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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.
slide3
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
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