Chapter 12 Section 12.1: Three-Dimensional Coordinate Systems We locate a point on a number line as one coordinate, in the plane as an ordered pair, and in space as an ordered triple. So we call number line as one dimensional, plane as two dimensional, and space as three dimensional co – ordinate system. In three dimensional, there is origin (0, 0, 0) and there are three axes – x -, y - , and z – axis. X – and y – axes are horizontal and z – axis is vertical. These three axes divide the space into eight equal parts, called the octants. In addition, these three axes divide the space into three coordinate planes. – The xy-plane contains the x- and y-axes. The equation is z = 0. – The yz-plane contains the y- and z-axes. The equation is x = 0. – The xz-plane contains the x- and z-axes. The equation is y = 0. If P is any point in space, let: – a be the (directed) distance from the yz-plane to P. – b be the distance from the xz-plane to P. – c be the distance from the xy-plane to P. Then the point P by the ordered triple of real numbers (a, b, c), where a, b, and c are the coordinates of P. – a is the x-coordinate. – b is the y-coordinate. – c is the z-coordinate. – Thus, to locate a point (a, b, c) in space, start from the origin (0, 0, 0) and move a units along the x-axis. Then, move b units parallel to the y-axis. Finally, move c units parallel to the z-axis. The three dimensional Cartesian co – ordinate system follows the right hand rule. Examples: Plot the points (2,3,4), (2, -3, 4), (-2, -3, 4), (2, -3, -4), and (-2, -3, -4). The Cartesian product  x  x  = {(x, y, z) | x, y, z in } is the set of all ordered triples of real numbers and is denoted by 3 . Note: 1. In 2 – dimension, an equation in x and y represents a curve in the plane 2 . In 3 – dimension, an equation in x, y, and z represents a surface in space 3 . 2. When we see an equation, we must understand from the context that it is a curve in the plane or a surface in space. For example, y = 5 is a line in 2 �but it is a plane in 3 � ������ 3. in space, if k, l, & m are constants, then – x = k represents a plane parallel to the yz-plane ( a vertical plane). – y = k is a plane parallel to the xz-plane ( a vertical plane). – z = k is a plane parallel to the xy-plane ( a horizontal plane). – x = k & y = l is a line. – x = k & z = m is a line. – y = l & z = m is a line. – x = k, y = l and z = m is a point. Examples: Describe and sketch y = x in 3 Example: Solve: Which of the points P(6, 2, 3), Q(-5, -1, 4), and R(0, 3, 8) is closest to the xz – plane? Which point lies in the yz – plane? Distance between two points in space: We simply extend the formula from 2 to . 3 . The distance |p1 p2 | between the points P1(x1,y1, z1) and P2(x2, y2, z2) is: 2 2 21 2 2 1 ...

1. Chapter 12
Section 12.1: Three-Dimensional Coordinate Systems
We locate a point on a number line as one coordinate, in the
plane as an ordered pair, and in
space as an ordered triple. So we call number line as one
dimensional, plane as two
dimensional, and space as three dimensional co – ordinate
system.
In three dimensional, there is origin (0, 0, 0) and there are three
axes – x -, y - , and z – axis. X –
and y – axes are horizontal and z – axis is vertical. These three
axes divide the space into eight
equal parts, called the octants. In addition, these three axes
divide the space into three
coordinate planes.
– The xy-plane contains the x- and y-axes. The equation is z =
0.
– The yz-plane contains the y- and z-axes. The equation is x =
0.
– The xz-plane contains the x- and z-axes. The equation is y =
0.
If P is any point in space, let:
– a be the (directed) distance from the yz-plane to P.
– b be the distance from the xz-plane to P.
– c be the distance from the xy-plane to P.

2. Then the point P by the ordered triple of real numbers (a, b, c),
where a, b, and c are the
coordinates of P.
– a is the x-coordinate.
– b is the y-coordinate.
– c is the z-coordinate.
– Thus, to locate a point (a, b, c) in space, start from the origin
(0, 0, 0) and move a
units along the x-axis. Then, move b units parallel to the y-axis.
Finally, move c
units parallel to the z-axis.
The three dimensional Cartesian co – ordinate system follows
the right hand rule.
Examples:
Plot the points (2,3,4), (2, -3, 4), (-2, -3, 4), (2, -3, -4), and (-
2, -3, -4).

