# Chapter 11: Three-Dimensional Analytic Geometry and Vectors ```Chapter 11: Three-Dimensional Analytic Geometry
and Vectors
Section 11.1: Three-Dimensional Coordinate System
Definition: In order to represent points in space, we first choose a fixed point O (the origin)
and three directed lines through O perpendicular to each other, called the coordinate
axes, and labeled the x-axis, y-axis, and z-axis. The xy, xz, and yz-planes are called the
coordinate planes. These planes divide space into octants.
Definition: Points in space are represented by ordered triples of real numbers (x, y, z). The
set of of all ordered triples of real numbers is called the three-dimensional rectangular
coordinate system or space and denoted by
R3 = {(x, y, z)|x, y, z ∈ R}.
𝑧
𝑧
𝑦
𝑦
𝑥
𝑥
Coordinate Axes
First Octant
Theorem: (Distance Formula)
The distance between the points P1 = (x1 , y1 , z1 ) and P2 = (x2 , y2 , z2 ) is given by
p
|P1 P2 | = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 .
Example: Find the lengths of the sides of the triangle with vertices A = (5, 5, 1), B = (3, 3, 2),
and C = (1, 4, 4). Determine whether the triangle is isosceles, right, both, or neither.
The lengths of the sides are
p
√
√
|AB| = (5 − 3)2 + (5 − 3)2 + (1 − 2)2 = 4 + 4 + 1 = 9 = 3,
p
√
√
√
|AC| = (5 − 1)2 + (5 − 4)2 + (5 − 4)2 = 16 + 1 + 1 = 18 = 3 2,
p
√
√
|BC| = (3 − 2)2 + (3 − 4)2 + (2 − 4)2 = 1 + 1 + 4 = 5.
The triangle is neither isosceles nor right.
Example: Find the lengths of the sides of the triangle with vertices A = (2, 1, 0), B = (3, 3, 4),
and C = (5, 4, 3). Determine whether the triangle is isosceles, right, both, or neither.
The lengths of the sides are
p
√
√
|AB| = (3 − 2)2 + (3 − 1)2 + (4 − 0)2 = 1 + 4 + 16 = 21,
p
√
√
|AC| = (5 − 2)2 + (4 − 1)2 + (3 − 0)2 = 9 + 9 + 9 = 3 3,
p
√
√
|BC| = (5 − 3)2 + (4 − 3)2 + (3 − 4)2 = 4 + 1 + 1 = 6.
The triangle is right since
|AB|2 + |BC|2 = |AC|2 ,
but is not isosceles.
Theorem: (Equation of a Sphere)
An equation of the sphere centered at (x0 , y0 , z0 ) with radius r is
(x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r2 .
Example: Find an equation of the sphere with center (5, 3, 2) and radius 3. What is the
intersection of this sphere with the coordinate planes?
An equation of the sphere is
(x − 5)2 + (y − 3)2 + (z − 2)2 = 9.
The intersection with the xy-plane (z = 0) is a circle with center (5, 3, 0) and radius
(x − 5)2 + (y − 3)2 + 4 = 9
(x − 5)2 + (y − 3)2 = 5.
√
5
The intersection with the xz-plane (y = 0) is the point (5, 0, 2)
(x − 5)2 + 9 + (z − 2)2 = 9
(x − 5)2 + (z − 2)2 = 0.
The sphere does not intersect the yz-plane (x = 0). Indeed,
25 + (y − 3)2 + (z − 2)2 + 4 = 9
(y − 3)2 + (z − 2)2 = −16.
There are no real solutions of this equation.
Example: Find an equation of the sphere with center (1, 2, 3) and passing through (−1, 1, −2).
The radius of the sphere is
p
√
√
r = (1 + 1)2 + (2 − 1)2 + (3 + 2)2 = 4 + 1 + 25 = 30.
Thus, an equation of the sphere is
(x − 1)2 + (y − 2)2 + (z − 3)2 = 30.
Example: Show that the equation
x2 + y 2 + z 2 − 6x + 8y − 4z = 20
represents a sphere and find its center and radius.
By completing the square for each variable, we obtain
(x2 − 6x) + (y 2 + 8y) + (z 2 − 4z) = 20
(x2 − 6x + 9) + (y 2 + 8y + 16) + (z 2 − 4z + 4) = 20 + 29
(x − 3)2 + (y + 4)2 + (z − 2)2 = 49.
The center is (3, −4, 2) and the radius is r = 7.
Theorem: (Midpoint Formula)
The midpoint of the line segment from (x1 , y1 , z1 ) to (x2 , y2 , z2 ) is
x1 + x2 y1 + y2 z1 + z2
,
,
.
2
2
2
Example: Find an equation of the sphere whose diameter has endpoints (2, 1, 4) and (4, 3, 10).
The center of the sphere is the midpoint of the diameter:
2 + 4 1 + 3 4 + 10
,
,
= (3, 2, 7).
2
2
2
Thus, the radius of the sphere is
p
√
√
r = (3 − 2)2 + (2 − 1)2 + (7 − 4)2 = 1 + 1 + 9 = 11.
An equation of the sphere is
(x − 3)2 + (y − 2)2 + (z − 7)2 = 11.
Note: In two dimensions, the graph of an equation involving x and y is a curve in R2 . In
three dimensions, the graph of an equation involving x, y, and z is a surface in R3 .
Example: Describe the following regions in the indicated dimension:
(a) y = 2 in R2 and R3
In R2 , the equation y = 2 represents a horizontal line passing through (0, 2). In R3 ,
the equation y = 2 represents a plane parallel to the xz-plane passing through (0, 2, 0).
(b) x2 + y 2 ≤ 1 in R2 and R3
In R2 , the inequality represents the closed disk of radius 1 centered at (0, 0). In R3 , the
inequality represents a solid cylinder of radius 1 which is unbounded along the z-axis.
(c) yz = 0 in R3
This equation represents the union of the xy-plane (z = 0) and the xz-plane (y = 0).
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