THREE DIMENSIONAL GEOMETRY INTRODUCTION
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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
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