Contents
1
Euclidean Geometry in 3-Dimensional Space
1.4 The Cross Product . . . . . . . . . . . . . .
1.4.1
Properties of the Cross Product .
1.4.2 Triple Products . . . . . . . . . . .
1.4.3 Torque . . . . . . . . . . . . . . . .
1.4.4 Exercises . . . . . . . . . . . . . .
1.4.5 Solutions to Exercises . . . . . . .
Calculus III
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©2014-16 J. E. Franke, J. R. Griggs and L. K. Norris
CONTENTS
Calculus I
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CONTENTS
2
©2014-16 J. E. Franke, J. R. Griggs and L. K. Norris
Chapter 1
Euclidean Geometry in 3-Dimensional Space
Comments and suggestions are welcome!
1.4
The Cross Product
⃗
We now define a multiplication of vectors A⃗ and B,
the result of which is a new vector, called the cross
⃗
product of vectors A⃗ and B⃗ and denoted A⃗ × B.
⃗ this new vector
Given non-zero vectors A⃗ and B,
⃗
⃗
A × B has the following properties:
● A⃗ × B⃗ is orthogonal to both A⃗ and B⃗ as shown
in Figure 1. Its direction is determined using the
right-hand rule: starting with the fingers on the
right hand pointing in the direction of A⃗ we curve
⃗ then the thumb points in the
the fingers toward B,
⃗ as illustrated in Figure 2 and
direction of A⃗ × B,
discussed in Section 1.
Figure 1
● The magnitude of the cross product of vectors A⃗
and B⃗ is given by
⃗ = ∥A⃗ ∥ ∥B⃗ ∥ sin θ
∥A⃗ × B∥
(1)
⃗
where θ is the angle between vectors A⃗ and B.
Figure 2
Calculus III
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©2014-16 J. E. Franke, J. R. Griggs and L. K. Norris
CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Accordingly we have
Definition 1. Cross Product
⃗ B⃗ by
The cross product of vectors A⃗ and B⃗ in V3 is defined for non-zero vectors A,
A⃗ × B⃗ = (∥A⃗ ∥ ∥B⃗ ∥ sin θ) n̂
(1)
where θ is the angle between vectors A⃗ and B⃗ and n̂ is the unit vector orthogonal to both A⃗
⃗ whose direction is determined by the right-hand rule. If one or both of the vectors A,
⃗ B⃗
and B,
are zero, then the cross product is defined to be
A⃗ × 0⃗ = 0⃗ × B⃗ = 0⃗ × 0⃗ = 0⃗
Note that A⃗ × B⃗ is purposefully left undefined for vectors in V2 .
There is an alternative and useful formula for the cross product of two vectors in terms of the
components of the vectors. We state it as a corollary to the definition above.
Theorem 1. Cross Product in Terms of Components
The cross product of vector A⃗ = ⟨a1 , a2 , a3 ⟩ and vector B⃗ = ⟨b1 , b2 , b3 ⟩ is the vector
A⃗ × B⃗ = ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩
(2)
Proof. The proof takes several steps. We first show that the vector on the right-hand side of Equation
⃗ Then we show that it is orthogonal to both A⃗ and B.
⃗ This leaves
2 has the same magnitude as A⃗ × B.
⃗ We end by showing that when using the standard
only two possibilities, namely it is A⃗ × B⃗ or -A⃗ × B.
ˆ j,ˆ k̂, Equation 2 gives iˆ × jˆ = k̂, which is the correct answer using the right-hand rule.
basis vectors i,
To show that the magnitudes of the two vectors are equal, we need the trigonometric identity
sin2 θ + cos2 θ = 1 and the formula for cos θ in terms of the dot product
cos θ =
A⃗ ⋅ B⃗
∥A⃗ ∥ ∥B⃗ ∥
Combining them we can express sin2 θ as follows.
