# 9.4. The Scalar Product. Introduction. Prerequisites. Learning Style. Learning Outcomes

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1 The Scalar Product 9.4 Introduction There are two kinds of multiplication involving vectors. The first is known as the scalar product or dot product. This is so-called because when the scalar product of two vectors is calculated the result is a scalar. The second product is known as the vector product. When this is calculated the result is a vector. The definitions of these products may seem rather strange at first, but they are widely used in applications. In this Block we consider only the scalar product. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... calculate, from its definition, the scalar product of two given vectors calculate the scalar product of two vectors given in cartesian form use the scalar product to find the angle between two vectors use the scalar product to test whether two vectors are perpendicular 1 that a vector can be represented as a directed line segment 2 how to express a vector in cartesian form 3 how to find the modulus of a vector Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. Definition of the Scalar Product Consider the two vectors a and b shown in Figure??. b θ a Figure 1: Two vectors subtend an angle θ. Note that the tails of the two vectors coincide and that the angle between the vectors has been labelled θ. Their scalar product, denoted by a b, is defined as the product a b cos θ. It is very important to use the dot in the formula. The dot is the specific symbol for the scalar product, and is the reason why the scalar product is also known as the dot product. You should not use a sign in this context because this sign is reserved for the vector product which is quite different. The angle θ is always chosen to lie between 0 and π, and the tails of the two vectors should coincide. Figure?? shows two incorrect ways of measuring θ. b b θ a a θ Figure 2: θ should not be measured in these ways. We can remember this formula as: scalar product : a b = a b cos θ the modulus of the first vector, multiplied by the modulus of the second vector, multiplied by the cosine of the angle between them. Clearly b a = b a cos θ and so a b = b a. Thus we can evaluate a scalar product in any order: the operation is said to be commutative. Engineering Mathematics: Open Learning Unit Level 1 2

3 Example Vectors a and b are shown in the figure below. Vector a has modulus 6 and vector b has modulus 7 and the angle between them is 60. Calculate a.b. b 60 a The angle between the two vectors is 60. Hence a b = a b cos θ = (6)(7) cos 60 =21 The scalar product of a and b is equal to 21. Note that when finding a scalar product the result is always a scalar. Example Find i i where i is the unit vector in the direction of the positive x axis. Because i is a unit vector its modulus is 1. Also, the angle between i and itself is zero. Therefore i.i = (1)(1) cos 0 =1 So the scalar product of i with itself equals 1. It is easy to verify that j.j = 1 and k.k =1. Example Find i j where i and j are unit vectors in the directions of the x and y axes. Because i and j are unit vectors they each have a modulus of 1. The angle between the two vectors is 90. Therefore i j = (1)(1) cos 90 =0 That is i j =0. More generally, the following results are easily verified: i i = j j = k k =1 i j = i k = j k =0 3 Engineering Mathematics: Open Learning Unit Level 1

4 Even more generally, whenever any two vectors are perpendicular to each other their scalar product is zero because the angle between the vectors is 90 and cos 90 =0. The scalar product of perpendicular vectors is zero. 2. A Formula for Finding the Scalar Product We can use the previous results to obtain a formula for finding a scalar product when the vectors are given in cartesian form. We consider vectors in the xy plane. Suppose a = a 1 i + a 2 j and b = b 1 i + b 2 j. Then a b = (a 1 i + a 2 j) (b 1 i + b 2 j) = a 1 i (b 1 i + b 2 j)+a 2 j (b 1 i + b 2 j) = a 1 b 1 i i + a 1 b 2 i j + a 2 b 1 j i + a 2 b 2 j j Now, using the previous boxed results we can simplify this to give the following formula: if a = a 1 i + a 2 j and b = b 1 i + b 2 j then a b = a 1 b 1 + a 2 b 2 Thus to find the scalar product of two vectors their i components are multiplied together, their j components are multiplied together and the results are added. Example If a =7i +8j and b =5i 2j, find the scalar product a b. We use the previous boxed formula and multiply corresponding components together, adding the results. a b = (7i +8j) (5i 2j) = (7)(5) + (8)( 2) = = 19 The formula readily generalises to vectors in three dimensions as follows: if a = a 1 i + a 2 j + a 3 k and b = b 1 i + b 2 j + b 3 k then a b = a 1 b 1 + a 2 b 2 + a 3 b 3 Engineering Mathematics: Open Learning Unit Level 1 4

