THREE DIMENSIONAL GEOMETRY

INTRODUCTION

TWO-DIMENSIONAL (2-D) COORDINATE SYSTEMS

To locate a point in a plane, two numbers are necessary.

We know that any point in the plane can be represented as an ordered pair (a, b) of real numbers—where a is the x-coordinate and b is the y-coordinate.

For this reason, a plane is called two-dimension

THREE-DIMENSIONAL (3-D) COORDINATE SYSTEMS To locate a point in space,

three numbers are required.

We represent any point in space by an ordered triple (a, b, c) of real numbers

THREE-DIMENSIONAL (3-D) COORDINATE SYSTEMS In order to represent points in

space, we first choose:

A fixed point O (the origin)

Three directed lines through O that are perpendicular to each other

COORDINATE AXES The three lines are called the

coordinate axes. They are labeled:

x-axisy-axisz-axis

COORDINATE AXESWe draw the orientation of the axes as shown.

COORDINATE PLANES The three coordinate axes

determine the three coordinate planes.

i. The xy-plane contains the x- and y-axes.

ii. The yz-plane contains the y- and z-axes.

iii. The xz-plane contains the x- and z-axes.

OCTANTS

These three coordinate planes divide space into eight parts, called octants.

The first octant, in the foreground, is determined by the positive axes

3-D COORDINATE SYSTEMS- EXAMPLE

1) Look at any bottom corner of a room The wall on your left is in the xz-plane.

2) The wall on your right is in the yz-plane.

3) The floor is in the xy-plane.4) and call the corner the origin.

3-D COORDINATE SYSTEMS

Now, if P is any point in space

We represent the point P by the ordered triple of real numbers (a, b, c).

We call a, b, and c the coordinates of P.

a is the x-coordinate.b is the y-coordinate.c is the z-coordinate.

DISTANCE FORMULA IN THREE DIMENSIONS

The distance |P1P2| between the points P1(x1,y1, z1) and P2(x2, y2, z2) is:

2 2 21 2 2 1 2 1 2 1( ) ( ) ( )PP x x y y z z

PROOF OF DISTANCE FORMULA

To see why this formula is true, we construct a rectangular box as shown, where:

P1 and P2 are opposite vertices.

The faces of the box are parallel to the coordinate planes

PROOF

If A(x2, y1, z1) and B(x2, y2, z1) are the vertices of the box, then

|P1A| = |x2 – x1|

|AB| = |y2 – y1|

|BP2| = |z2 – z1|

PROOF

Triangles P1BP2 and P1AB are right-angled.

So, two applications of the Pythagorean Theorem give:

|P1P2|2 = |P1B|2 + |BP2|2

|P1B|2 = |P1A|2 + |AB|2

PROOF

I. Combining those equations, we get:

II. |P1P2|2 = |P1A|2 + |AB|2 + |BP2|2

III. = |x2 – x1|2 + |y2 – y1|2 + |z2 – z1|2

IV. = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2

Therefore,2 2 2

1 2 2 1 2 1 2 1( ) ( ) ( )PP x x y y z z

EXAMPLE OF DISTANCE FORMULA The distance from the point P(2, –1,

7) to the point Q(1, –3, 5) is:

2 2 2(1 2) ( 3 1) (5 7)

1 4 43

PQ

SECTION FORMULA

INTERNAL DIVISION EXTERNAL DIVISION

MID POINT FORMULA

SOLVED EXAMPLE OF MID POINT FORMULA

CENTROID OF A TRIANGLE

The Centroid of a Triangle is usually represented by G

Therefore

G=(x1+x2+x3/3, y1+y2+y3/3,z1+z2+z3/3)

FORMULA FOR CENTROID OF A TRIANGLE