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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.

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Section 9 3 the dot product
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
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.


Section 9 3 the dot product 1335506
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 d1
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 d2
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
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
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
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
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 d2
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 d3
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
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
Component Form (cont’d)

  • Here are some examples:


Example1
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.


Example2
Example

  • Find the angle between

  • Solution Let θ be the required angle. Since


Solution cont d
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


Example3
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
Properties

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


Projections
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 d1
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 d2
Projections (cont’d)

  • To summarize:

  • For example we find the scalar projection and vector projection of


Solution
Solution

  • Sinceprojection of b onto a is

  • The vector projection is this scalar projection times the unit vector in the direction of a:



Review
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