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8.01
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Fall 2009
Review Module: Cross Product
We shall now introduce our second vector operation, called the “cross product” that takes any
two vectors and generates a new vector. The cross product is a type of “multiplication” law that
turns our vector space (law for addition of vectors) into a vector algebra (a vector algebra is a
vector space with an additional rule for multiplication of vectors). The first application of the
cross product will be the physical concept of torque about a point P that can be described
mathematically by the cross product between two vectors: a vector from P to where the force
acts, and the force vector.
Definition: Cross Product
r
r
Let A and B be two vectors. Since any two non-parallel vectors form a plane, we define
r
r
the angle θ to be the angle between the vectors A and B as shown in Figure 1. The
r r
r
r
magnitude of the cross product A × B of the vectors A and B is defined to be product of
r
r
the magnitude of the vectors A and B with the sine of the angle θ between the two
vectors,
r r
A × B = ABsin(θ ) ,
(1)
r
r
where A and B denote the magnitudes of A and B , respectively. The angle θ between
the vectors is limited to the values 0 ≤ θ ≤ π insuring that sin(θ ) ≥ 0 .
Figure 1 Cross product geometry.
r
r
The direction of the cross product is defined as follows. The vectors A and B form a plane.
Consider the direction perpendicular to this plane. There are two possibilities: we shall choose
r r
one of these two (shown in Figure 1) for the direction of the cross product A × B using a
convention that is commonly called the “right-hand rule”.
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Right-hand Rule for the Direction of Cross Product
r
r
The first step is to redraw the vectors A and B so that their tails are touching. Then draw an arc
r
r
starting from the vector A and finishing on the vector B . Curl your right fingers the same way
r r
as the arc. Your right thumb points in the direction of the cross product A × B (Figure 2).
Figure 2 Right-Hand Rule.
r r
You should remember that the direction of the cross product A × B is perpendicular to the plane
r
r
formed by A and B .
We can give a geometric interpretation to the magnitude of the cross product by writing the
magnitude as
r r
A × B = A ( B sin θ ) .
(2)
r
r
The vectors A and B form a parallelogram. The area of the parallelogram is equal to the height
times the base, which is the magnitude of the cross product. In Figure 3, two different
representations of the height and base of a parallelogram are illustrated. As depicted in Figure
r
3(a), the term B sin(θ ) is the projection of the vector B in the direction perpendicular to the
r
vector B . We could also write the magnitude of the cross product as
r r
A × B = ( A sin(θ )) B .
(3)
r
Now the term A sin(θ ) is the projection of the vector A in the direction perpendicular to the
r
vector B as shown in Figure 3(b).
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Figure 3(a) and 3(b) Projection of vectors and the cross product
The cross product of two vectors that are parallel (or anti-parallel) to each other is zero since the
angle between the vectors is 0 (or π ) and sin(0) = 0 (or sin(π ) = 0 ). Geometrically, two parallel
vectors do not have a unique component perpendicular to their common direction.
Properties of the Cross Product
(1) The cross product is anti-commutative since changing the order of the vectors cross product
changes the direction of the cross product vector by the right hand rule:
r r
r r
A × B = −B × A .
(4)
r
r
(2) The cross product between a vector c A where c is a scalar and a vector B is
r r
r r
c A × B = c ( A × B) .
(5)
r
r r
r
A × c B = c ( A × B) .
(6)
Similarly,
r
r
r
(3) The cross product between the sum of two vectors A and B with a vector C is
r r r r r r r
( A + B) × C = A × C + B × C
(7)
r r r
r r r r
A × (B + C) = A × B + A × C .
(8)
Similarly,
Vector Decomposition and the Cross Product
We first calculate that the magnitude of cross product of the unit vector î with ĵ :
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| ˆi × ˆj |=| ˆi || ˆj |sin(π / 2) = 1 ,
(9)
since the unit vector has magnitude | ˆi |=| ˆj |= 1 and sin(π / 2) = 1 . By the right hand rule, the
direction of ˆi × ˆj is in the +kˆ as shown in Figure 4. Thus ˆi × ˆj = kˆ .
Figure 4 Cross product of ˆi × ˆj
We note that the same rule applies for the unit vectors in the y and z directions,
ˆj × kˆ = ˆi , kˆ × ˆi = ˆj .
(10)
Note that by the anti-commutatively property (1) of the cross product,
ˆj × ˆi = −kˆ , ˆi × kˆ = −ˆj
(11)
The cross product of the unit vector î with itself is zero because the two unit vectors are parallel
to each other, ( sin(0) = 0 ),
| ˆi × ˆi |=| ˆi || ˆi | sin(0) = 0 .
(12)
The cross product of the unit vector ĵ with itself and the unit vector k̂ with itself are also zero
for the same reason.
