This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Dot and Cross Products
John E. Gilbert, Heather Van Ligten, and Benni Goetz
Dot Product: so far we’ve added vectors, subtracted them, and multiplied by a scalar, but now
it’s time to ‘multiply’ two vectors. There are two diﬀerent products, one producing a scalar, the other a
vector. Both, however, have important applications to geometry as well as physics and engineering.
The angle between vectors u and v is the angle θ
v shown in the ﬁgure to the right by ﬁrst arranging them
so that they have the same tail. Notice that there are
really two choices of θ , one smaller than π , the other
larger than π (unless both equal π ). By convention the θ smaller one is always chosen, so that 0 ≤ θ ≤ π .
Notice that the angle between any two of the unit
coordinate vectors i, j, and k is 1
π
2 u because they are mutually perpendicular. Deﬁnition: The Dot Product u · v of vectors u and v is the
scalar deﬁned by
u · v = uv cos θ
where θ is the angle between u and v. Since cos( 1 π ) = 0,
2
vectors u, v are perpendicular when u · v = 0 and u, v = 0.
A number of properties follow immediately from this deﬁnition and the perpendicular vectors i , j , k: Properties:
1.
2.
3. u· v = v · u, u · u = u2 , u · (v + w ) = u · v + u · w ,
i ·j = 0, j· k = 0, k· i = 0. The previously listed properties provide a convenient algebraic way of computing the dot product of
vectors
u= u1 , u2 , u3 = u1 i + u2 j + u3 k, v= v1, v2 , v3 = v1 i + v2 j + v3 k . For by expanding using also Properties 1, 2, and 3, we get
u · v = (u1 i + u2 j + u3 k) · (v1 i + v2 j + v3 k) = u1 v1 + u2 v2 + u3 v3 .
Example 1: determine the dot product of a · b = a1 b1 + a2 b2 + a3 b3 . the vectors
So when
a= −3, 2, −3 , b= 2, 1, −3 .
a= Solution : the dot product, a · b, of vectors
a= a1 , , a2 , a3 , b= b1 , b2 , b3 −3, 2, −3 , b= 2, 1, −3 we see that
a · b = (−3)(2) + (2)(1) + (−3)(−3) = 5 . is given by Example 2: determine the dot product of
the vectors a, b when a · b = a b cos θ
where θ is the angle between them. When a = 4, b = 5 and the angle between a, b is π/3.
Solution : the dot product is deﬁned in
coordinatefree form by a = 4 , b = 5 and θ = π/3, therefore,
a · b = 20 cos π
= 10 .
3 Projections, Components: the geometric deﬁnition of dot product helps us express the projection
of one vector onto another as well as the component of one vector in the direction of another.
The dashed red vector to the right in the direction of v is v called the projection, projv (u), of u onto v. Thus
projv (u) = u cos θ v
v  = cos
u
 u·v
v.
v 2 θ θ The length of this vector is called the component of u in
the direction of v and is given in terms of a dot product by u u·v
.
compv (u) = u cos θ =
v 
Important Special case: for any vector v = a, b, c ,
v = (v · i) i + (v · j) j + (v · k) k = a i + b j + c k . −→
−
−
→
Determine the projection of AD onto AB . Example 3: the box shown in Solution : since the unit cube has sidelength z = 1, A A = (0, 0, 1) , B = (1, 0, 0) , D = (0, 1, 0) .
So if
O Dy B
x
is the unit cube having one corner at the origin O
and the coordinate planes for three of its faces . −→
−
u = AD = 0, 1, −1 , and −
→
v = AB = 1, 0, −1 ,
−→
−
−
→
then the projection of AD on AB is the vector
projv (u) = 1
u·v
v=
1, 0, −1 .
2
v 
2 These are extremely important ideas  you’ll meet them again when you learn about Fourier
coeﬃcients and Fourier series, for instance, where trig functions (harmonics) play the role of the unit
coordinate vectors. Also, the coeﬃcient f (n) (c)/n! in a Taylor series is a ‘component’ of a a vector
f (x) in the direction of of the basis function xn . Cross Product: now we want ‘multiplication’ of vectors to produce a vector, u × v, not a scalar.
Such a vector product occurs many times in geometry as well as in engineering and physics.
