- 93 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Section 9.3 The Dot Product' - adamina

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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.

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 vectordirection, as shown on the next slide:

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.

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, soθ = 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)

- 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 to5i– 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) toQ(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 containingwith representationprojection of b onto a and is denoted by projab.

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.

Projections (cont’d)

- To summarize:
- For example we find the scalar projection and vector projection of

Solution

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

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

Download Presentation

Connecting to Server..