Vectors in Three
Dimensions
Pre-Calculus Chapter 8 Sections 3 – 5
Vector in 3D

Vectors in 3D space are described by
ordered triple pairs P(x1, y1, z1).
 Three number lines that intersect at the
zero points
 To show this on paper we have the x-axis
appearing to come out of the paper

To locate a point first find x1 on the x-axis, y1 on the
y-axis and z1 on the z-axis.
 Imagine a plane perpendicular to the x-axis
at x1 and the same for the other points.
 The three planes will intersect at point P.
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 1
•
Locate the point at (–4, 3,2).
z
(–4, 3, 2)
y
x
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Pre-Calculus Chapter 8 Sections 3 – 5
Ordered Triples
•
Ordered triples can be used to represent vectors.
• The geometric interpretation is the same as ordered pairs.
• A directed line from the origin O to P(x, y, z) is called
vector OP corresponding to vector (x, y, z)
•
An extension of the formula for distance give us the
distance from the origin to the point (x, y, z) is
x2  y 2  z 2
• So the magnitude of vector (x, y, z) is
x2  y 2  z 2
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 2
•
Write the ordered triple that represents the vector
from A(–2, –5, 0) to B(3, 1, 8).
AB  3,1,8   2,5,0
 3   2,1   5,8  0
 5,6,8
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Pre-Calculus Chapter 8 Sections 3 – 5
Unit Vectors in 3D
•
Three unit vectors are used as components of vectors
in space.
•

 
The unit vectors on the x-, y-, z-axes are i , j ,and k


•

Vector a  a1 , a2, a3  can
be 


written as a  a1i  a2 j  a3k
respectively, where i  1,0,0, j  0,1,0, and k  0,0,1
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 3
Write GH as the sum of unit vectors for G(–2, –5, 4)
and H(1, 5, 6).
First express GH as an ordered
 Then
  triple.
write the sum of the unit vectorsi , j , and k .
GH  1,5,6   2,5,4
 1   2,5   5,6  4  3, 10, 2



 3i  10 j  2k .
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Perpendicular Vectors
Pre-Calculus Chapter 8 Sections 3 – 5
Perpendicular Vectors


 Let a and b be perpendicular vectors, and
let BA be a vector between their terminal
points as shown
 
 The magnitudes of a , b and BA must satisfy the
Pythagorean Theorem. BA 2  a 2  b 2
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Pre-Calculus Chapter 8 Sections 3 – 5
Inner Products and Dot Products

Compare
the resulting
equation with the original one.
2
2
2
BA  a  b if and only if a1b1 + a2b2 = 0.

The expression a1b1 + a2b2 = 0 is frequently used in the
study of vectors. It is called the inner product of


a and b .

Two vectors are perpendicular if and
only
if their inner




product is zero. For vector s a and b , a b  0
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Pre-Calculus Chapter 8 Sections 3 – 5
 
a  b  a1b1  a2b2
Example 1


Find each inner product if x  2,5, y  4,1, and

z  10,4. Are any pairs of vectors perpendicular?
 
x  y  24   51
 8  5  3; not perpendicu lar
 
x  z  210   54  20  20  0; perpendicu lar
 
y  z  410  14
 40  4  44; not perpendicu lar
The inner products of vectors in space is similar to
that of vectors in a plane.
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Pre-Calculus Chapter 8 Sections 3 – 5
Vectors in Space

Example 2


Find the inner product of v and w if v  2,3,4 and



Are
w  8, 3, 2.
v and w perpendicular?
 
v  w  28   33   42
 
v  w  16  9  8  1


Since the inner product is –1, v and w are not perpendicular.
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Application
Pre-Calculus Chapter 8 Sections 3 – 5
Lenny Montana was a former 5-year world
heavyweight wrestling champion. Suppose
Lenny and a tag team partner are each pulling
horizontally on the arms of an opponent. Lenny exert a force of 180
pounds due north while his partner exerts a force of 125 due east.
Example 1
a. Draw a labeled diagram that represents the forces.
b. Determine the resultant force exerted on the opponent.
c. Determine the angle the resultant makes with the east-west axis.
180
2
2  2
N
tan  
f  f1  f 2
125

180 lb f
180 lb  2
1 180
2
2

  tan




f

125

180
f2
125
θ
W
S
125 lb

f1
E

f  219 lb
  55.2 north of east
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Pre-Calculus Chapter 8 Sections 3 – 5
Justin work for a package delivery service.
Suppose that he is pushing a cart full of
packages weighing 125 pounds up a ramp 10 feet long at an incline of
20°. Find the work done by gravity as the cart moves the length of the
ramp. Assume that friction is not a factor.
Example 2
y
O
Let OQ represent the force of gravity or weight.
P(x, y)
The weight has a force of 125 lbs and it’s 


d  10
direction is down. The unit vector is F  0i  125 j .
y  3.42
20°
The application of force is OP and it’s magnitude
x  9.39
x
is 10 ft.

 
Write OP as d  xi  yj use trig to find x and y.
Q (0, –125)
x
cos 20 
10
x  10 cos 20
y
10
y  10 sin 20
sin 20 
x  9.39
y  3.42



So d  9.39i  3.42 j
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 2
Justin work for a package delivery service.
Suppose that he is pushing a cart full of packages weighing 125
pounds up a ramp 10 feet long at an incline of 20°. Find the work done
by gravity as the cart moves the length of the ramp. Assume that
friction is not a factor.
y
P(x, y)
10
20°
O x  9.39
Q (0, –125)
y  3.42
x
Apply the formula for determining the work done
by gravity.
   


 
W  F  d F  0i  125 j d  9.39i  3.42 j




W  0i  125 j  9.39i  3.42 j 
W  0  427.5 or  427.5 ft  lb
Work done by gravity is negative when a object
is lifted.
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 3
Ms. Davis is hanging a sigh for her restaurant
The sign is supported by two lightweight support bars as shown in the
diagram. If the bars make a 30° angles with each other and the sign
weighs 200 pounds, what are the magnitudes of the forces exerted by
the sign on each support bar?



F1 represents force on bar 1, F2 on bar 2, and Fw
weight of the sign.

200
sin 30  
F2

200
F2 
sin 30
 400
cos 30 
F1
400

F1  400 cos 30
 346
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Pre-Calculus Chapter 8 Sections 3 – 5
Example 4
A lighting system for a theater is supported
equally by two cables suspended from the ceiling. The cables form a
140° angle with each other. If the lighting system weighs 950
pounds, what is the force exerted by each of the cables on the
lighting system?
140°
x lb
x lb
20°
20°
950 lb
475 lb
x lb
475
sin 20 
x
475
x
sin 20
20°
950 lb
 1388.81 or 1389 lbs
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Pre-Calculus Chapter 8 Sections 3 – 5
Daily Assignment
8 Sections 3 – 5
 Text Book
 Chapter
 Pg 503
○ #13 – 33 Odd;
 Pg 509
○ #11 – 19 Odd;
 Pg 517
○ #15, 17, 19;
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