Section 9.3 The Dot Product

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Presentation transcript:

Section 9.3 The 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.

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.

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 vector direction, as shown on the next slide:

Work (cont’d)

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

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:

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.

a ∙ b = |a||b| cos(π/3) = 4 ∙ 6 ∙ ½ = 12 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

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, so θ = π/2.

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:

Perpendicular Vectors (cont’d)

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.

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|

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:

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

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

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

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

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

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

Projections The next slide shows representations the same initial point P. If S is the foot of the perpendicular from R to the line containing with representation projection of b onto a and is denoted by proja b.

Projections (cont’d)

Projections (cont’d) The scalar projection of b onto a is the length |b|cos θ of the vector projection. This is denoted by compa b, and can also be computed by taking the dot product of b with the unit vector in the direction of a.

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

Solution Since projection of b onto a is The vector projection is this scalar projection times the unit vector in the direction of a:

Solution (cont’d)

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