Vector Dot Product

9 Key excerpts on "Vector Dot Product"

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  University Physics Volume 1

  • William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
  • 2016(Publication Date)
  • Openstax

    (Publisher)

  There are two kinds of products of vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors. Chapter 2 | Vectors 77 Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used to define work and energy relations. For example, the work that a force (a vector) performs on an object while causing its displacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kind of multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, as its name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations, a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot to force (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar product is a scalar quantity and a vector product is a vector quantity. The Scalar Product of Two Vectors (the Dot Product) Scalar multiplication of two vectors yields a scalar product. Scalar Product (Dot Product) The scalar product A → · B → of two vectors A → and B → is a number defined by the equation (2.27) A → · B → = AB cos φ, where φ is the angle between the vectors (shown in Figure 2.27). The scalar product is also called the dot product because of the dot notation that indicates it. In the definition of the dot product, the direction of angle φ does not matter, and φ can be measured from either of the two vectors to the other because cos φ = cos (−φ) = cos (2π − φ) . The dot product is a negative number when 90° < φ ≤ 180° and is a positive number when 0° ≤ φ < 90° . Moreover, the dot product of two parallel vectors is A → · B → = AB cos 0° = AB , and the dot product of two antiparallel vectors is A → · B → = AB cos 180° = −AB .

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  A Student's Guide to Vectors and Tensors

  • Daniel A. Fleisch(Author)
  • 2011(Publication Date)
  • Cambridge University Press

    (Publisher)

  2 Vector operations If you were tracking the main ideas of Chapter 1 , you should realize that vectors are representations of physical quantities – they’re mathematical tools that help you visualize and describe a physical situation. In this chapter, you can read about a variety of ways to use those tools to solve problems. You’ve already seen how to add vectors and how to multiply vectors by a scalar (and why such operations are useful); this chapter contains many other “vector oper-ations” through which you can combine and manipulate vectors. Some of these operations are simple and some are more complex, but each will prove useful in solving problems in physics and engineering. The first section of this chapter explains the simplest form of vector multiplication: the scalar product. 2.1 Scalar product Why is it worth your time to understand the form of vector multiplication called the scalar or “dot” product? For one thing, forming the dot product between two vectors is very useful when you’re trying to find the projection of one vector onto another. And why might you want to do that? Well, you may be interested in knowing how much work is done by a force acting on an object. The first instinct of many students is to think of work as “force times distance” (which is a reasonable starting point). But if you’ve ever taken a course that went a bit deeper than the introductory level, you may remember that the definition of work as force times distance applies only to the special case in which the force points in exactly the same direction as the displacement of the object. In the more general case in which the force acts at some angle to the direction of the displacement, you have to find the component of the force along the displacement. That’s one example of exactly what the dot product can do for you, and you’ll find more in the problems at the end of this chapter. 25

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  Calculus: Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  The question arises: is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, which we now define. Another is the cross product, which is discussed in the next section. ■ The Dot Product of Two Vectors To find the dot product of vectors a and b we multiply corresponding components and add. 1 Definition of the Dot Product If a - k a 1 , a 2 , a 3 l and b - k b 1 , b 2 , b 3 l, then the dot product of a and b is the number a  b given by a  b - a 1 b 1 1 a 2 b 2 1 a 3 b 3 The dot product of two vectors is a real number, not a vector. For this reason, the dot product is sometimes called the scalar product (or inner product). Although Defini- tion 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors is defined in a similar fashion: k a 1 , a 2 l  k b 1 , b 2 l - a 1 b 1 1 a 2 b 2 EXAMPLE 1 k2, 4l  k3, 21l - 2s3d 1 4s21d - 2 k21, 7, 4l  k 6, 2, 2 1 2 l - s21ds6d 1 7s2d 1 4(2 1 2 ) - 6 si 1 2 j 2 3 kd  s2 j 2 kd - 1s0d 1 2s2d 1 s23ds21d - 7 ■ 12.3 2. By interpreting dyydx as the slope of a tangent at sx, yd, show that dy dx - s s xd a where a - | T0 | yst d, a constant. 3. Differentiate both sides of the differential equation in Problem 2 and use Equation 8.1.6 to obtain the second-order differential equation d 2 y dx 2 - 1 a Î1 1 S dy dx D 2 with initial conditions ys0d - 0 (the curve passes through the origin) and y 9s0d - 0 (the tangent at the origin is horizontal). Solve this equation by first substituting z - dyydx and then solving the resulting first-order differential equation. Conclude that the equation of the curve is y - a cosh x a 2 a 4. Graph y - a coshs xyad 2 a for a - 1 2 , a - 1, and a - 3. How does the value of a affect the shape of the curve? ; Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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  Calculus