ˆj × ˆj = 0,
kˆ × kˆ = 0 .
(13)
With these properties in mind we can now develop an algebraic expression for the cross product
r
in terms of components. Let’s choose a Cartesian coordinate system with the vector B pointing
r
r
along the positive x-axis with positive x-component Bx . Then the vectors A and B can be
written as
r
A = Ax ˆi + Ay ˆj + Az kˆ
(14)
and
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r
B = Bx ˆi ,
(15)
respectively. The cross product in vector components is
r r
ˆ × B ˆi .
A × B = ( Ax ˆi + Ay ˆj + Az k)
x
(16)
This becomes,
r r
A × B = ( Ax ˆi × Bx ˆi ) + ( Ay ˆj × Bx ˆi ) + ( Az kˆ × Bx ˆi )
= Ax Bx (ˆi × ˆi ) + Ay Bx (ˆj × ˆi ) + Az Bx (kˆ × ˆi )
.
(17)
= − Ay Bx kˆ + Az Bx ˆj
The vector component expression for the cross product easily generalizes for arbitrary vectors
and
r
A = Ax ˆi + Ay ˆj + Az kˆ
(18)
r
B = Bx ˆi + By ˆj + Bz kˆ ,
(19)
r r
A × B = ( Ay Bz − Az By ) ˆi + ( Az Bx − Ax Bz ) ˆj + ( Ax By − Ay Bx ) kˆ .
(20)
to yield
Example 1: Angular Momentum for a point particle an example of cross product
r
Consider a point particle of mass m moving with a velocity v (Figure 5).
Figure 5 A point particle and its position and velocity vectors.
r
r
The linear momentum of the particle is p = m v . Consider a point S located anywhere in space.
r
Let rS , m denote the vector from the point S to the location of the object.
r
Definition: The angular momentum L S about the point S of a point particle is defined to be the
vector cross product of the vector from the point S to the location of the object with the
momentum of the particle,
r
r
r
L S = rS , m × p .
(21)
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Magnitude: The magnitude of the angular momentum about S is given by
r
r
r
L S = rS , m p sin θ ,
(22)
r
r
where θ is the angle between the vectors rS , m and p , and lies within the range [0 ≤ θ ≤ π ] . The
SI units for angular momentum are ⎡⎣ kg ⋅ m 2 ⋅ s −1 ⎤⎦ .
Direction of the Angular Momentum
Remember that, as in all cross products, the direction of the angular momentum is perpendicular
r
r
to the plane formed by rS , m and p , and is given by the right hand rule.
Figure 6 The Right Hand Rule.
There are two ways to visualize the magnitude of the angular momentum.
Figure 7 Vector diagram for angular momentum.
Define the moment arm, r⊥ , as the perpendicular distance from the point S to the line defined by
the direction of the momentum. Then
r
r⊥ = rS , m sin θ .
(23)
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Hence the magnitude of the angular momentum is the product of the moment arm with the
magnitude of the momentum.
r
r
L S = r⊥ p .
(24)
Alternatively, define the perpendicular momentum, p⊥ , to be the magnitude of the component of
r
the momentum perpendicular to the line defined by the direction of the vector rS , m . Thus
r
p⊥ = p sin θ .
(25)
Thus the magnitude of the angular momentum is the product of the distance from S to the point
object with the perpendicular momentum,
r
r
L S = rS , m p⊥ .
(26)
Example 2: Torque as a Cross Product
r
r
r
Let a force FP with magnitude F = FP act at a point P . Let rS , P be the vector from the point
r
r
S to a point P , with magnitude r = rS , P . Let F⊥ = F sin θ be the component of the force FP
that is perpendicular to the line passing from the point S to P . The angle between the vectors
r
r
rS , P and FP is θ with [0 ≤ θ ≤ π ] (Figure 8).
Figure 8 Torque about a point S
r
r
Definition: The torque τ S about a point S , due to a force FP , acting at the point P , is the
r
r
vector cross product of the vectors rS , P and FP ,
r
r
r
τ S = rS , P × FP .
(27)
Torque is a vector quantity with magnitude and direction given as follows:
(1) Magnitude: The magnitude of the torque about S is
τ S = r F⊥ = rF sin θ .
(28)
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The SI units for torque are [N ⋅ m] .
(2) Direction: The direction of the torque is perpendicular to the plane formed by the vectors
r
r
rS , P and FP (for [0 < θ < π ] ), and by definition points in the direction of the unit normal vector
to the plane n̂1 as shown in Figure 9.
Figure 9 Direction of torque about a point S
r
The magnitude of the cross product is the area of the parallelogram defined by vectors rS , P and
r
FP , the height times the base. Figure 10 shows the two different ways of defining height and
base,
Area = τ S = r⊥ F = r F⊥
(29)
Figure 10 Area of the torque parallelogram.