4. is the set of all ordered triples of
real numbers and is denot
Note:
1. In 2 – dimension, an equation in x and y represents a curve
–
dimension, an equation in x, y, and z represents a surface in
2. When we see an equation, we must understand from the
context that it is a curve in the
� �
������
3. in space, if k, l, & m are constants, then
– x = k represents a plane parallel to the yz-plane ( a vertical
plane).
– y = k is a plane parallel to the xz-plane ( a vertical plane).
– z = k is a plane parallel to the xy-plane ( a horizontal plane).
– x = k & y = l is a line.
– x = k & z = m is a line.
– y = l & z = m is a line.
– x = k, y = l and z = m is a point.

5. Example:
Solve:
Which of the points P(6, 2, 3), Q(-5, -1, 4), and R(0, 3, 8) is
closest to the xz – plane? Which point
lies in the yz – plane?

6. Distance between two points in space:
|p1 p2 | between the points
P1(x1,y1, z1) and P2(x2, y2, z2) is: 2 2 21 2 2 1 2 1 2 1( ) ( ) (
)P P x x y y z z= − + − + − .
Example: Find the distance from (3, 7, -5) to
a) The xy – plane
b) The yz – plane
c) The xz – plane
d) The x – axis
e) The y – axis
f) The z – axis.

7. Equation of a Sphere:
An equation of a sphere with center C(x0, y0, z0) and radius r
is:
(x – x0)2 + (y – y0)2 + (z – z0)2 = r2
In particular, if the center is the origin O, then an equation of
the sphere is:
x2 + y2 + z2 = r2
Examples:
1. Find an equation of the sphere with center (2, -6, 4) and
radius 5. Describe the
intersection with each of the coordinate planes.

8. 2. Find an equation of the sphere that passes through the origin
and whose center is
(1, 2, 3).
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Describe in words the region represented by the equation or
inequality:
3.

9. 4.
5.
Describe the given set with a single equation or a pair of
equations:
6.

10. 7.
Write inequalities to describe the set:
8.

11. 9.
Find the center and radius of the sphere:
10.

12. Section – 12.2 Vectors
A vector is a quantity that has both magnitude and direction. A
quantity that has only
magnitude is called a scalar. A vector is represented by a
directed line segment (an arrow).
The arrow shows the direction and the length gives the
magnitude.
We denote a vector by a boldface letter (v) or by an arrow
above the letter ( v
).
Example: velocity is a vector quantity and speed is a scalar
quantity.
The following picture shows the velocity vectors in plane and
space.
Definitions:
1. Equal vectors: two vectors are equal if both have the same
magnitude and direction.
2. Zero vector (o): a vector with magnitude 0.
3. Unit vector (u): a vector with magnitude 1.
4. Negative of a vector v (– v): a vector with same length of v
and opposite direction.

13. Vector addition (triangle law):
If u and v are vectors positioned so the initial point of v is at
the terminal point of u,
then
the sum u + v is the vector from the initial point of u to the
terminal point of v.
That is, when we combine two vectors, the effective vector has
the magnitude and
direction of the vector u + v.
Vector addition is commutative: that is u + v = v + u. this can
be easily seen from the
parallelogram law illustrated below.
Also vector addition is associative.
Subtraction of vectors: to subtract v from u, we simply add the
negative of v.
That is u – v = u + (-v)
Scalar multiplication of vectors:

14. The scalar multiple cv of a vector v is the vector with the same
direction of v and c times
its length. Some examples are below:
To study more about vectors, it is essential to place vectors with
a common initial point. So in
rectangular coordinate system, we fix the initial point of all
vectors as the origin. Then we can
represent a vector by simply using the terminal point. This form
of the vector is called the
position vector. So for convenient purpose and for studying
vectors we fix all vectors as position
vectors.
To find the position vector of any vector in the plane, or space,
we simply subtract the initial
point from the terminal point.
The position vector of PQ
with initial point (x1, y1) and terminal point (x2, y2) is (x2 –
x1, y2 –
y1). Similarly in space the position vector of PQ
with initial point (x1, y1, z1 ) and terminal point
(x2, y2, z2 ) is (x2 – x1, y2 – y1, z2 – z1 ).
The position vector with terminal point (x, y) can be written as
< x, y>, which is called the
component form. See definition below:

15. Also we have the following:
The picture below illustrates addition geometrically.
Furthermore, If we call i = ‹1, 0, 0›
j = ‹0, 1, 0›
k = ‹0, 0, 1›
(These are standard vectors in the direction of the coordinate
axes with length I unit).
Then u = <u1, u2, u3> can be written as u1 i + u2 j + u3 k . In
two dimension it is
u = <u1, u2> = u1 i + u2 j
Length (Magnitude) of a vector:
The length of a vector u = <u1, u2, u3> is given by 2 2 21 2 3u
u u u= + + .
In plane, 2 21 2u u u= +
In this notes, we use both |u| and ||u|| notations for magnitude

16. of u.
Unit vector u in the direction of the vector v:
Examples:
Find a +b, 2a+3b, |a|, and | a – b |
1. a = 4i + j, b = i – 2j.
2. a = 2i – 4j + 4k, b = 2j – k.

