Mathematical Investigations IV
Name:
Vectors
Getting To the Point
Cross Products
Computing the Cross Product
Another version of “multiplication” of vectors is called the cross product. The result is a vector,
and it is calculated as follows:
iˆ
ĵ k̂
y1 z1 ˆ
z1 x1
x1 y1
i +
ĵ +
k̂
v1  v2 = x1 y1 z1 =
y2 z2
z2 x 2
x2 y2
x2 y2 z2
Alternatively, one may change the second determinant on the right to give:
iˆ
ĵ k̂
y1 z1 ˆ
x1 z1
x1 y1
i –
ĵ +
k̂
v1  v2 = x1 y1 z1 =
y2 z2
x 2 z2
x2 y2
x2 y2 z2
1.
Given v = < 3, 0, 0> and w = < 0, 4, 0>, find v  w. Sketch v, w, and v  w.
2.
Now, let v =
3 iˆ + ĵ and w = iˆ +
3 ĵ . Find v  w and sketch v, w, and v  w.
In general, if v and w both lie in the positive x-y plane, then the vector v  w has the same
direction as what vector?
Vectors 9x.1
Rev. F08
Mathematical Investigations IV
Name:
3.
Let v =<0, 2, 0> and w = <0, 1, 1>. Find v  w.
What plane do both v and w lie in?
The vector v  w has the same direction as what vector?
4.
Let v = 2 iˆ + 2 k̂ and w = iˆ +
3 k̂ . Find v  w.
What plane do both v and w lie in?
The vector v  w has the same direction as what vector?
5.
Use the results from problems #1- #4 to make a conjecture about the relationship between the
plane containing two vectors and the direction of the cross product of the two vectors.
6.
Now consider the vectors p = < –1, 0, 1 > and q = < 0, –1, 1 >. Find p  q and sketch p, q,
and p  q.
Does your conjecture from problem #5 still seem to be correct for these vectors?
Vectors 9x.2
Rev. F08
Mathematical Investigations IV
Name:
You are now ready for the right hand rule. Ask about it.
Another interesting feature of the cross product relates it to the area of the
parallelogram made by two vectors v and w, with tails lined up together and angle 
between them. Note that we are not even thinking of proving this very useful fact at this
time.
v
h

w
Write the altitude h as a function of | v | and of .
The area of the parallelogram = | w | h =
______________________
(in terms of | v |, | w |, and .)
The magnitude of the cross product
equals
the area of the parallelogram
formed by v and w.
That is,
| v  w | = | v | | w | sin 
In summary:
The cross product produces a vector.
*
Direction: perpendicular to the plane containing v and w
*
Magnitude: | v | | w | sin , where  is the angle between v and w.
Vectors 9x.3
Rev. F08
Mathematical Investigations IV
Name:
Problems
1.
Suppose v = <5, 3, 1> and w = <2, 5, 6>,
a.
Find v  w and w  v.
b.
Find | v  w |.
c.
Find the unit vector in the direction of v  w.
(How can you find the unit vector in the direction of w  v without extra work?)
2.
Find <1, 2, 1>  <–2, –1, 3>.
3.
Find <0, 2, –3>  <1, 1, 1>
Vectors 9x.4
Rev. F08
Mathematical Investigations IV
Name:
4.
What do you get when you cross a vector with itself? (Try it using v or w from the
previous problem). Could you have found this result without using a specific example?
What reasoning would you use?
5.
a.
Find a such that p  q is perpendicular to the x-y plane, where
p = < 2, 4, a – 4 > and q = <–1, –3, 0 >.
b.
What is (p  q) q for the value of a determined in part a?
Does this make sense? Why or why not?
c.
6.
What is q  q ?
Find c such that the parallelogram formed by r = < c, 2, –5 > and s = < –1, 3, 1 > has an area
of 40 units.
Vectors 9x.5
Rev. F08
Mathematical Investigations IV
Name:
7.
Use the cross product to find the area of the triangle formed by the vectors
<3, 0, 0> and <5, 4, 0> if both originate at the origin? (Hint: What is the relationship
between the parallelogram formed and the triangle formed by two vectors?)
8.
Suppose m = <1, 1, 1> and w = < –2, 3, –6>.
a.
Sketch m and w, and use the dot product to
find the angle between them.
Make sure your answer seems reasonable.
b.
Find | m  w |.
c.
Now use the formula | m  w | = | m | | w | sin to find the angle between the vectors
m and w.
d.
What is the range of y = sin-1x? Why might this matter when considering whether or not
to use the formula | m  w | = | m | | w | sin to find the angle between two vectors?
Vectors 9x.6
Rev. F08