The Cross Product of Two Vectors In Space

Report
The Cross Product of Two
Vectors In Space
Written by Dr. Julia Arnold
Associate Professor of Mathematics
Tidewater Community College, Norfolk Campus, Norfolk, VA
With Assistance from a VCCS LearningWare Grant
In this lesson you will learn
•how to find the cross product of two vectors
•how to find an orthogonal vector to a plane defined by two vectors
•how to find the area of a parallelogram given two vectors
•how to find the volume of a parallelepiped given three vectors
Objective 1
Finding the cross product of two
vectors
The Cross Product
The cross product of two vectors, denoted as
the dot product, represents a vector.
ab
, unlike
The cross product is defined to be for
a  a1 ,a2 , a3 and b  b1 ,b2 , b3
a  b  a2b3  a3b2 , a3b1  a1b3 , a1b2  a2b1
You are probably wondering if there is an easy way to
remember this.
The easy way is to use determinants of size 3 x 3.
a  a1 ,a2 , a3 and b  b1 ,b2 , b3
a  b  a2b3  a3b2 , a3b1  a1b3 , a1b2  a2b1
Let’s set up a 3 x 3 determinant as follows:
1. First use the unit vectors
i , j, and k
as the first row of the determinant.
2. Use row 2 for the components of a and row 3 for the
components of b.
i
j
k
a1
a2
a3  i a2b3   j a3b1   k a1b2   i a3b2   j a1b3   k a2b1  
b1
b2
b3

a2b3  a3b2 i  a3b1  a1b3  j  a1b2  a2b1 k

Find the cross product for the vectors below. Do the problem
before clicking again.
a  2,4,5
i
j
k
2
4
5 
1
2 1
and b  1,2,1


i 4 1  j 51  k 2 2  i 5 2  j 2 1  k 41 
 4  10i  51  2 j   4  4 k
6i  7 j  8k
Since the cross product is determined by using determinants, we
can understand the algebraic properties of the Cross Product
which are:
•
u  v   v  u Which would come from the fact that if you
interchange two rows of a determinant you
negate the determinant.
 
•
•

   
u v w  uv  uw
   
 
c u  v  cu  v  u  cv
•
u  0  0 u  0
•
uu  0
•
u  v w  u  v  w
   
Objective 2
find an orthogonal vector to a
plane defined by two vectors
Now that you can do a cross product the next step is to see why
this is useful.
Let’s look at the 3 vectors from the last problem
a  2,4,5
, b  1,2,1 and a  b  6,7,8
What is the dot product of
a  2,4,5
with a  b  6,7,8
And
b  1,2,1 with a  b  6,7,8
If you answered 0 in both cases, you?would be
correct.
Recall that whenever two non-zero vectors are
perpendicular, their dot product is 0. Thus the cross
product creates a vector perpendicular to the vectors
a and b. Orthogonal is another name for
perpendicular.
   
Show that u  v  w  u  v  w is true for
u  1,2,3 , v   1,1,0 , and w  5,0,1
Solution:
u  1,2,3 , v   1,1,0 , and w  5,0,1
The left hand side
i
j
1 1
k
0  i  0  0  (5k  j  0)  i  j  5k
5
1
0
1,2,3  1,1,5  1  2  15  14
The right hand side
i
j k
1 2 3  0  3 j  k  (2k  0  3i )  3i  3 j  k
1 1 0
3,3,1  5,0,1  15  0  1  14
Geometric Properties of the Cross Product
Let u and v be nonzero vectors and let
theta) be the angle between u and v
1. u v is orthogonal to both u and v
u v
u

 (the Greek letter
v
Remember: orthogonal means
perpendicular to
Geometric Properties of the Cross Product
Let u and v be nonzero vectors and let
theta) be the angle between u and v
2.
u v
u

v
u v  u

(the Greek letter
v sin
Proof:
Let u  u1 ,u 2 , u 3 and v  v1 , v 2 , v 3
u  v  u 2 v 3  u 3 v 2 , u 3 v1  u1v 3 , u1v 2  u 2 v1
uv 
u 2 v 3  u 3 v 2 2  u 3 v1  u1v 3 2  u1v 2  u 2 v1 2
This is the left hand side
Let u  u1 ,u 2 , u 3 and v  v1 ,v 2 , v 3
u v sin   u12  u 2 2  u 3 2 v12  v 2 2  v 3 2 1  cos2 
Re call cos  
uv
u  v
2
so cos2  
2
u v
u
u  v
2
u v sin   u v 1 
2
u
2
v

