7.3 Dot Product

INTRODUCTION

In this and the following section, we shall consider two kinds of products between vectors that originated in the study of mechanics and electricity and magnetism. The first of these products, known as the dot product, is studied in this section.

Component Form of the Dot Product

The dot product, defined next, is also known as the inner product, or scalar product. The dot product of two vectors a and b is denoted by a · b and is a real number, or scalar, defined in terms of the components of the vectors.

DEFINITION 7.3.1 Dot Product of Two Vectors

In 2-space the dot product of two vectors a = 〈a1, a2〉 and b = 〈b1, b2〉 is the number

a · b = a1b1 + a2b2. (1)

In 3-space the dot product of two vectors a = 〈a1, a2, a3〉 and b = 〈b1, b2, b3〉 is the number

a · b = a1b1 + a2b2 + a3b3. (2)

EXAMPLE 1 Dot Product Using (2)

If a = 10i + 2j − 6k and b = −i + 4j − 3k, then it follows from (2) that

a · b = (10)(−) + (2)(4) + (−6)(−3) = 21. ≡

EXAMPLE 2 Dot Products of the Basis Vectors

Since i = 〈1, 0, 0〉, j = 〈0, 1, 0〉, and k = 〈0, 0, 1〉, we see from (2) that

i · j = j · i = 0, j · k = k · j = 0, and k · i = i · k = 0. (3)

Similarly, by (2)

i · i = 1, j · j = 1, and k · k = 1. (4) ≡

Properties

The dot product possesses the following properties.

THEOREM 7.3.1 Properties of the Dot Product

  1. a · b = 0 if a = 0 or b = 0
  2. a · b = b · a ← commutative law
  3. a · (b + c) = a · b + a · c ← distributive law
  4. a · (kb) = (ka)· b = k(a · b), k a scalar
  5. a · a ≥ 0
  6. a · a =
PROOF:

All the properties can be proved directly from (2). We illustrate by proving parts (iii) and (vi).

To prove part (iii) we let a = 〈a1, a2, a3〉, b = 〈b1, b2, b3〉, and c = 〈c1, c2, c3〉. Then

To prove part (vi) we note that

≡

Notice that (vi) of Theorem 7.3.1 states that the magnitude of a vector can be written in terms of the dot product:

Alternative Form

The dot product of two vectors a and b can also be expressed in terms of the lengths of the vectors and the angle between them. If the vectors a and b are positioned in such a manner that their initial points coincide, then we define the angle between a and b as the angle θ that satisfies

THEOREM 7.3.2 Alternative Form of Dot Product

The dot product of two vectors a and b is

(5)

where θ is the angle between the vectors.

This more geometric form is what is generally used as the definition of the dot product in a physics course.

PROOF:

Suppose θ is the angle between the vectors a = a1i + a2j + a3k and b = b1i + b2j + b3k. Then the vector

c = b − a = (b1 − a1)i + (b2 − a2)j + (b3 − a3)k

is the third side of the triangle indicated in FIGURE 7.3.1. By the law of cosines we can write or

. (6)

Using ,

and ,

the right-hand side of equation (6) simplifies to a1b1 + a2b2 + a3b3. Since the last expression is (2) in Definition 7.3.1, we see that = a · b. ≡

2 vectors a and b have a common initial point. The angle between the vectors is theta. Vector c joins the end points of a and b.

FIGURE 7.3.1 Vector c in the proof of Theorem 7.3.2

Orthogonal Vectors

If a and b are nonzero vectors, then and are both positive numbers. In this case it follows from (5) that the sign of the dot product a · b is the same as the sign of cos θ . In FIGURE 7.3.2 we see various orientations of two vectors for angles that satisfy Observe from Figure 7.3.2 and (5) that a · b > 0 for θ = 0, a · b > 0 when θ is acute, a · b = 0 for θ = π/2, a · b < 0 when θ is obtuse, and a · b < 0 for θ = π. In the special case θ = π/2 we say that the vectors are orthogonal or perpendicular. Moreover, if we know that a · b = 0, then the only angle for which this is true and satisfying 0 ≤ θ ≤ π is θ = π/2. This leads us to the next result.

