Section 9.3 The Dot Product

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# Section 9.3 The Dot Product - PowerPoint PPT Presentation

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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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