2
v
u v
2
u
u v
2
 
v  uv
2

2
u
2
 
v  uv
Now all we need to do is show that the stuff under the radical is
the same as the square of the magnitude of the cross product.
In other words, equate the radicands.
2
Let u  u1 , u 2 , u3 and v  v1 , v 2 , v3
 u2v3  u3v 2    u3v1  u1v 3    u1v2  u2v 1 
uv 
u
2
2
2
2

2
2
u
v  (u12  u22  u32 )(v12  v22  v32 ) 
2
 
v  uv
2
u12v12  u12v22  u12v32  u22v12  u22v22  u22v32  u32v12  u32v22  u32v32
(u  v) 2    u1v1  u2v2  u3v3  u1v1  u2v2  u3v3  
u12 v12  2u1v1u2 v2  2u1v1u3v3  u22v22  2u2v2u3v3  u32 v32
u
2
2
v  (u  v) 2  u12 v12  u12v22  u12v32  u22v12  u22v22  u22v32  u32v12  u32v22  u32v32
u12 v12  2u1v1u2v2  2u1v1u3v3  u22v22  2u2v2u3v3  u32v32 
u12v22  u12v32  u22v12  u22v32  u32v12  u32v22  2u1v1u2v2  2u1v1u3v3  2u2v2u3v3 
u12v22  2u1v1u2 v2  u22v12  u12 v32  2u1v1u3v3  u32v12  u22v32  2u2 v2u3v3  u32v22 
 u1 v2  u2 v1   u1 v3  u3 v1   u2v3  u3 v2 
2
2
2
Which concludes the proof
(Sometimes proofs are not
hard but some do require
patience.)
3.
u  v  0 if and only if u and v are multiples of each other
4.
u  v  area of the paralleogram having
u and v as adjacent sides.
Proof: The area of a parallelogram is base
times height. A = bh
||u||

y
||v||
sin  = y/||u||
||u||sin  = y = height
||v|| y = ||v|| ||u||sin  =

u v
Example problem for property 1
Find 2 unit vectors perpendicular to a= 2 i - j + 3k and
b = -4i + 2j - k.
Solution
The two given vectors define a plane.
The Cross Product of the vectors is perpendicular to
the plane and is proportional to one of the desired unit
vectors.
To make its length equal to one, we simply divide by its
magnitude:
.
Find 2 unit vectors perpendicular to a= 2 i - j + 3k
and b = -4i + 2j - k.
Solution
i
j k
2  1 3  (i  12 j  4k )  (6i  2 j  4k )  (5i  10 j )
 4 2 1
 5i  10 j  25  100  125  5 5
Divide the vector by its m agnitude


5
10
1
2
 5
2 5
 5i  10 j 
i
j
i
j
i
j
5
5
5 5
5 5
5 5
5
5
1
For a second unit vector simply multiply the answer by -1
5
2 5
i
j
5
5
Objective 3
find the area of a parallelogram
given two vectors
Example for property 4
4.
u  v  area of the paralleogram having
u and v as adjacent sides.
Area of a Parallelogram via the Cross Product
z
Show that the following 4 points define a parallelogram and then find the
A
area.
A(4,4,6), B(4,14,6), C(1,11,2), D(1,1,2)
x
B
C
y
D
Solution:
First we find the vectors of two
adjacent sides:
AB  4  4, 4  14,6  6  0,  10,0
AD  4  1, 4  1,6  2  3, 3,4
Now find the vectors associated with the opposite sides:
DC  1  1,1  11,2  2  0,  10,0  AB
DB  1  4,11 14,2  6   3,  3,4   AD
This shows that opposite sides are associated with the same vector,
hence parallel. Thus the figure is of a parallelogram.
Continued
The area is equal to the magnitude of the cross product of
vectors representing two adjacent sides:
Area = |AB XAD|
i
j k
0  10 0   40 i  0 j  0k  0 i  0 j  30k  40 i  0 j  30k

3
3


4
 40 i  30k (40) 2  302  1600 900  2500  50
The area of the parallelogram is 50 square units.
Area of a Triangle via the Cross Product
Since the area of a triangle is based on the area of a parallelogram, it
follows that the area would be ½ of the cross product of vectors of two
adjacent sides.
Find the area of the triangle whose vertices are P(4,4,6), Q(5,16,-2) R(1,1,2)
Area parallelogram) = |PQ x PR|. The area of the triangle is half of this.
z
P
R
Q
x
Area of a Triangle via the Cross Product Continued
P(4,4,6), Q(5,16,-2) R(1,1,2)
Area parallelogram) = |PQ x PR|
But what if we choose RP and RQ? Would the result be the same?
Let’s do both and see!
z
P
R
PQ  5  4,16  4,2  6  1,12,8
PR  1  4,1  4,2  6   3,3,4
Q
x
RP  4  1,4  1,6  2  3,3,4
RQ  5  1,16  1,2  2  4,15,4
PQ  5  4,16  4,2  6  1,12,8
RP  4  1,4  1,6  2  3,3,4
PR  1  4,1  4,2  6   3,3,4
RQ  5  1,16  1,2  2  4,15,4
i
j
k



