Section 13.3: The Dot Product 1

Objectives 1. Compute the dot product algebraically and geometrically. (1,5,7,9) 2. State 5 properties of the dot product. 3. Find the included angle between two nonzero vectors. (11,15,19,21,23,25,27) 4. Compute the direction cosines of a nonzero vector. (31,35,41,43,45,55) 5. Compute the work done by moving an object along a straight path. (49,51)

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Assignments 1. Read Section 13.3 2. Problems: 1,5,9,11,17,23,27,35,39,41,46,49,53,55 3. Read Section 13.4

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Lecture Notes

We have discussed addition and multiplication of a vector by a scalar. In this section, we introduce one way to multiply vectors, known as the dot product (or inner product). In Section 13-4, we’ll use the cross product to multiply vectors. One of the main differences between these multiplication methods is that the dot product returns a scalar quantity while the cross product returns a vector quantity. Both are useful. The dot product is defined as the sum of the products of the respective entries. Given a =< a1 , a2 , a3 > and b =< b1 , b2 , b3 >, the dot product a · b is a · b = a 1 b1 + a 2 b2 + a 3 b3 .

3.1

Algebraic properties

The properties of the dot product are: 1. a · a = |a|2 This property states that square of the magnitude of the vector a is the dot product of a with itself. 2. a · b = b · a (commutativity) 3. a · (b + c) = a · b + a · c (distributivity) 1

4. (ca) · b = c (a · b) = a · (cb) 5. 0 · a = 0

3.2

Geometrical intepretation

The dot product can also be interpreted with regard to the angle between the vectors a and b. The dot product is given as a · b = |a| |b| cos θ, where θ is the angle between a and b. We can show this relationship by examining the triangle formed by the vectors a, b, and a − b. This triangle is drawn in your text, so I won’t draw it again here. The proof of the geometrical relationship between the dot product and θ is also given in the text. We’ll go over it in class, but I won’t repeat the argument in these notes. Also, note that if θ is the angle between a and b, where a 6= 0 and b 6= 0, then cos θ =

a·b . |a| |b|

Class Questions 1. What happens if θ = π/2? 2. What does it mean for θ = π or θ = 0? 3.2.1

Example 1: Problems 13.2.6 and 13.2.10

Finding the dot product is easy. Given a =< 2, 6, −3 > and b =< 8, −2, −1 >, a · b = 16 − 12 + 3 = 7. We can use the second relationship to find a · b given vector lengths and the angle between them. Note that a·b |a| |b| ⇒ a · b = cos θ |a| |b| . cos θ =

Hence, given |a| = 4 and |b| = 10, where θ = 120, we calculate a·b = cos 120(4)(10) = −20.

3.3

Orthogonal vectors

Two vectors a and b are orthogonal if they meet to form a right angle. In particular, π this means that the angle between them, θ, is . Thus we have the following lemma. 2 Lemma: Two vectors a and b are orthogonal if and only if a · b = 0.

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3.3.1

Example 2: Problems 13.2.24 and 13.2.28

The vectors a =< 2, −4 > and b =< 4, 2 > are orthogonal since a · b = 0. The vectors a = 2i + 6j − 4k =< 2, 6, −4 > and b = −3i − 9j + 6k =< −3, −9, 6 > have a dot product equal to −6 − 54 − 24 6= 0, so they are not orthogonal. However, −3 a = b, so they are parallel. There’s nothing in the dot product that notice that 2 would give you this information.

3.4

Direction Angles

The vector a makes an angle with each of the x−, y−, and z−axes on the interval from [0, π]. These angles are known as the direction angles, and we’ll denote the direction angle of a with the x−axis as α, the direction angle of a with the y−axis as β, and the direction angle of a with the z−axis as γ. The direction cosines are the cosines of these angles. Note that the angle α is the angle between a and i. Thus, we have a·i |a| |i| a1 = . |a|

cos α =

Class Exercise: Find cos β and cos γ. Note that a22 a23 a21 + + |a|2 |a|2 |a|2 |a||2 = |a|2 = 1.

cos2 α + cos2 β + cos2 γ =

Also, since a1 = |a| cos α, a2 = |a| cos β, and a3 = |a| cos γ, we have a =< |a| cos α, |a| cos β, |a| cos γ > . From this, you get that 1 a =< cos α, cos β, cos γ > . |a| Thus these direction cosines give you the normalized version of a; that is, they are the components of the unit vector in the direction of a.

3.5

Projections

Dot products are also used to define projections of one vector onto another. A projection of a vector is defined similarly to point projections. The projection of b 3

onto a is found by places the starting points of b and a together and then dropping a perpendicular line segment from the head of b onto a. The resulting vector is the vector projection. Figure 1 shows the vector projection of b onto a, which is denoted by proja b. Projection of b on a

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Figure 1: Projection of b onto a The scalar projection of the vector b onto a is the magnitude of the vector projection. This value is actually |b| cos θ, which can be easily seen using trigonometry. This scalar projection is denoted by compa b. It is actually the component of b along a. Since the dot product is given as a · b = |a||b| cos θ = |a| (|b| cos θ) , the dot product is actually a times compa b. Thus we have these expressions for the scalar and vector projections, defined in terms of the dot product: a·b 1. compa b = |a| µ ¶ a·b a 2. proja b = |a| |a| 3.5.1

Example 3: Problem 13.3.42

Given a =< −1, −2, 2 > and b =< 3, 3, 4 >, we have a · b = −1 and |a| = 3. Thus, −1 compa b = 3 and −1 < −1, −2, 2 > proja b = 3À ¿3 1 2 −2 = , , . 9 9 9 4

3.6

Use of projections

Projection vectors can be used to calculate work. The work done by a constant force F in moving an object through a distance d is W = F d, but this is onlly true if the force is along the same line as the motion of the object. If the force is in another direction described by the vector F, we will need to use projection vectors to determine the amount of work done. The vector describing the motion of the object will be called the displacement vector, D. Thus, the action of the force in the direction of the motion is the projection of F onto D. After finding the projection, the work can be calculated as before. The size of the force F in the direction of D is given by |F| cos θ, where θ is the angle between F and D. Thus the work W , a scalar quantity, is given by W = |F| cos θ|D| = F · D. (Note that |D| is the distance the object moved, d.) Class Question: Does this contradict the earlier notion of work, when the force and the object moved in the same direction? 3.6.1

Example 4: Problem 13.3.52

In this problem, we know |F| = 50, |D = 100, and θ = 30. Thus the calculation of W is straightforward, and W = cos 30(50)(100).

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