In all 5 graphs, vector b goes to the right. The angle theta between vector b and vector a is marked. Both vectors have a common initial point. Graph a. vectors a and b both go to the right. Caption. Same direction. Theta equals 0. cos of theta equals 1. Graph b. Vector a goes up and to the right, forming an acute angle theta with vector b. Caption. Acute angle. Theta is greater than 0 and less than pi over 2. Cos of theta is greater than 0. Graph c. Vector a goes up, forming a right angle with vector b. Caption. Orthogonal. Theta equals pi over 2. Cos of theta equals 0. Graph d. Vector a goes up and to the left, forming an obtuse angle with b. Caption. Obtuse angle. Theta is greater than pi over 2 and less than pi. Cos of theta is less than 0. Graph e. Vector a goes to the left, and forms a flat angle with vector b. Caption. Opposite direction. Theta equals pi. Cos of theta equals negative 1.

FIGURE 7.3.2 Angles between vectors a and b

THEOREM 7.3.3 Criterion for Orthogonal Vectors

Two nonzero vectors a and b are orthogonal if and only if a · b = 0.

Since 0 · b = 〈0, 0, 0〉 · 〈b1, b2, b3〉 = 0(b1) + 0(b2) + 0(b3) = 0 for every vector b, the zero vector 0 is considered to be orthogonal to every vector.

EXAMPLE 3 Orthogonal Vectors

If a = −3i − j + 4k and b = 2i + 14j + 5k, then

a · b = (−3)(2) + (−1)(14) + (4)(5) = 0.

From Theorem 7.3.3, we conclude that a and b are orthogonal. ≡

In Example 1, from a · b ≠ 0 we can conclude that the vectors a = 10i + 2j − 6k and b = −i + 4j − 3k are not orthogonal. From (3) of Example 2, we see what is apparent in Figure 7.2.8, namely, the vectors i, j, and k constitute a set of mutually orthogonal unit vectors.

Angle Between Two Vectors

By equating the two forms of the dot product, (2) and (5), we can determine the angle between two vectors from

(7)

EXAMPLE 4 Angle Between Two Vectors

Find the angle between a = 2i + 3j + k and b = −i + 5j + k.

SOLUTION

From = , = , a · b = 14, we see from (7) that

,

and so ≡

Direction Cosines

For a nonzero vector a = a1i + a2j + a3k in 3-space, the angles α, β, and γ between a and the unit vectors i, j, and k, respectively, are called direction angles of a. See FIGURE 7.3.3. Now, by (7),

which simplify to

We say that cos α, cos β, and cos γ are the direction cosines of a. The direction cosines of a nonzero vector a are simply the components of the unit vector a:

Since the magnitude of a is 1, it follows from the last equation that

cos2α + cos2β + cos2γ = 1.

A vector a is graphed in the x y z coordinate system. The vector forms an angle alpha with the positive x axis, an angle beta with the positive y axis, and an angle gamma with the positive z axis.

FIGURE 7.3.3 Direction angles α, β, and γ

EXAMPLE 5 Direction Cosines/Angles

Find the direction cosines and direction angles of the vector a = 2i + 5j + 4k.

SOLUTION

From = , we see that the direction cosines are

The direction angles are

≡

Observe in Example 5 that

Component of a on b

The distributive law and (3) enable us to express the components of a vector a = a1i + a2j + a3k in terms of the dot product:

a1 = a · i, a2 = a · j, a3 = a · k. (8)

Symbolically, we write the components of a as

compia = a · i, compja = a · j, compka = a · k. (9)

We shall now see that the results indicated in (9) carry over to finding the component of a on an arbitrary vector b. Note that in either of the two cases shown in FIGURE 7.3.4,

compba = cos θ. (10)

In Figure 7.3.4(b), compba < 0 since π/2 < θ ≤ π. Now, by writing (10) as

we see that (11)

In other words:

To find the component of a on a vector b, we dot a with a unit vector in the direction of b.

Two vectors a and b have a common initial point. The angle between the vectors is theta. Graph a. Theta is an acute angle. The orthogonal projection of a on b is at a distance absolute value of a times cos theta from the initial point. Graph b. Theta is an obtuse angle. The orthogonal projection of a on b is on the line of b but on the other side, at a distance absolute value of a times cos theta from the initial point.

FIGURE 7.3.4 Component of a on b

EXAMPLE 6 Component of a Vector on Another Vector

Let a = 2i + 3j − 4k and b = i + j + 2k. Find compba and compab.

SOLUTION

We first form a unit vector in the direction of b:

Then from (11) we have

By modifying (11) accordingly, we have

Therefore,

and ≡

Projection of a onto b

As illustrated in FIGURE 7.3.5, the projection of a vector a in any of the directions determined by i, j, k is simply the vector formed by multiplying the component of a in the specified direction with a unit vector in that direction; for example,

projia = (compia)i = (a · i)i = a1i

and so on. FIGURE 7.3.6 shows the general case of the projection of a onto b:

(12)

A position vector a is graphed in 3-space. Its projection on the axes are vectors labeled proj subscript i a, proj subscript j a, and proj subscript k a. These vectors are along the unit vectors.