1 12  8   48 i  24 j  3k  24 i  4 j  36k  72 i  28 j  33k
3 3 4
1
1
1
1
 72 i  28 j  33k 
(72) 2  282  332 
5184 784  1089 
7075
2
2
2
2
i
j
k



3 3 4   12 i  16 j  45k  60 i  12 j  12k  72 i  28 j  33k
4 15  4
Since this is the same vector, the magnitude would
be the same also.
z
P
R
Q
x
The area of the triangle is ½ 7057  42 square units.
The triple scalar product is defined as:
 
u  vw
For u  u1 i  u 2 j  u3 k , v  v1 i  v2 j  v3 k , and w  w1 i  w2 j  w3 k
 
u1 u 2
u  v  w  v1
v2
u3
v3
w1 w2 w3
Objective 4
how to find the volume of a
parallelepiped given three vectors
The triple scalar product is defined as: AB ·(AC x AD).
A parallelepiped is a 6 sided figure whose sides are parallelograms.
The volume of the parallelepiped
can be found using the absolute
value of the triple scalar product
and 3 adjacent vectors.
Why would the triple scalar
product be the volume of the
figure?
Why would the triple scalar
product be the volume of the
figure?
B
Since the base is a parallelogram
we could represent its area by
the cross product ||AC x AD||
Recall that the area of a
parallelepiped is the base times
the height.
The height would be equivalent to:
C
A
D
proj AC AD AB
The triple scalar product is defined as: AB ·(AC x AD).
base times the height
B
||AC x AD|| proj AC AD AB
AC  AD projAC AD AB
AC  AD
C
A
D

  AB  AC  AD
AB  AC  AD
AC  AD
The absolute value insures a
positive answer for the volume.
Volume of a Parallelepiped via the Scalar Triple Product:
Find the volume of the parallelepiped with adjacent edges
AB, AC, and AD, where the points are A(4, -3, -2), B(2, 0, 5),
C(-3, 2, 1), and D(1, 3, 2).
The volume is given by the scalar triple product: AB · (AC X AD).
First we need the three vectors:
AB = [2 - 4]i + [0 - (-3)]j + [5 - (-2)]k = -2i + 3j + 7k.
AC = [-3 - 4]j + [2 - (-3)]j + [1 - (-2)]k = -7i + 5j + 3k
AD = [1 - 4]i + [3 - (-3)]j + [2 - (-2)]k = -3i + 6j + 4k
First, find the cross product:
= i[20 - 18] - j[-28 - (-9)] + k[-42 - (-15)] = 2i + 19j - 27k
Now form the dot product to get the volume:
Volume = |AB · (2i + 19j - 27k)|
= | (-2i + 3j + 7k) · (2i + 19j - 27k)|
= | 4 + 57 - 189 | = 136 cubic units
From:http://www.jtaylor1142001.net/calcjat/Solutions/VCro
ssProduct/VCPVolParallelepiped.htm
Finally, for all of you potential physicists, a real world
application of the cross product.
Torque is defined by Webster as “a twisting or wrenching
effect or moment exerted by a force acting at a distance
on a body, equal to the force multiplied by the
perpendicular distance between the line of action of the
force and the center of rotation at which it is exerted”.
M
If a vector force F is applied at
point B of the vector AB where
both vectors are in the same plane,
then M is the moment of the force
F about the point B.
||M|| will measure the tendency of
the vector AB to rotate
counterclockwise according to the
right hand rule about an axis
directed along the vector M.
A great example of torque is tightening a bolt with a
wrench.
A few observations can be made:
The farther away from the bolt that the
force is applied the greater the
magnitude of the torque. Thus, a longer
wrench produces greater torque for a
given amount of force.
Second, we can see
that the angle 
which produces the
largest torque
would be when
 = 90o.
Recall:
force lengthvector  force length sin 
where  is the angle between them.
Example: Suppose you have a 12 inch wrench and you
apply a 20 lb force at an angle of 30 degrees. What is
the torque in foot-pounds at the bolt? What is the
maximum torque that can be applied to this bolt?
Solution:
force lengthvector  force length sin 
where  is the angle between them.

1
201sin   20   10 foot pounds
6
2
The maximum torque is applied when  

201sin   201  20 foot pounds
2

2
For comments on this presentation you may email the author
Dr. Julia Arnold at [email protected]

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