  Multivariable

  • William G. McCallum, Deborah Hughes-Hallett, Daniel E. Flath, Andrew M. Gleason, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, Guadalupe I. Lozano, Jerry Morris, David Mumford, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Ayse Arzu Sahin, Adam H. Spiegler, Jeff Tecosky-Feldman, Thomas W. Tucker, Aaron(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  Definition of the Dot Product The dot product links geometry and algebra. We already know how to calculate the length of a vector from its components; the dot product gives us a way of computing the angle between two vectors. For any two vectors   =  1   +  2   +  3   and   =  1   +  2   +  3   , shown in Figure 13.26, we define a scalar as follows: The following two definitions of the dot product, or scalar product,   ⋅   , are equivalent: • Geometric definition   ⋅   = ‖   ‖‖   ‖ cos  where  is the angle between   and   and 0 ≤  ≤ . • Algebraic definition   ⋅   =  1  1 +  2  2 +  3  3 . Notice that the dot product of two vectors is a number, not a vector. Why don’t we give just one definition of   ⋅   ? The reason is that both definitions are equally important; the geometric definition gives us a picture of what the dot product means and the algebraic definition gives us a way of calculating it. How do we know the two definitions are equivalent—that is, they really do define the same thing? First, we observe that the two definitions give the same result in a particular example. Then we show why they are equivalent in general.      Figure 13.26: The vectors   and   1 2 2        Figure 13.27: Calculating the dot product of the vectors  =   and   = 2   + 2   geometrically and algebraically gives the same result Example 1 Suppose   =   and   = 2   + 2   . Compute   ⋅   both geometrically and algebraically. Solution To use the geometric definition, see Figure 13.27. The angle between the vectors is ∕4, or 45 ◦ , and the lengths of the vectors are given by ‖   ‖ = 1 and ‖   ‖ = 2 √ 2. Thus,   ⋅   = ‖   ‖‖   ‖ cos  = 1 ⋅ 2 √ 2 cos (  4 ) = 2. Using the algebraic definition, we get the same result:   ⋅   = 1 ⋅ 2 + 0 ⋅ 2 = 2.

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  Precalculus: Mathematics for Calculus, International Metric Edition

  • James Stewart, Lothar Redlin, Saleem Watson(Authors)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  CHAPTER 9 ■ Review 671 The Dot Product of Vectors (p. 640) If u  8 a 1 , a 2 9 and v  8 b 1 , b 2 9 , then their dot product is u # v  a 1 b 1  a 2 b 2 If u is the angle between u and v, then u # v  0 u 0 0 v 0 cos u The angle u between u and v satisfies cos u  u # v 0 u 0 0 v 0 The vectors u and v are perpendicular if and only if u # v  0 The component of u along v (a scalar) and the projection of u onto v (a vector) are given by comp v u  u # v 0 v 0 proj v u  a u # v 0 v 0 2 b v ¨ v u v u ¨ proj v u comp v u The work W done by a force F in moving along a vector D is W  F # D Three-Dimensional Coordinate Geometry (p. 648) A coordinate system in space consists of a fixed point O (the origin) and three directed lines through O that are perpendicular to each other, called the coordinate axes and labeled the x-axis, y-axis, and z-axis. The coordinates of a point P1 a, b, c 2 determine its location relative to the coordinate axes. O b a c P(a, b, c) y x z The distance between the points P1 x 1 , y 1 , z 1 2 and Q1 x 2 , y 2 , z 2 2 is given by the Distance Formula: d1 P, Q 2  "1 x 2  x 1 2 2  1 y 2  y 1 2 2  1 z 2  z 1 2 2 The equation of a sphere with center C1 h, k, l 2 and radius r is 1 x  h 2 2  1 y  k 2 2  1 z  l 2 2  r 2 Vectors in Three Dimensions (p. 653) A vector in space is a line segment with a direction. We sketch a vector as an arrow to indicate the direction.

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  Mathematical Methods for Physicists

  A Concise Introduction

  • Tai L. Chow(Author)
  • 2000(Publication Date)
  • Cambridge University Press

    (Publisher)

