Lesson 1: Coordinates and Distance

Section 9.1Coordinates and Distance

Math 21a

February 4, 2008

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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

Axes and Coordinates in spaceAxesOrientationCoordinate lines and planes

DistanceThe Pythagorean TheoremSimple curves and surfaces

Outline

Dimensions

Axes in Flatland

Axes in Spaceland

Mirror-image axes

Our convention is only to choose axes like those on the right.

Mirror-image axes

The right-hand rule

Placing points—Flatland

Example

Place the point P(3, 4) in the plane.

Solution

Example

Solution

Example

Solution

Example

Solution

Example

Solution

Placing points—spaceland

Example

Place the point P(3, 4, 5) in space.

Solution

4| | | |

Example

Solution

4| | | |

Example

Solution

4| | | |

Example

Solution

4| | | |

Example

Solution

4| | | |

Meet the Mathematician: Rene Descartes

I French, 1596–1650

I Philosopher andmathematician

I Cogito ergo sum

I Cartesian coordinatesystem

Coordinate lines in flatland

Example

Draw the line x = 3.

Solution

(3, 0)

Example

Solution

(3, 0)

Example

Solution

(3, 0)

Coordinate planes in spaceland

Example

Draw the plane x = 3.

Solution

(3, 0, 0)

Example

Solution

(3, 0, 0)

Example

Solution

(3, 0, 0)

Outline

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:

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

1 2√5

Example

Solution

1 2√5

Example

Solution

Example

Solution

1 2√5

Example

Solution

1 2√5

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.

A curve

In flatland, the set (or locus) of all points which are a fixeddistance from a fixed point is a circle.

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

(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).

Example

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25

= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9

(x − 2)2 + (y + 4) + (z − 5)2 = 9

=⇒ |(x , y , z)(2,−4, 5)| = 3

Example

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25

= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9

(x − 2)2 + (y + 4) + (z − 5)2 = 9

=⇒ |(x , y , z)(2,−4, 5)| = 3

Example

x2 + y2 + z2 − 4x + 8y − 10z + 36 = 0

0 = x2 − 4x + 4 + y2 + 8y + 16 + z2 − 10z + 25 + 36− 4− 16− 25

= (x − 2)2 + (y + 4)2 + (z − 5)2 − 9

(x − 2)2 + (y + 4) + (z − 5)2 = 9

=⇒ |(x , y , z)(2,−4, 5)| = 3

Lesson 1: Coordinates and Distance

Education

Transcript of Lesson 1: Coordinates and Distance