Section 9.3 The Dot Product

1. Section 9.3The Dot Product • Goals • Introduce the dot product of two vectors and explain its significance to work. • Discuss the dot product and perpendicularity. • Give properties of the dot product. • Introduce projections.

2. Introduction • So far we have added two vectors and multiplied a vector by a scalar. • 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 consider in this section.

3. Work • Recall that the work done by a constant force F in moving an object through a distance d is W = Fd. • However this applies only when the force is directed along the line of motion. • Suppose, however, that the constant force is a vectordirection, as shown on the next slide:

4. Work (cont’d)

5. Work (cont’d) • If the force moves the object from P to Q, then the displacement vector is • The work W done by F is defined as the magnitude of the displacement, |D| , multiplied by the magnitude of the applied force in the direction of the motion, namely

6. Work (cont’d) • Thus • We use this expression to define the dot product of two vectors even when they don’t represent force or displacement:

7. Remarks • This product is called the dot product because of the dot in the notation a ∙ b. • a ∙ b is a scalar, not a vector. • Sometimes the dot product is called the scalar product. • In the example of finding work done, it makes no sense for θ to be greater than π/2, but in our general definition we allow θ to be any angle from 0 to π.

8. Example • If the vectors a and b have lengths 4 and 6, and the angle between them is π/3, find a ∙ b. • Solution According to the definition, a ∙ b = |a||b| cos(π/3) = 4∙ 6∙ ½ = 12

9. Perpendicular Vectors • Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is θ = π/2. • For such vectors we have a ∙ b = |a||b| cos(π/2) = 0 • Conversely, if a ∙ b = 0, then cos θ = 0, soθ = π/2.

10. Perpendicular Vectors (cont’d) • Since the zero vector 0 is considered to be perpendicular to all vectors, we have • Further, by properties of the cosine, a ∙ b is • positive for θ < π/2, and • negative for θ > π/2, as the next slide illustrates:

11. Perpendicular Vectors (cont’d)

12. Perpendicular Vectors (cont’d) • We can think of a ∙ b as measuring the extent to which a and b point in the same general direction. • The dot product a ∙ b is… • positive if a and b point in the same general direction, • 0 if they are perpendicular, and • negative if they point in generally opposite directions.

13. Perpendicular Vectors (cont’d) • In the extreme cases where… • a and b point in exactly the same direction, we have θ = 0, so cos θ = 1 and a ∙ b = |a||b| • a and b point in exactly opposite directions, then θ = π, so cos θ = –1 and a ∙ b = –|a||b|

14. Component Form • Suppose we are given two vectors in component form: • We want to find a convenient expression for a ∙ b in terms of these components. • An application of the Law of Cosines gives the result on the next slide:

15. Component Form (cont’d) • Here are some examples:

16. Example • Show that 2i + 2j – k is perpendicular to5i– 4j + 2k. • Solution Since (2i + 2j – k)∙ (5i– 4j + 2k) = 2(5) + 2(– 4) + (– 1)(2) = 0, these vectors are perpendicular.

17. Example • Find the angle between • Solution Let θ be the required angle. Since

18. Solution (cont’d) and since a ∙ b = 2(5) + 2(– 3) + (– 1)(2) = 0, the definition of dot product gives • So the angle between a and b is

19. Example • A force is given by a vector F = 3i + 4j + 5k and moves a particle from P(2, 1, 0) toQ(4, 6, 2). Find the work done. • Solution The displacement vector is

20. Properties • The dot product obeys many of the laws that hold for ordinary products of real numbers:

21. Projections • The next slide shows representations the same initial point P. • If S is the foot of the perpendicular from R to the line containingwith representationprojection of b onto a and is denoted by projab.

22. Projections (cont’d)

23. Projections (cont’d) • The scalar projection ofb onto a is the length|b|cos θ of the vectorprojection. • This is denoted bycompab, and can alsobe computed by taking the dot product of b with the unit vector in the direction of a.

24. Projections (cont’d) • To summarize: • For example we find the scalar projection and vector projection of

25. Solution • Sinceprojection of b onto a is • The vector projection is this scalar projection times the unit vector in the direction of a:

26. Solution (cont’d)

27. Review • Definition of dot product in terms of work done • Orthogonality and the dot product • The dot product in component form • Properties of the dot product • Projections