  Note that division by a vector is not defined: expressions such as k = A or B = A are meaningless. There are several ways of multiplying two vectors, each of which has a special meaning; two types are defined. The scalar product The scalar (dot or inner) product of two vectors A and B is a real number defined (in geometrical language) as the product of their magnitude and the cosine of the (smaller) angle between them (Figure 1.6): A B AB cos 0 † : 1 : 4 † It is clear from the definition (1.4) that the scalar product is commutative: A B ˆ B A ; 1 : 5 † and the product of a vector with itself gives the square of the dot product of the vector: A A ˆ A 2 : 1 : 6 † If A B ˆ 0 and neither A nor B is a null (zero) vector, then A is perpendicular to B . 5 THE SCALAR PRODUCT Figure 1.6. The scalar product of two vectors. We can get a simple geometric interpretation of the dot product from an inspection of Fig. 1.6: B cos † A ˆ projection of B onto A multiplied by the magnitude of A ; A cos † B ˆ projection of A onto B multiplied by the magnitude of B : If only the components of A and B are known, then it would not be practical to calculate A B from definition (1.4). But, in this case, we can calculate A B in terms of the components: A B ˆ A 1 ^ e 1 ‡ A 2 ^ e 2 ‡ A 3 ^ e 3 † B 1 ^ e 1 ‡ B 2 ^ e 2 ‡ B 3 ^ e 3 † ; 1 : 7 † the right hand side has nine terms, all involving the product ^ e i ^ e j . Fortunately, the angle between each pair of unit vectors is 90 8 , and from (1.4) and (1.6) we find that ^ e i ^ e j ˆ ij ; i ; j ˆ 1 ; 2 ; 3 ; 1 : 8 † where ij is the Kronecker delta symbol ij ˆ 0 ; if i 6 ˆ j ; 1 ; if i ˆ j : ( 1 : 9 † After we use (1.8) to simplify the resulting nine terms on the right-side of (7), we obtain A B ˆ A 1 B 1 ‡ A 2 B 2 ‡ A 3 B 3 ˆ X 3 i ˆ 1 A i B i : 1 : 10 † The law of cosines for plane triangles can be easily proved with the application of the scalar product: refer to Fig.

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  What Every Engineer Should Know about MATLAB® and Simulink®

  • Adrian B. Biran(Author)
  • 2010(Publication Date)
  • CRC Press

    (Publisher)

  Given two vectors, V 1 , V 2 , and knowing that the angle between them is α, their scalar , also called dot product , is defined as p = V1 · V 2 = V 1 · V 2 cos α (2.1) Vectors and matrices 67 The result is a scalar, hence the first name of the product. The dot, ‘ · ’, used to mark this product explains the other name. The operation is performed in MATLAB by the function dot . As an example let us define two vectors and calculate their dot product: V1 = [ 3.4641; 2 ]; V2 = [ 2.5; 4.3301 ]; p = dot(V1, V2) p = 17.3205 To verify that the result corresponds to the definition given above, calculate the lengths of the two vectors (using the function norm ) and the angle between them. norm(V1) ans = 4.0000 norm(V2) ans = 5.0000 beta = atand(V1(2)/V1(1)); gamma = atand(V2(2)/V2(1)); alpha = gamma -beta; p = norm(V1)*norm(V2)* cosd(alpha) We obtain the same result as the one calculated with the command dot . It is easy to prove that, if the vector V 1 has the components V 1 x , V 1 y , and the vector V 2 , the components V 2 x , V 2 y , then the product V 1 · V 2 equals V 1 x V 2 x + V 1 y V 2 y . Let us check this in MATLAB: p1 = V1(1)*V2(1) + V1(2)*V2(2) p1 = 17.3205 p -p1 ans = 0 The generalization to three and more dimensions is straightforward. The scalar product enables us to ‘measure’ lengths and angles. The length can be a displacement or the magnitude of a vector. For example consider the vector V 1 with the components defined above; its magnitude can be calculated as 68 What every engineer should know about MATLAB and Simulink V 1 = V 1 · V 1 = V 2 1 x + V 2 1 y With the values exemplified above, V1 = [ 3.4641; 2 ]; sqrt(dot(V1, V1)) ans = 4.0000 We recovered the same value as that obtained with the MATLAB built-in function norm . The use of the latter is simpler; however, it is important to show that the task can be carried on by calling the dot product. Let us show now how the dot product enables us to ‘measure’ angles.