2
A⃗ ⋅ B⃗
)
sin θ = 1 − cos θ = 1 − (
∥A⃗ ∥ ∥B⃗ ∥
2
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©2014-16 J. E. Franke, J. R. Griggs and L. K. Norris
CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Equation 1 gives us the magnitude of the cross product, thus
⃗ 2 = (∥A⃗ ∥ ∥B⃗ ∥ sin θ)2
∥A⃗ × B∥
= (∥A⃗ ∥)2 (∥B⃗ ∥)2 sin2 θ
2
⎛
⎞
A⃗ ⋅ B⃗
2
2
⃗
⃗
)
= (∥A ∥) (∥B ∥) 1 − (
⎝
∥A⃗ ∥ ∥B⃗ ∥ ⎠
= (∥A⃗ ∥)2 (∥B⃗ ∥)2 − (∥A⃗ ∥)2 (∥B⃗ ∥)2 (
⃗ 2
(A⃗ ⋅ B)
)
(∥A⃗ ∥)2 (∥B⃗ ∥)2
⃗ 2
= (∥A⃗ ∥)2 (∥B⃗ ∥)2 − (A⃗ ⋅ B)
⃗ 2 = a 2 + a2 + a 2 , ∥B∥
⃗ 2 = b2 + b2 + b2
Let A⃗ = ⟨a1 , a2 , a3 ⟩ and B⃗ = ⟨b1 , b2 , b3 ⟩. Then substituting ∥A∥
1
2
3
1
2
3
and A⃗ ⋅ B⃗ = a1 b1 + a2 b2 + a3 b3 in the last line above yields
⃗ 2 = (a12 + a22 + a32 )(b12 + b22 + b32 ) − (a1 b1 + a2 b2 + a3 b3 )2
∥A⃗ × B∥
Expanding out the right-hand side one can show (using Maple for example) that the result is identical
to
(a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2 = ∥ ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩ ∥2
Thus the cross product and the vector on the right-hand side of Equation 2 have the same magnitude.
The next step in our proof is to show that the vector on the right-hand side of Equation 2 is
⃗ We do this by verifying that the dot products of both A⃗ and B⃗ with the
orthogonal to both A⃗ and B.
vector are zero.
A⃗ ⋅ ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩ = ⟨a1 , a2 , a3 ⟩ ⋅ ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩
= a1 (a2 b3 − a3 b2 ) + a2 (a3 b1 − a1 b3 ) + a3 (a1 b2 − a2 b1 )
X
X
X
a1
a2
b3 − X
a2
a1
b3 + X
=
a1X
a3X
bX2 + a2
a3
b
a3X
a1X
bX2 − a3X
a2
b
X
X
X1 − X1
=0
The identity B⃗ ⋅ ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩ = 0 is proved in a similar manner. Thus our
vector ⟨a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ⟩ is either in the direction of A⃗ × B⃗ or in the opposite
direction.
We demonstrate that it is in the correct direction with one example. The right-hand rule applied
ˆ j,ˆ k̂ yields
to the standard basis vectors i,
iˆ × jˆ = k̂
Calculating the right-hand side of Equation 2 yields the same result.
iˆ × jˆ = ⟨1, 0, 0⟩ × ⟨0, 1, 0⟩ = ⟨0 ⋅ 0 − 0 ⋅ 1, 0 ⋅ 0 − 1 ⋅ 0, 1 ⋅ 1 − 0 ⋅ 0⟩ = ⟨0, 0, 1⟩ = k̂
Calculus I
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Example 1. Compute the cross product of vector u⃗ = ⟨1, 2, 3⟩ with vector v⃗ = ⟨−2, 7, 11⟩.
Solution: Using Formula (2) we find
u⃗ × v⃗ = ⟨2 ⋅ 11 − 3 ⋅ 7 , 3 ⋅ (−2) − 1 ⋅ 11 , 1 ⋅ 7 − 2 ⋅ (−2)⟩
= ⟨22 − 21 , −6 − 11 , 7 + 4⟩
= ⟨1, −17, 11⟩
Remark: The solution in this example was found by simply using Formula (2) to compute the
components of the cross product. Formula (2) is somewhat difficult to remember (although not
impossible), but there is a standard computational tool that one may use to compute the cross
product based on the idea of a determinant of a 3 × 3 matrix.
Determinant rule for computing cross products
The value of the cross product of vectors A⃗ = ⟨a1 , a2 , a3 ⟩ and B⃗ = ⟨b1 , b2 , b3 ⟩ can be computed by formally expanding the following 3 × 3 “determinant”:
⎛ iˆ jˆ k̂ ⎞
a a
a a
a a
⃗
⃗
A× B = det ⎜ a1 a2 a3 ⎟ = det ( 2 3 )⋅ iˆ−det ( 1 3 )⋅ jˆ+det ( 1 2 )⋅ k̂ (3)
b2 b3
b1 b3
b1 b2
⎝ b1 b2 b3 ⎠
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
We recall that a 3 × 3 determinant is computed in terms of three 2 × 2 determinants, so we review
both.