FIGURE 7.3.5 Projections of a onto i, j, and k

Two vectors a and b have a common initial point. The unit vector along b is labeled 1 over absolute value of b, times b. The projection of a on b is at a distance proj subscript b a from the initial point.

FIGURE 7.3.6 Projection of a onto b

EXAMPLE 7 Projection of a Vector on Another Vector

Find the projection of a = 4i + j onto the vector b = 2i + 3j. Graph.

SOLUTION

First, we find the component of a and b. Since = , we find from (11) that

Thus, from (12),

The graph of this vector is shown in FIGURE 7.3.7. ≡

2 vectors a and b are graphed on an x y coordinate plane. They are position vectors with a = 4 i + 1 j and b = 2 i + 3 j. The projection of a on b is a position vector 22 over 13 i + 33 over 13 j.

FIGURE 7.3.7 Projection of a onto b in Example 7

Physical Interpretation of the Dot Product

When a constant force of magnitude F moves an object a distance d in the same direction of the force, the work done is simply W = Fd. However, if a constant force F applied to a body acts at an angle θ to the direction of motion, then the work done by F is defined to be the product of the component of F in the direction of the displacement and the distance that the body moves:

See FIGURE 7.3.8. It follows from Theorem 7.3.2 that if F causes a displacement d of a body, then the work done is

W = F · d. (13)

A force F acts on an object to move it a distance d. The force is at an angle theta from the horizontal. Its horizontal component is absolute value of F times cos theta.

FIGURE 7.3.8 Work done by a force F

EXAMPLE 8 Work Done by a Constant Force

Find the work done by a constant force F = 2i + 4j if its point of application to a block moves from P1(1, 1) to P2(4, 6). Assume that is measured in newtons and is measured in meters.

SOLUTION

The displacement of the block is given by

It follows from (13) that the work done is

W = (2i + 4j) · (3i + 5j) = 26 N-m. ≡

7.3 Exercises Answers to selected odd-numbered problems begin on page ANS-16.

In Problems 1–12, a = 〈2, −3, 4〉, b = 〈−1, 2, 5〉, and c = 〈3, 6, −1〉. Find the indicated scalar or vector.

  1. a · b
  2. b · c
  3. a · c
  4. a · (b + c)
  5. a · (4b)
  6. b · (a − c)
  7. a · a
  8. (2b) · (3c)
  9. a · (a + b + c)
  10. (2a) · (a − 2b)
  12. (c · b) a

In Problems 13 and 14, find a · b if the smaller angle between a and b is as given.

  1. = 10, = 5, θ = π/4
  2. = 6, = 12, θ = π/6
  3. Determine which pairs of the following vectors are orthogonal:
    1. 〈2, 0, 1〉
    2. 3i + 2j − k
    3. 2i − j − k
    4. i − 4j + 6k
    5. 〈1, −1, 1〉
    6. 〈−4, 3, 8〉
  4. Determine a scalar c so that the given vectors are orthogonal.
    1. a = 2i − cj + 3k, b = 3i + 2j + 4k
    2. a = 〈c, , c〉, b = 〈−3, 4, c〉
  5. Find a vector v = 〈x1, y1, 1〉 that is orthogonal to both a = 〈3, 1, −1〉 and b = 〈−3, 2, 2〉.
  6. A rhombus is an oblique-angled parallelogram with all four sides equal. Use the dot product to show that the diagonals of a rhombus are perpendicular.
  7. Verify that the vector

    is orthogonal to the vector a.
  8. Determine a scalar c so that the angle between a = i + cj and b = i + j is 45°.

In Problems 21–24, find the angle θ between the given vectors.

  1. a = 3i − k, b = 2i + 2k
  2. a = 2i + j, b = −3i − 4j
  3. a = 〈2, 4, 0〉, b = 〈−1, −1, 4〉
  4. a = 〈, , 〉, b = 〈2, −4, 6〉

In Problems 25–28, find the direction cosines and direction angles of the given vector.

  1. a = i + 2j + 3k
  2. a = 6i + 6j − 3k
  3. a = 〈1, 0, −〉
  4. a = 〈5, 7, 2〉
  5. Find the angle between the diagonal of the cube shown in FIGURE 7.3.9 and the edge AB. Find the angle between the diagonal of the cube and the diagonal .
    The vertices of a cube graphed in a coordinate system are as follows. Vertex A is on the positive y axis, point B on the x y plane, point C on the positive x axis, and point D is diagonally across A on the x z plane above C.