The direction of the cross product is defined by the right hand rule and so the cross
r
r
product of rS , P and FP points out of the page in Figure 10.
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Problems:
r
r r
r
Problem 1 Given two vectors, A = 2 ˆi + −3 ˆj + 7 kˆ and B = 5ˆi + ˆj + 2kˆ , find A × B .
Solution:
r r
A × B = ( Ay Bz − Az By ) ˆi + ( Az Bx − Ax Bz ) ˆj + ( Ax By − Ay Bx ) kˆ
= ((−3)(2) − (7)(1)) ˆi + ((7)(5) − (2)(2)) ˆj + ((2)(1) − (−3)(5)) kˆ
= −13 ˆi + 31 ˆj + 17 kˆ
Problem 2 Prove the law of sines using the cross product. (Hint: Consider the area of a triangle
r r r
r r
r
formed by three vectors A , B , and C , where A + B + C = 0 .)
r r r
r r r r
r r r r
Solution: Since A + B + C = 0 , we have that 0 = A × ( A + B + C) = A × B + A × C . Thus
r r
r r
r r
r r
A × B = − A × C or A × B = A × C .
r r
r r
r r
r r
From the figure we see that A × B = A B sin γ and A × C = A C sin β . Therefore
r
r r
r r
r
A B sin γ = A C sin β , and hence B / sin β = C / sin γ . A similar argument shows that
r
r
B / sin β = A / sin α proving the law of sines.
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r
r
Problem 3 Find a unit vector perpendicular to A = ˆi + ˆj − kˆ and B = −2ˆi − ˆj + 3kˆ .
r
r r
r
Solution: The cross product A × B is perpendicular to both A and B . Therefore the unit vectors
r
r r r r
r
nˆ = ± A × B / A × B are perpendicular to both A and B . We first calculate
r r
A × B = ( Ay Bz − Az By ) ˆi + ( Az Bx − Ax Bz ) ˆj + ( Ax By − Ay Bx ) kˆ
= ((1)(3) − ( −1)(−1)) ˆi + ((−1)(2) − (1)(3)) ˆj + ((1)(−1) − (1)(2)) kˆ .
= 2 ˆi − 5 ˆj − 3 kˆ
We now calculate the magnitude
r r
1/ 2
A × B = ( 22 + 52 + 32 ) = (38)1/ 2 .
Therefore the perpendicular unit vectors are
r r r r
nˆ = ± A × B / A × B = ± 2 ˆi − 5 ˆj − 3 kˆ /(38)1/ 2 .
(
)
r r
Problem 4 Show that the volume of a parallelpiped with edges formed by the vectors A , B ,
r r r
r
and C is given by A ⋅ (B × C) .
Solution: The volume of a parallelpiped is given by area of the base times height. If the base is
r
r r
r
formed by the vectors B and C , then the area of the base is given by the magnitude of B × C .
r r r r
The vector B × C = B × C nˆ where n̂ is a unit vector perpendicular to the base. The projection of
r
the vector A along the direction n̂ gives the height of the parallelpiped. This projection is given
r
r
by taking the dot product of A with a unit vector and is equal to A ⋅ nˆ = height . Therefore
r r r
r r r
r r r
A ⋅ (B × C) = A ⋅ B × C nˆ = B × C A ⋅ nˆ = (area )(height ) = (volume) .
(
) (
)
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r
Problem 5 Let A be an arbitrary vector and let n̂ be a unit vector in some fixed direction.
r
r
r
Show that A = A ⋅ nˆ nˆ + nˆ × A × nˆ .
(
) (
)
r
r
Solution: Let A = An nˆ + A⊥ eˆ where An is the component A in the direction of n̂ , ê is the
r
r
direction of the projection of A in a plane perpendicular to n̂ , and A⊥ is the component of A in
r
the direction of ê . Since eˆ ⋅ nˆ = 0 , A ⋅ nˆ = An . Note that
r
nˆ × A = nˆ × An nˆ + A⊥ eˆ = nˆ × A⊥ eˆ = A⊥ (nˆ × eˆ ) .
(
)
The unit vector nˆ × eˆ lies in the plane perpendicular to n̂ and is also perpendicular to ê .
Therefore ( nˆ × eˆ ) × nˆ is also a unit vector that is must be parallel to ê (by the right hand rule. So
r
nˆ × A × nˆ = A⊥ eˆ . Thus
(
)
r
r
r
A = A ⋅ nˆ nˆ + nˆ × A × nˆ = An nˆ + A⊥ eˆ .