17. 3. Find the unit vector in the direction of the vector v = < -4, 2,
4>.
4. Find a vector that has the same direction as < -2, 4, 2 > but
has length 6.

18. 5.
Applications:
Force:
Since force has both magnitude and direction, it is usually
represented by a vector. When two
or more forces are acting on an object the resulting force is the
vector sum of the forces.

19. Examples:
1. A 100 lb weight hangs from two wires making angles 50 and
32 degrees with the
horizontal (see the picture below). Find the forces (tensions) F1
and F2 on each wire and
their magnitudes.
500
320
F1
F2
Solution
:
We express F1 and F2 in terms of their horizontal and vertical
components.

20. F1 = - F1 cos 50 i + F1 sin 50 j and F2 = F2 cos
32 i + F2 sin 32 j
And F1 + F2 = - ( -100) j = 100 j
So we have –F
1
cos 50° + F
2
cos 32° = 0
100 lb
F
1

21. sin 50° + F
2
sin 32° = 100
Solving for F1 from the first equation and substituting in the
second, we get,
1
1
cos 50
sin 50 sin 32 100
cos 32
F
F
°
°+ ° =
°
1

22. 100
85.64 lb
sin 50 tan 32 cos 50
F = ≈
°+ ° °
and
1
2
cos 50
64.91 lb
cos 32
F
F
°
= ≈
°

23. Substituting we have F
1
≈ –55.05 i + 65.60 j and F
2
≈ 55.05 i + 34.40 j
2. If a child pulls a sled through the snow on a level path with a
force of 50N exerted at an
angle of 38 degrees above the horizontal, find the horizontal
and vertical components of
the force.

24. 3.

25. 4.

26. 5.

27. 12.3 The Dot Product
Here we discuss the multiplication of vectors. There are two
types of products – dot product
(which is a scalar), and cross product (a vector).
In space the angle θ between two vectors u = < u1, u2, u3 >
and v = < v1, v2, v3 > is given by
�� = ������−1 ���1��1+��2��2+��3��3
‖��‖‖��‖
�
This can easily be proved by law of cosines.
This numerator is defined as the dot product of the two vectors.

28. So we have the angle between the vectors u and v is given by
Note:
The dot product u · v is:
– Positive, if u and v point in the same general direction
– Zero, if they are perpendicular
– Negative, if they point in generally opposite directions.
Note:

29. If u and v are in the same direction, then θ = 0, cosθ = 1, and u
. v = ||u|| ||v||
And if they are in opposite direction, then θ = π, cosθ = -1,
and u . v = - ||u|| ||v||
Properties of Dot Products:
Vector Projection:
Vector Projection of u = PQ
onto another nonzero vector v = PS
is the vector PR
determined by drawing a perpendicular from Q to the line PS .

30. It is denoted as projv u (read as the vector projection of u on to
v).
If we think of u as the force vector, then projv u is the effective
force in the direction of v.
If the angle θ between u and v is acute , then projv u has length
|| u || cos θ and direction
v / (||v||) and if θ is obtuse, then since cos θ < 0, projv u has
length – || u || cos θ and
direction – v / (||v||).
And we have
The number || u || cos θ is called the scalar component of u in
the direction of v.
Hence we have:

31. Examples:
Find:
1.

32. 2.

33. Work:
We have learned that the work done by a constant force F
to move an object a distance d is
W = Fd. This is true when the force is applied in the direction
of motion.
Now consider a force vector F is applied at an angle θ to the
direction of the direction of
motion of the object, the resulting displacement in the direction
of the motion is a

34. vector, d.
Then work W = F . d = (||F|| cos θ) || d||.
Examples:
1. A wagon is pulled a distance of 100 m along a horizontal
path by a constant force of 80
N. The handle of the wagon is held at an angle of 35° above the
horizontal. Find the
work done by the force