    FIGURE 7.3.9 Diagonal in Problem 29

  6. Show that if two nonzero vectors a and b are orthogonal, then their direction cosines satisfy

    cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 = 0.

  7. An airplane is 4 km high, 5 km south, and 7 km east of an airport. See FIGURE 7.3.10. Find the direction angles of the plane.
    The 3 axes of a coordinate system are labeled east, south, and up. The airport is located at the origin. An airplane is 7 east, 5 south and 4 up from the origin.

    FIGURE 7.3.10 Airplane in Problem 31

  8. Determine a unit vector whose direction angles, relative to the three coordinate axes, are equal.

In Problems 33–36, a = 〈1, −1, 3〉 and b = 〈2, 6, 3〉. Find the indicated number.

  1. compba
  2. compab
  3. compa(b − a)
  4. comp2b(a + b)

In Problems 37 and 38, find the component of the given vector in the direction from the origin to the indicated point.

  1. a = 4i + 6j, P(3, 10)
  2. a = 〈2, 1, −1〉, P(1, −1, 1)

In Problems 39–42, find projba.

  1. a = −5i + 5j, b = −3i + 4j
  2. a = 4i + 2j, b = −3i + j
  3. a = −i − 2j + 7k, b = 6i − 3j − 2k
  4. a = 〈1, 1, 1〉, b = 〈−2, 2, −1〉

In Problems 43 and 44, a = 4i + 3j and b = −i + j. Find the indicated vector.

  1. proj(a+b)a
  2. proj(a−b)b
  3. A sled is pulled horizontally over ice by a rope attached to the front of the sled. A 20-lb force acting at an angle of 60° with the horizontal moves the sled 100 ft. Find the work done.
  4. Find the work done if the point at which the constant force F = 4i + 3j + 5k is applied to an object moves from P1(3, 1, −2) to P2(2, 4, 6). Assume is measured in newtons and is measured in meters.
  5. A block with weight w is pulled along a frictionless horizontal surface by a constant force F of magnitude 30 newtons in the direction given by a vector d. See FIGURE 7.3.11. Assume is measured in meters.
    1. What is the work done by the weight w?
    2. What is the work done by the force F if d = 4i + 3j?
    A force F acts on a block and moves it a distance d. The force F is horizontal, and the weight w of the block is vertically down.

    FIGURE 7.3.11 Block in Problem 47

  6. A constant force F of magnitude 3 lb is applied to the block shown in FIGURE 7.3.12. F has the same direction as the vector a = 3i + 4j. Find the work done in the direction of motion if the block moves from P1(3, 1) to P2(9, 3). Assume distance is measured in feet.
    A block on a ramp is graphed on a coordinate plane. The ramp begins at the origin and goes up and to the right through (6, 2). A force F moves a block from (3, 1) to (9, 3). The force acts along a direction up and to the right of the ramp.

    FIGURE 7.3.12 Block in Problem 48

  7. In the methane molecule CH4, the hydrogen atoms are located at the four vertices of a regular tetrahedron. See FIGURE 7.3.13. The distance between the center of a hydrogen atom and the center of a carbon atom is 1.10 angstroms (1 angstrom = 10−10 meter), and the hydrogen–carbon–hydrogen bond angle is θ = 109.5°. Using vector methods only, find the distance between two hydrogen atoms.
    A methane molecule is a tetrahedron with 4 hydrogen atoms at the vertices and a carbon atom in the center. The angle between 2 C H bonds is theta.

    FIGURE 7.3.13 Molecule in Problem 49

Discussion Problems

  1. Use the dot product to prove the Cauchy–Schwarz inequality: |a · b| ≤ .
  2. Use the dot product to prove the triangle inequality a + b ≤ + . [Hint: Consider property (vi) of the dot product.]
  3. Prove that the vector n = ai + bj is perpendicular to the line whose equation is ax + by + c = 0. [Hint: Let P1(x1, y1) and P2(x2, y2) be distinct points on the line.]
  4. Use the result of Problem 52 and FIGURE 7.3.14 to show that the distance d from a point P1(x1, y1) to a line ax + by + c = 0 is
    A line labeled a x + b y + c equals 0 is graphed on an x y coordinate plane. It goes up and to the right. A vector n from point P subscript 2 on the line is perpendicular to the line. The coordinates of a point P subscript 1 located at a distance d from the line and of point P subscript 2 are (x subscript 1, y subscript 1) and (x subscript 2, y subscript 2)

    FIGURE 7.3.14 Distance d in Problem 53