(
) (
)
Problem 6 Angular Momentum of a Point-like Particle
A particle of mass m = 2.0 kg moves as shown in the sketch with a uniform velocity
r
v = 3.0 m ⋅ s −1 ˆi + 3.0 m ⋅ s −1 ˆj . At time t , the particle passes through the point
r
r0,m = 2.0 m ˆi + 3.0 m ˆj . Find the direction and the magnitude of the angular momentum about the
origin at time t .
Solution: Choose Cartesian coordinates with unit vectors shown in the figure above. The angular
r
momentum vector L 0 of the particle about the origin is given by:
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r
r
r r
r
L 0 = r0,m × p = r0,m × m v
(
)
(
= 2.0 m ˆi + 3.0 m ˆj × (2kg) 3.0 m ⋅ s −1ˆi + 3.0 m ⋅ s −1ˆj
r
= 0 + 12 kg ⋅ m 2 ⋅ s −1 kˆ − 18 kg ⋅ m 2 ⋅ s −1 (−kˆ ) + 0
= − 6 kg ⋅ m 2 ⋅ s −1 kˆ .
)
In the above, the relations
r r r r r
r r r r r r
i × j = k , j × i = −k , i × i = j × j = 0
were used.
Problem 7 Angular Momentum of a Point Particle Undergoing Circular Motion
Consider a point particle of mass m moving in a circle of radius R with velocity
r
v = Rω θˆ , with angular speed ω > 0 . Find the direction and magnitude of the angular
momentum about the center of a circle in terms of the radius of the circle R, the mass of the
moving object m, and the angular speed ω .
r
Solution: The velocity of the particle is given by v = Rω θˆ . The vector from the center of the
r
circle (the point S) to the object is given by rS ,m = R rˆ . The angular momentum about the center
of the circle is the cross product
r
r
r r
r
L S = rS ,m × p = rS ,m × mv = Rmv kˆ = RmRω kˆ = mR 2ω kˆ .
r
The magnitude is L S = mR 2ω , and the direction is in the + kˆ -direction.
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Problem 8 Torque and Static Equilibrium: the Ankle
A person of mass m = 75 kg is crouching with their weight evenly distributed on both tiptoes.
The forces on the skeletal part of the foot are shown in the diagram. In this position, the tibia acts
r
r
on the foot with a force F of an unknown magnitude F = F and makes an unknown angle β
with the vertical. This force acts on the ankle a horizontal distance s = 4.8 cm from the point
where the foot contacts the floor. The Achilles tendon also acts on the foot and is under
r
considerable tension with magnitude T ≡ T and acts at an angle α = 370 with the horizontal as
shown in the figure. The tendon acts on the ankle a horizontal distance b = 6.0 cm from the
point where the tibia acts on the foot. You may ignore the weight of the foot. Let g = 9.8 m ⋅ s -2
be the gravitational constant. Compute the torque about the point of action of the tibia on the
ankle due to (i) the normal force of the floor; (ii) the tendon force; (iii) the force of the tibia. If
the torque about the point of action of the tibia on the ankle is zero, what is the magnitude of the
tendon force?
Solution: We shall first calculate the torque due to the force of the Achilles tendon on the ankle.
r
The tendon force has the vector decomposition T = T cos α ˆi + T sin α ˆj .
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r
The vector from the point S to the point of action of the force is given by rS ,T = −bˆi . Therefore
r
the torque due to the force of the tendon T on the ankle about the point where the tibia exerts a
force on the foot is then
r
r
r
τ S ,T = rS ,T × T = −bˆi × (T cos α ˆi + T sin α ˆj) = −bT sin α kˆ
The torque diagram for the normal force is shown in the figure below;
The vector from the point S to the point where the normal force acts on the foot is given by
r
rS , N = ( sˆi − hˆj) . Since the weight is evenly distributed on the two feet, the normal force on one
foot is equal to half the weight, or N = (1 2 ) mg . The normal force is therefore given by
r
N = N ˆj = (1/ 2) mg ˆj . Therefore the torque of the normal force about the point S is
r
r
τ S , N = rS , N × N ˆj = ( sˆi − hˆj) × N ˆj = s N kˆ = (1/ 2) s mg kˆ
r
The force F that the tibia exerts on the ankle will make no contribution to the torque about this
r
r
point S since the tibia force acts at the point S and therefore the vector rS , F = 0 .
If the torque about the point S is zero then the two vector contributions to the torque about the
point S add up to zero
r
−b T sin α kˆ + (1/ 2) s mg kˆ = 0 .
We can use this torque condition to solve for the tension in the tendon,
T=
(1/ 2) s mg (1/ 2)(4.8 × 10−2 m)(75 kg)(9.8 m ⋅ s −2 )
=
= 4.9 × 102 N .
o
−2
b sin α
(6.0 ×10 m) sin(37 )
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