DotCrossProductText(1) - Dot and Cross Products John E....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = |u||v| 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 = |u|2 , 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 coordinate-free 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 side-length 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 right-handed 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 right-handed system. Question: for a right-handed system {a, b, c}, Answer: Switching a, b reverses direction of c. (i) is {b, a, c} a right-handed system? (i) NO, {b, a, c} is not right-handed (ii) is {b, a, −c} a right-handed system? (ii) YES, {b, a, −c} is right-handed. Definition: The Cross Product, u × v, of vectors u and v is the vector defined by u × v = (|u||v| 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 right-handed 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 right-handed system: Properties: 1. 2. 3. u × v = −v × u, u × u = 0, |u × v| = |u||v| 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 non-zero 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 = |u||v| 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

Ask a homework question - tutors are online