5 Example If a =5i +3j 2k and b =8i 9j +11k, find a b. Corresponding components are multiplied together and the results are added. a b = (5)(8) + (3)( 9)+( 2)(11) = = 9 Note again that the result is a scalar: there are no i s, j s, or k s in the answer. Now do this exercise If p =4i 3j +7k and q =6i j +2k, find p q. Corresponding components are multiplied together and the results are added. Now do this exercise If r =3i +2j +9k find r r. Show that this is the same as r 2. Answer Answer For any vector r we have r 2 = r r 3. Resolving One Vector Along Another The scalar product can be used to find the component of a vector in the direction of another vector. Consider Figure?? which shows two arbitrary vectors a and n. Let ˆn be a unit vector in the direction of n. P a O ˆn θ Q n projection of a onto n Figure 3: 5 Engineering Mathematics: Open Learning Unit Level 1

6 Study the figure carefully and note that a perpendicular has been drawn from P to meet n at Q. The distance OQ is called the projection of a onto n. Simple trigonometry tells us that the length of the projection is a cos θ. Now by taking the scalar product of a with the unit vector ˆn we find a ˆn = a ˆn cos θ = a cos θ since ˆn =1 We conclude that a ˆn is the component of a in the direction of n Example The figure below shows a plane containing the point A with position vector a. The vector ˆn is a unit vector perpendicular to the plane (such a vector is called a normal vector). Find an expression for the perpendicular distance of the plane from the origin. ˆn ˆn l A O a From the diagram note that the perpendicular distance l of the plane from the origin is the projection of a onto ˆn and is thus a ˆn. 4. Using the Scalar Product to Find the Angle Between Two Vectors We have two distinct ways of calculating the scalar product of two vectors. From the first Key Point of this block a b = a b cos θ whislst from the last of Section 2 we have a b = a 1 b 1 + a 2 b 2 + a 3 b 3. Both methods of calculating the scalar product are entirely equivalent and will always give the same value for the scalar product. We can exploit this correspondence to find the angle between two vectors. The following example illustrates the procedure to be followed. Engineering Mathematics: Open Learning Unit Level 1 6

7 Example Find the angle between the vectors a =5i +3j 2k and b =8i 9j +11k. The scalar product of these two vectors has already been found in the example on page 4 to be 9. The modulus of a is ( 2) 2 = 38. The modulus of b is 8 2 +( 9) = 266. Substituting these into the formula for the scalar product we find a b = a b cos θ 9 = cos θ from which so that cos θ = = θ = cos 1 ( ) = In general, the angle between two vectors can be found from the following formula: The angle θ between vectors a, b is such that: cos θ = a b a b More exercises for you to try 1. If a =2i 5j and b =3i +2j find a b and verify that a b = b a. 2. Find the angle between p =3i j and q = 4i +6j. 3. Use the definition of the scalar product to show that if two vectors are perpendicular, their scalar product is zero. 4. If a and b are perpendicular, simplify (a 2b) (3a +5b). 5. If p = i +8j +7k and q =3i 2j +5k, find p q. 6. Show that the vectors 1 i + j and 2i j are perpendicular The work done by a force F in moving a body through a displacement r is given by F r. Find the work done by the force F =3i +7k if it causes a body to move from the point with coordinates (1, 1, 2) to the point (7, 3, 5). 8. Find the angle between the vectors i j k and 2i + j +2k. Answer 7 Engineering Mathematics: Open Learning Unit Level 1

8 5. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. DERIVE has routines for dealing with two- and threedimensional vectors. To use vectors you can enter them by keying Author:Vector. When requested for the dimension respond with either 2 or 3 as appropriate. You will then be able to enter your vectors. Alternatively, and this is probably easier, use Author:Expression and enclose your vector within square brackets. For example the vector 2i +3j 4k would be entered as [2, 3, 4]. The product of a scalar with a vector is written in an obvious way. The vector 6(2i +3j 4k) would be keyed in as Author:Expression 6[2, 3, 4]. DERIVE responds with [12, 18, 24]. The modulus of a vector is obtained using the abs command. For example to find 2i+3j 4k you would key in Author:Expression abs[2, 3, 4]. DERIVE responds with 29. The scalar product can be obtained in DERIVE by using the. notation. To obtain (2i +3j 4k).( 1i +3j +5k) you would key in Author:Expression [2, 3, 4].[ 1, 3, 5]. DERIVE responds with 13. As a useful exercise use DERIVE to obtain the solutions to Exercises 1,5,6 and 8. Engineering Mathematics: Open Learning Unit Level 1 8

9 End of Block Engineering Mathematics: Open Learning Unit Level 1

10 41 Back to the theory Engineering Mathematics: Open Learning Unit Level 1 10

11 94. r = = 94 hence r 2 = r r. Back to the theory 11 Engineering Mathematics: Open Learning Unit Level 1

12 , 4. 3a 2 10b units Back to the theory Engineering Mathematics: Open Learning Unit Level 1 12

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