Because a vector has direction, a convention has to be
adopted. If a, b are vectors arranged so that they have the
same tail, then vectors {a, b, c} are said to form a c righthanded system when c is perpendicular to the plane b containing a, b and points in the direction shown to the
right. Since there could be two directions for c to point and
still be perpendicular to the plane containing a and b, the a right hand convention amounts to specifying which direction
we’ll choose. Notice that {i, j, k} is a righthanded system. Question: for a righthanded system {a, b, c}, Answer: Switching a, b reverses direction of c. (i) is {b, a, c} a righthanded system? (i) NO, {b, a, c} is not righthanded (ii) is {b, a, −c} a righthanded system? (ii) YES, {b, a, −c} is righthanded. Deﬁnition: The Cross Product, u × v, of vectors u and v is the vector deﬁned
by
u × v = (uv sin θ)n
where θ is the angle between u and v, and n is the unit vector such that {u, v, n}
forms a righthanded system. Since sin 0 = sin π = 0, vectors u, v are parallel
when u × v = 0 and u, v = 0.
A number of properties follow immediately from this deﬁnition and the fact that {i, j, k} is a
righthanded system: Properties:
1.
2.
3. u × v = −v × u, u × u = 0, u × v = uv sin θ, u × (v + w ) = u × v + u × w ,
i ×j = k, j ×k = i, k ×i = j. Computing Cross Products: the previously listed properties provide good algebraic ways of
computing the cross product of vectors
a= a1 , a2 , a3 = a1 i + a2 j + a3 k , b= b1 , b2 , b3 = b1 i + b2 j + b3 k . For by Properties 1, 2, and 3 on the previous slide,
a × b = (a2 b3 − a3 b2 ) i − (a1 b3 − a3 b1 ) j + (a1 b2 − a2 b1 ) k ,
which, in turn, may (or may not!) look like something to do with determinants.
Recall that for a 2 × 2 determinant,
ab = ad − bc , cd
while for a 3 × 3 determinant,
XY Z
=X a1 a2 a3
b1 b2 b3 a2 a3 b2 b3 −Y a1 a3
b1 b3 +Z a1 a2
b1 b2 . This says that
a×b = a2 a3
b2 b3 i− a1 a3 a1 a2 j+ b1 b3 b1 b2 k, which means that the cross product can be written algebraically as the 3 × 3 determinant
i
a×b = j k a1 a2 a3 . b1 b2 b3 The two products of vectors can be combined: the (scalar) triple product of vectors a, b and c is a · (b × c) = a1 b2 b3
c2 c3 − a2 b1 b3
c1 c3 + a3 b1 b2
c1 c2 a1 a2 a3
= b1 b2 b3
c1 c2 c3 There’s also a vector triple product a × (b × c), but we won’t have need of it. . Example 4: determine all unit vectors v Now orthogonal to i a = 3i+ j + 4k, a×b = b = 3i+ 2j+ 6k. jk 314
326 Solution : the nonzero vectors orthogonal to
= a, b are all of the form
v = λ(a × b), λ=0 26 i− 34
36 j+ a × b = −2 i − 6 j + 3 k, 32 k. a × b = 7 . Thus the unit vectors orthogonal to a, b are a×b
v=±
.
a × b v=± 63
2
.
i+ −
7
77 Applications of Cross Products:
1. When vectors u, v are adjacent sides of a
parallelogram as shown to the right, the height of the
parallelogram is v sin θ. Now the parallelogram has
v sin θ v
area = base x height = uv sin θ = u × v .
θ u
u So the length of the cross product of u and v is the
area of the parallelogram these vectors span.
2. The vectors a, b and c shown respectively in blue,
red and green to the right form adjacent edges of a
parallelepiped. Now by 1., the base has area = a × b, a×b while its
height = compa×b (c) = 31 So with λ a scalar. This means the only unit
vectors orthogonal to a, b are 14 c c · (a × b)
.
a × b Thus the parallelepiped has b
θ
a volume = (area of base) x height = c · (a × b) . 3. When we push down on a bike pedal with a foot as
shown to the right, we produce a turning eﬀect. The
F torque, τ , is deﬁned to be the cross product θ τ = r×F
of the position vector r and the force F. The direction r of τ is the axis of rotation (the crankshaft). Example 5: compute the volume of the
parallelopiped with adjacent edges OP , OQ and So its volume is given by scalar triple
product OR determined by vertices 1 −1 −2
a · (b × c ) = 1 1 −1 1 P (1, −1, −2), Q(1, 1, 1), R(1, 3, 4), 3 4 where O is the origin.
Solution: the parallelopiped is determined by
−
→
a = OP = 1, −1, −2 ,
−
→
b = OP = 1, 1, −1 ,
−
→
c = OP = 1, 3, 4 . = 1 −1
3 4 − (−1) 1 −1
1 4 −2 Consequently, the parallelepiped has
volume = 8 units . 11
13 . ...
View
Full
Document
 Spring '11
 Gilbert
 Vectors, Scalar, Dot Product

Click to edit the document details