Definition 2. Determinant
The determinant of the 2 by 2 matrix (
det (
a b
) is the number
c d
a b
) = ad − bc
c d
(4)
⎛ a b c ⎞
The determinant of the 3 by 3 matrix ⎜ d e f ⎟ is the number, computed recursively using
⎝ g h i ⎠
(4), given by
⎛ a b c ⎞
e f
d
) − b det (
det ⎜ d e f ⎟ = a det (
h i
g
⎝ g h i ⎠
f
d e
) + c det (
)
i
g h
(5)
We use this new tool to redo the Example 1.
Example 2. Compute the cross product of vector u⃗ = ⟨1, 2, 3⟩ with vector v⃗ = ⟨−2, 7, 11⟩
Solution: Using the expansion formulas (4) and (5) we find
⎛
u⃗ × v⃗ = det ⎜
⎝
iˆ jˆ k̂ ⎞
1 2 3 ⎟
−2 7 11 ⎠
2 3
1 3
1 2
) ⋅ iˆ − det (
) ⋅ jˆ + det (
) ⋅ k̂
7 11
−2 11
−2 7
= 1 ⋅ iˆ − 17 ⋅ jˆ + 11 ⋅ k̂
= ⟨1, −17, 11⟩
= det (
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4.1
1.4. THE CROSS PRODUCT
Properties of the Cross Product
● The cross product is anti-symmetric:
A⃗ × B⃗ = −B⃗ × A⃗
(6)
● The cross product satisfies the Jacobi Identity:
⃗ + C⃗ × (A⃗ × B)
⃗ + B⃗ × (C⃗ × A)
⃗ = 0⃗
A⃗ × (B⃗ × C)
(7)
● The cross product also satisfies the following associative and distributive rules: Let α be any real
⃗ B⃗ and C⃗ be any three vectors. Then:
number and A,
⃗ = A⃗ × B⃗ + A⃗ × C⃗
A⃗ × (B⃗ + C)
⃗ × C⃗ = A⃗ × C⃗ + B⃗ × C⃗
(A⃗ + B)
⃗ × B⃗ = A⃗ × (α B)
⃗ = α(A⃗ × B)
⃗
(α A)
(8)
(9)
(10)
●The cross product can tell us when 2 vectors are parallel.
Theorem 2. Two non-zero vectors A⃗ and B⃗ are parallel if and only if their cross product is
the zero vector:
A⃗ × B⃗ = 0⃗
Proof. Assuming first that the two vectors A⃗ and B⃗ are parallel, we know that the angle between
vectors A⃗ and B⃗ is θ = 0 or θ = π, and sin(0) = sin(π) = 0. So if vectors A⃗ and B⃗ are parallel, then
⃗ = ∥A⃗ ∥ ∥B⃗ ∥ sin(θ) = 0 Ô⇒ A⃗ × B⃗ = 0⃗
∥A⃗ × B∥
⃗
the last equality following from the fact that the only vector with magnitude 0 is 0.
⃗ = 0. Since vectors A⃗ and B⃗ are
Conversely, suppose that A⃗ × B⃗ = 0⃗ which implies that ∥A⃗ × B∥
⃗
⃗
assumed non-zero, we know that ∥A ∥ ≠ 0 and ∥B ∥ ≠ 0. Then using Formula (1) we have
⃗ = ∥A⃗ ∥ ∥B⃗ ∥ sin(θ) Ô⇒ sin(θ) = 0
0 = ∥A⃗ × B∥
Since θ must lie in the range 0 ≤ θ ≤ π
Figure 3: Graph of sin θ
we see from Figure 3 that sin(θ) = 0 Ô⇒ θ = 0 or θ = π and hence vectors A⃗ and B⃗ are parallel.
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
⃗ ∥A⃗ × B∥
⃗ = ∥A⃗ ∥ ∥B⃗ ∥ sin(θ), gives
Lemma 1. The magnitude of the cross product of A⃗ and B,
the area of the parallelogram spanned by the vectors A⃗ and B⃗ .
To see why this is true consider the diagrams below showing vectors A⃗ and B⃗ on the left and the
parallelogram spanned by these two vectors on the right.
Figure 4
Figure 5
To compute the area of the parallelogram we recall the technique illustrated in the following two
diagrams: we cut off a right triangle on the right hand side of the parallelogram and glue it back
⃗ and height ∥B∥
⃗ sin θ, where θ is the angle
on on the left side to construct a rectangle of length ∥A∥
⃗
⃗
between A and B.
