution
We can complete the square:
0 = x 2 − 4x + 4 + y 2 + 8y + 16 + z 2 − 10z + 25 + 36 − 4 − 16 − 25
= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9
50. Munging an equation to see its surface
Example
Find the surface is represented by the equation
x 2 + y 2 + z 2 − 4x + 8y − 10z + 36 = 0
Solution
We can complete the square:
0 = x 2 − 4x + 4 + y 2 + 8y + 16 + z 2 − 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
51. Munging an equation to see its surface
Example
Find the surface is represented by the equation
x 2 + y 2 + z 2 − 4x + 8y − 10z + 36 = 0
Solution
We can complete the square:
0 = x 2 − 4x + 4 + y 2 + 8y + 16 + z 2 − 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).