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  Calculus

  Multivariable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  Note that the dot product of two vectors is a scalar. 674 CHAPTER 11 Three-Dimensional Space; Vectors Example 1 3, 5 · −1, 2 = 3(−1) + 5(2) = 7 2, 3 · −3, 2 = 2(−3) + 3(2) = 0 1, −3, 4 · 1, 5, 2 = 1(1) + (−3)(5) + 4(2) = −6 Here are the same computations expressed another way: (3i + 5 j) · (−i + 2 j) = 3(−1) + 5(2) = 7 (2i + 3 j) · (−3i + 2 j) = 2(−3) + 3(2) = 0 (i − 3 j + 4k) · (i + 5 j + 2k) = 1(1) + (−3)(5) + 4(2) = −6 Technology Mastery Many calculating utilities have a built-in dot product opera- tion. If your calculating utility has this capability, use it to check the computations in Example 1. Algebraic Properties of the Dot Product The following theorem provides some of the basic algebraic properties of the dot product. Note the difference between the two zeros that appear in part (e) of Theorem 11.3.2— the zero on the left side is the zero vector (boldface), whereas the zero on the right side is the zero scalar (lightface). Theorem 11.3.2 If u, v, and w are vectors in 2- or 3-space and k is a scalar, then: (a) u · v = v · u (b) u · (v + w) = u · v + u · w (c) k(u · v) = (ku) · v = u · (kv) (d) v · v = v 2 (e) 0 · v = 0 We will prove parts (c) and (d) for vectors in 3-space and leave some of the others as exercises. Proof (c) Let u = u 1 , u 2 , u 3  and v = v 1 , v 2 , v 3 . Then k(u · v) = k(u 1 v 1 + u 2 v 2 + u 3 v 3 ) = (ku 1 )v 1 + (ku 2 )v 2 + (ku 3 )v 3 = (ku) · v Similarly, k(u · v) = u · (kv). Proof (d) v · v = v 1 v 1 + v 2 v 2 + v 3 v 3 = v 2 1 + v 2 2 + v 2 3 = v 2 . The following alternative form of the formula in part (d) of Theorem 11.3.2 provides a useful way of expressing the norm of a vector in terms of a dot product: v = √ v · v (1) Angle Between Vectors Suppose that u and v are nonzero vectors in 2-space or 3-space that are positioned so their initial points coincide. We define the angle between u and v to be the angle θ determined by the vectors that satisfies the condition 0 ≤ θ ≤ π (Figure 11.3.1).

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  Advanced Engineering Mathematics

  • Lawrence Turyn(Author)
  • 2013(Publication Date)
  • CRC Press

    (Publisher)

  6 Geometry, Calculus, and Other Tools 6.1 Dot Product, Cross Product, Lines, and Planes In this chapter, we study many geometric and differential calculus results that have many useful and powerful applications to problems of engineering and science. 6.1.1 Dot Product and Cross Product We have been using the dot product, which is familiar also from its physical applications. For example, the work , W , is given by W = F • u || F || || u || cos θ , where F is a constant force u is the displacement θ is the angle between the vectors F and u satisfying 0 ≤ θ ≤ π . Algebraically, for two vectors r 1 , r 2 in R 3 , r 1 • r 2 = ⎡ ⎣ x 1 y 1 z 1 ⎤ ⎦ • ⎡ ⎣ x 2 y 2 z 2 ⎤ ⎦ = x 1 x 2 + y 1 y 2 + z 1 z 2 . Denote ˆ ı ⎡ ⎣ 1 0 0 ⎤ ⎦ , ˆ j ⎡ ⎣ 0 1 0 ⎤ ⎦ , ˆ k ⎡ ⎣ 0 0 1 ⎤ ⎦ . (6.1) The cross product between vectors r 1 = x 1 ˆ ı + y 1 ˆ j + z 1 ˆ k and r 2 = x 2 ˆ ı + y 2 ˆ j + z 2 ˆ k is defined by r 1 × r 2 = ˆ ı ˆ j ˆ k x 1 y 1 z 1 x 2 y 2 z 2 = y 1 z 1 y 2 z 2 ˆ ı − x 1 z 1 x 2 z 2 ˆ j + x 1 y 1 x 2 y 2 ˆ k = ( y 1 z 2 − z 1 y 2 ) ˆ ı + ( z 1 x 2 − x 1 z 2 ) ˆ j + ( x 1 y 2 − y 1 x 2 ) ˆ k . 457 458 Advanced Engineering Mathematics Theorem 6.1 (Properties of the cross product) (a) x × y is perpendicular to both x and y . (b) || x × y || = || x || || y || sin θ , where θ is the angle between the vectors x and y such that 0 ≤ θ ≤ π . (c) x × y = 0 if, and only if, { x , y } is linearly dependent, that is, x and y are parallel or one of them is the zero vector. (d) y × x = − x × y . (e) ˆ ı × ˆ j = ˆ k , ˆ j × ˆ k = ˆ ı , ˆ k × ˆ ı = ˆ j . (f) ˆ ı × ˆ ı = 0 , ˆ j × ˆ j = 0 , ˆ k × ˆ k = 0 . Example 6.1 A force F acting on a lever arm at a position r relative to an axis applies the torque τ r × F , as depicted in Figure 6.1. Example 6.2 A charge q moving with a velocity v in a magnetic flux density B experiences the Lorentz force F = q v × B . Example 6.3 Find a unit vector normal to both of the vectors r 1 = ˆ ı + 2 ˆ j + 3 ˆ k , and r 2 = ˆ j − ˆ k .

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