Figure 7
Figure 6
Then since the area of the parallelogram is the same as the area of the new rectangle, we have
area = base × height
= ∥A⃗ ∥ ⋅ (∥B⃗ ∥ sin θ)
= ∥A⃗ ∥∥B⃗ ∥ sin θ
⃗
= ∥A⃗ × B∥
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4.2
1.4. THE CROSS PRODUCT
Triple Products
The Scalar Triple Product
⃗ The absolute value of this particular combiWe first consider the scalar triple product A⃗ ⋅ (B⃗ × C).
⃗ B⃗
nation of dot and cross products yields the volume of the parallelepiped spanned by vectors A,
⃗
and C as shown in Figure 8.
Figure 8
⃗ The volume
From above we know that the area of the base of the parallelepiped is ∥B⃗ × C∥.
of the parallelepiped is the same as the volume of the rectangular solid formed by cutting off the
wedge-shaped portion on the right side of the parallelepiped and gluing it back on the left side as
shown in Figure 9 below.
Figure 9
The volume of the resulting rectangular solid is then given by the product of the area of the base
times the height h. The value of h is given by
h = ∥A⃗ ∥ ⋅ ∣ cos θ∣
(11)
⃗ (See Figure 10.) We need ∣ cos θ∣ because h is non-negative.
where θ is the angle between A⃗ and B⃗ × C.
Hence the volume of the parallelepiped is
⃗
Volume = ∥A⃗ ∥ ⋅ ∣ cos θ∣ ⋅ ∥B⃗ × C∥
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Figure 10
⃗ we may conclude from the
However since the angle θ is the angle between vectors A⃗ and B⃗ × C,
alternative dot product formula in Section 3 that
⃗
Volume = ∣A⃗ ⋅ (B⃗ × C)∣
(12)
⃗ is needed since volume must be non-negative.
where the absolute value of A⃗ ⋅ (B⃗ × C)
The Vector Triple Product
⃗ occurs in many applications in engineering physics, and the
The triple vector product A⃗ × (B⃗ × C)
following identity is often useful in such cases.
⃗ B⃗ and C⃗ be any three vectors in V3 . Then
Lemma 2. Let A,
⃗ = (A⃗ ⋅ C)
⃗ B⃗ − (A⃗ ⋅ B)
⃗ C⃗
A⃗ × (B⃗ × C)
(13)
The proof of this identity is a tedious but straightforward exercise using the definitions of the dot
and cross products and is left as an exercise for the student.
Example 3. Show that
ˆ k̂
iˆ × ( iˆ × k̂) = ( iˆ ⋅ k̂) iˆ − ( iˆ ⋅ i)
Solution: Using the right-hand rule for cross produces yields
iˆ × k̂ = − jˆ
ˆ = − k̂
iˆ × ( iˆ × k̂) = iˆ × (− j)
ˆ k̂ = 0 ⋅ iˆ − 1 ⋅ k̂ = − k̂
( iˆ ⋅ k̂) iˆ − ( iˆ ⋅ i)
Since both sides of the equation reduce to − k̂, they are equal.
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4.3
1.4. THE CROSS PRODUCT
Torque
Consider a force F⃗ acting on a rigid arm as represented by the vector ⃗r in Figure 11 below.
Figure 11
The arm is pinned at the point P and is free to rotate in a plane about P. In this particular case vectors
⃗r and F⃗ lie in the same plane. For example, one can think of ⃗r as representing a torque wrench as in
Figure 12 with the force F⃗ being applied by the operator’s hand.
Figure 12
We wish to compute the effect of the force F⃗ on the motion of our rigid arm ⃗r . To do so, we decompose
⃗
F⃗ with respect to ⃗r as illustrated in Figure 13. Note that θ is the angle between ⃗r and F.
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Figure 13
From the figure we see that only the component F⃗⊥ (with magnitude ∥F⃗ ∥ sin θ) affects the rotation
of ⃗r about P. In Newtonian physics, the vector quantity associated to rotational acceleration is called
torque.
