Math 152: 11.3 The Cross Product
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Warm-Up: Expand a 1ˆ
i + a 2 ĵ + a 3k̂ ⋅ b1ˆi + b2 ĵ + b 3k̂ .
Recall that the Dot Product of two vectors is a way of multiplying two vectors (either
in polar or rectangular component form) such that the product is a scalar. A typical
example of a physical quantity that results from a dot product of two vectors (force
and displacement) is the scalar quantity work, W, measured in joules.
Two vectors can also be multiplied in such a way as to yield another vector as the
product. A typical example of a physical quantity that results from a Cross Product
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of two vectors is torque, τ , measured in newton-metres, N ⋅ m .
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Definition: The cross product a × b = a b sinθ n̂ is the product of two vectors a
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and b that is itself a vector, perpendicular to both a and b . The magnitude of a × b
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is a b sinθ , where θ is the angle measured FROM a TO b (direction matters!
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Why?). The direction of a × b is that of the vector n̂ , the unit normal
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(perpendicular) vector to both a and b as determined by the right hand rule.
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Right Hand Rule: Let the fingers of your right hand curl in the direction from a to
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b . Then, the direction that your thumb points in is the direction of the cross
product.
Torque:
The vector quantity torque is the twisting or turning action (a vector) that is
produced when a force is applied to a radial arm (or radius), and has both magnitude
and direction. Common examples are seesaws, doors and wrenches, which rely on
rotation to magnify a small force to produce a large twisting motion (or torque!).
There must be a pivot point or axis of rotation or fulcrum.
© Raelene Dufresne 2012
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Math 152: 11.3 The Cross Product
Joules versus newton-metre:
The unit of newton-metre is dimensionally equivalent to that of joules, however the
different name reinforces the different vector product nature of the quantity each
unit measures, and according to the SI authority, are NOT interchangeable.
The distinction in using joules versus newton-metre illustrates the geometric
distinction between the scalar DOT product and the vector CROSS product: consider
that the displacement part of a joule in calculating work is parallel to the force
applied, whereas the displacement part of one newton-metre in calculating torque is
perpendicular to the force applied.
Exercise 1: A wrench 0.5 meters long is applied to a nut with a force of 80 N.
Because of limited space, the force must be exerted straight upwards. How much
torque is applied to the nut, and in which direction (into or out of the page) is the
torque applied?
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Derivation of the Cross Product: Let a = ⎡a 1, a 2, a 3 ⎤ and b = ⎡b1, b2, b 3 ⎤ , and let
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c = a × b = c 1, c 2, c 3 ⎤ such that (1) c is ⊥ to a , (2) c is ⊥ to b and (3) c = ab sinθ ,
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where θ is the angle measured FROM a TO b (direction matters … why?).
Exercise 2: Set up the equations required to be solved for c 1 , c 2 and c 3 given the
requirements of the definition of the cross product.
Exercise 3: Here is a simpler algebraic approach to the Cross Product Derivation.
a) Show, if c 3 = 1 , that c 2 =
a 1b 3 − a 3b1
and c 1 =
a 3b2 − a 2b 3
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a 2b1 − a 1b2
a 2b1 − a 1b2
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b) Determine k such that a × b = ⎡a 2b 3 − a 3b2, a 3b1 − a 1b 3, a 1b2 − a 2b1 ⎤ .
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Exercise 4: Calculate all of the cross products of the 3 standard basis vectors in  3 .
© Raelene Dufresne 2012
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Math 152: 11.3 The Cross Product
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Exercise 5: Show that a × b = ⎡a 2b 3 − a 3b2, a 3b1 − a 1b 3, a 1b2 − a 2b1 ⎤ by
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a) expanding a ˆ
i + a ĵ + a k̂ × b ˆi + b ĵ + b k̂ .
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b) calculating the determinant
ĵ
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k̂
a1 a2 a3 .
b1 b2 b 3
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i + 2ĵ − k̂ and b = 3ˆi − ĵ + 7k̂ .
Exercise 6: Determine a × b given a = ˆ
Properties of the Cross Product – Be able to explain/prove each one!
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1. a × b is orthogonal to both a and b .
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2. If either a or b is 0 , then a × b = 0 .
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3. Two nonzero vectors are parallel if and only if a × b = 0 .
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4. a × b = −b × a (Note: The cross product is NOT commutative: a × b ≠ b × a .)
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5. Commutativity of scalar in cross product: ca × b = c a × b = a × cb
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6. Left cross product: a × b + c = a × b + a × c
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7. Right cross product: a + b × c = a × c + b × c
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8. The cross product is NOT associative: a × b × c ≠ a × b × c .
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9. Geometric interpretation: The magnitude of a cross product a × b is the area of a
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parallelogram given by a and b .
© Raelene Dufresne 2012
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Math 152: 11.3 The Cross Product
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10. Scalar Triple Product: a ⋅ b × c
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a) The volume of a parallelepiped described by the vectors a, b, and c
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is given by the magnitude of a ⋅ b × c .
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b) If a ⋅ b × c = 0 , the three vectors a, b, and c are coplanar (lie in the same
plane).
c) Can be calculated as the determinant
a1 a2 a3
b1 b2 b 3 .
c1 c2 c3
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11. Vector Triple Product: a × b × c will be seen in Math 251 and physics.
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Exercise 7: Given the points P 1, 0, −1 , Q 2, 4,5 and R 3,1, 7 , determine
a) a vector orthogonal to the plane that contains these points.
b) the area of ΔPQR .
Exercise 8: Use the scalar triple product to determine the volume of the
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parallelepiped whose corner is formed by the vectors a = ⎡1, −1, 0 ⎤ , b = ⎡2, 0,3⎤ and
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c = 2,3, −2⎤ .
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© Raelene Dufresne 2012
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