Definition 3. (Torque) The torque τ⃗ produced by a force F⃗ acting on a rigid arm ⃗r is
τ⃗ = ⃗r × F⃗
(14)
∥τ⃗ ∥ = ∥⃗r ∥ ⋅ ∥F⃗ ∥ sin(θ)
(15)
with magnitude
One may interpret the Formula (15) in two ways. The first way groups sin θ with ∥F⃗ ∥ so that one
sees explicitly that only the component F⃗⊥ contributes to the torque:
∥τ⃗ ∥ = ∥⃗r ∥ (∥F⃗ ∥ sin θ) = ∥⃗r ∥ ⋅ ∥F⃗⊥ ∥
(16)
Another point of view introduces the concept of the lever arm whose length is the perpendicular
distance from the axis of rotation to the line of action of the force. Let L⃗ be the vector from P to the
⃗ Then ∥L⃗ ∥ is the length of the lever arm and
line of action of the force that is orthogonal to F.
L⃗ = ∥⃗r ∥ sin θ L̂
⃗
where L̂ is the unit vector in the direction of L.
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4. THE CROSS PRODUCT
Figure 14
In this context Formula (15) can be interpreted as saying that the magnitude of the torque is equal to
the magnitude of the force times the length of the lever arm:
∥τ⃗ ∥ = (∥⃗r ∥ sin(θ)) ∥F⃗ ∥ = ∥L⃗ ∥ ⋅ ∥F⃗ ∥
(17)
The magnitude of τ⃗ is the product of a length and a force so the units in the English system is
ft-lb and in the SI system is N-m or joules, j.
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CHAPTER 1. EUCLIDEAN GEOMETRY IN 3-DIMENSIONAL SPACE
1.4.4
1.4. THE CROSS PRODUCT
Exercises
Exercise # 1. Find two unit vectors orthogonal to both a⃗ = ⟨2, 0, −3⟩ and b⃗ = ⟨−1, 4, 2⟩.
Exercise # 2. Find the area of the parallelogram with vertices A(-2,1), B(0,4), C(4,2) and D(2,-1).
Exercise # 3. A torque wrench grips a bolt at the origin of coordinates, with the free end of the torque
wrench located at the point (L, 0, 0). The force on this end of the torque wrench points in the direction
of V⃗ = ⟨10, 0, 16⟩. Note that ⃗r = L⃗ in this problem. Compute the length of the lever arm ∥L⃗ ∥ needed to
produce a torque with magnitude 200 ft-lb if the magnitude of the force is 100 pounds.
Exercise # 4. We have seen the property that A⃗ × B⃗ is simultaneously orthogonal to A⃗ and B⃗ is very
important. Keeping this in mind, why is th cross product left undefined in R2 ?
1.4.5
Solutions to Exercises
⃗ (This
Solution 1. We use the fact that the cross product a⃗ × b⃗ is orthogonal to both vectors a⃗ and b.
is one of the main uses of the cross product!!!) So we first compute a⃗ × b⃗ and then make it into a unit
vector.
⎛ iˆ jˆ k̂ ⎞
⃗
a⃗ × b = det ⎜ 2 0 −3 ⎟ = [12, −1, 8]
⎝ −1 4 2 ⎠
1
c⃗. Computing
Now if c⃗ is any non-zero vector then one may create two unit vectors from c⃗, namely ±
∥c⃗∥
√
√
⃗ = 144 + 1 + 64 = 209. Hence two unit vectors orthogonal to
the magnitude of a⃗ × b⃗ we find ∥a⃗ × b∥
both a⃗ = [2, 0, −3] and b⃗ = [−1, 4, 2] are given by
1
±√
[12, −1, 8]
209
Solution 2. We use the fact (see top of page 660) that the area of a parallelogram spanned by (i.e.
⃗ Sketching the 4 points in the plane it is clear that
determined by) two vectors a⃗ and b⃗ is given by ∣ a⃗ × b∣.
⃗ = [2, 3, 0] and AD
⃗ = [4, −2, 0]. (There are other choices
the parallelogram is spanned by the vectors AB
as well for spanning vectors!!) Hence
⎛
⃗ × AD
⃗ = det ⎜
AB
⎝
iˆ jˆ k̂ ⎞
2 3 0 ⎟ = [0, 0, −16] Ô⇒ area = 16
4 −2 0 ⎠
Solution 3. Since this problem required us to find the lever arm, we use formula (17). Hence
200 = ∥F⃗ ∥ ⋅ ∥L⃗ ∥ = 100 ⋅ L Ô⇒ L = 2 feet
Calculus I
Last update: August 24, 2015
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©2014-16 J. E. Franke, J. R. Griggs and L. K. Norris