CHAPTER 3. VECTOR ALGEBRA Part 1: Addition and Scalar
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CHAPTER 3. VECTOR ALGEBRA Part 1: Addition and Scalar Multiplication for Vectors. §1. Basics. Geometric or physical quantities such as length, area, volume, tempera- ture, pressure, speed, energy, capacity etc. are given by specifying a single numbers. Such quantities are called scalars, because many of them can be measured by tools with scales. Simply put, a scalar is just a number. Quantities such as force, velocity, acceleration, momentum, angular velocity, electric or magnetic field at a point etc are vector quantities, which are represented by an arrow. If the ‘base’ and the ‘head’ of this arrow are B and H repectively, then we denote this vector by −−→BH: Figure 1. Often we use a single block letter in lower case, such as u, v, w, p, q, r etc. to denote a vector. Thus, if we also use v to denote the above vector −−→BH, then v = −−→BH.A vector v has two ingradients: magnitude and direction. The magnitude is the length of the arrow representing v, and is denoted by v . In case v = −−→BH, certainly we | | have v = −−→BH for the magnitude of v. The meaning of the direction of a vector is | | | | self–evident. Two vectors are considered to be equal if they have the same magnitude and direction. You recognize two equal vectors in drawing, if their representing arrows are parallel to each other, pointing in the same way, and have the same length 1 Figure 2. For example, if A, B, C, D are vertices of a parallelogram, followed in that order, then −→AB = −−→DC and −−→AD = −−→BC: Figure 3. Vectors are added according to the parallelogram law or the triangular law: Figure 4. Figure 5. 2 In figure 4, vectors u and v lying on the neighboring sides of a parallelogram are given by u = −→AB and v = −−→AD. Then their sum u + v lies on the diagonal, given by u + v = −→AC. In Figure 5, vectors u and v lying on two sides of a triangle ABC △ are given by u = −→AB and v = −−→BC. Then their sum u + v lies on the remaining side, given by u + v = −→AC. Thus we have −→AB + −−→BC = −→AC. The last identity can be generalized. Given a number of points, say A1,A2,...,An, we have −−−→A1A2 + −−−→A2A3 + + −−−−−→An−1An = −−−→A1An. ··· Figure 6. We only check this for five points, that is, for n = 5: −−−→A1A2 +−−−→A2A3 +−−−→A3A4 +−−−→A4A5 = −−−→A1A5. The quick way begins by observing that −−−→A1A2 +−−−→A2A3 = −−−→A1A3 and −−−−→A3,A4 +−−−→A4A5 = −−−→A3A5 and finishes with −−−→A1A2 + −−−→A2A3 + −−−−→A3,A4 + −−−→A4A5 = −−−→A1A3 + −−−→A3A5 = −−−→A1A5: Figure 7. 3 The ‘step by step’ way described next is slower; however, it can be easily adapted to the general case with arbitrary number of points: −−−→A1A2 + −−−→A2A3 = −−−→A1A3 −−−→A1A2 + −−−→A2A3 + −−−→A3A4 = −−−→A1A3 + −−−→A3A4 = −−−→A1A4 −−−→A1A2 + −−−→A2A3 + −−−→A3A4 + −−−→A4A5 = −−−→A1A4 + −−−→A4A5 = −−−→A1A5 The picture below gives the idea of this ‘step by step’ approach: Figure 8. The next example is crucial because it describes a general situation that frequently occurs. Example. Suppose that −→OA = u and −−→OB = v. We are asked to determine −→AB in terms of u and v. Figure 9. 4 First, we write down −→OA + −→AB = −−→OB. Hence −→AB = −−→OB −→OA = v u, − − which is the answer. The reader should consult Figure 9 and reading this answer and then keep the whole picture in mind. We denote by 0 the zero vector. It has the property that 0 + v = v and v + 0 = v. Geometrically it is represented by a point. For any point A, we have −→AA = 0. Clearly the magnitude of the zero vector is zero: 0 = 0. | | Exercise 1. Let A, B, C, D be the vertices of a parallelogram; see Figure XXX above. (a) Write −−→BD in terms of −→AB and −−→AD. (b) Write −−→BD in terms of −−→CB and −−→CD. (c) Write −−→BC in terms of −→AB and −→AC. Exercise 2. Let A, B, C, D, E, F be the vertices of a regular hexagon and let O be its center; see Figure 10 below. Let a = −→OA and b = −−→OB. Write −→OC and −→AE in terms of a and b. Figure 10. Exercise 3. Let A, B, C, D, E, F, G, H be the vertices of a cube and let u = −→AB v = −−→AD and w = −→AE; see Figure 11 below. Write −→FC −→AC and −→AG in terms of u, v and w. 5 Figure 11. Besides addition, another operations in vector algebra is scalar multiplication, that is, multiplying vectors by scalars. Recall that a scalar is just a number, such as 2. A vector v multiplied by 2, denoted by 2v is in the same direction as v but its length is doubled. (You can tell that this is the most natural way to define 2v.) What about ( 2)v? Well, you can rewrite it as 2( v), that is, v multiplied by 2. Now v is − − − − obtained by turning v in the opposite direction. Now it is clear that the magnitude of ( 2)v is doubled and its direction is opposite to v. − Figure 12. In general, given a scalar a and a vector v, the scalar multiplication av of v by a can be defined. The magnitude of av is determined by av = a v ; (for example, | | | || | 2v = 2 v = 2v and ( 2)v = 2 v = 2v ). The direction of av depends on the | | | | | | | − | | − | | | sign of a: if a is positive, then av has the same direction as v, and if a is negative, then av has the oppsite direction. Certainly, if a =0 or v = 0, then av is the zero vector. The two operations in vector algebra, namely, addition and scalar multiplication, give the so–called linear structure for a space of vectors (or simply called a vector space). The 6 theory based on the conceptual frame work of linear structure is called linear algebra, which is a basic mathematical course for almost all diciplines which need mathematics. Exercise 4. Let ∆ABC be a equilateral triangle with O as its center, and let b = −→AB and c = −→AC; see figure 13 below. Write −→AO in terms of b and c. (Hint: notice that −→OA + −−→OB + −→OC = 0. Figure 13. Problem 5. (a) Draw a picture to check the identity (u + v)=( u)+( v). (b) − − − Draw a picture to convince yourself the validity of the identity a(bfu + v)= au + av for a positive scalar a. (c) Explain why the last identity also holds for negative a. Problem 6. (a) Convince yourself that if vectors v and w are parallel and if v is nonzero, then there is a scalar a such that w = av. Why do we need to assume that v is nonzero here? (b) Let e1 and e2 be two nonzero vectors in the Euclidean plane. Assume that these two vectors are not parallel to each other. Convince yourself by drawing that any planar vector v can be written in the form a1e1 + a2e2 for some scalars a1, a2. Why the scalars are uniquely determined by v? Next we use Cartesian coordinates to study vectors. The reader is assumed to be familiar with the Cartesian coordinate system. Let us recall some of its salient features. Normally we draw the x–axis horizontally and the y–axis vertically. The intersection of these two axis is denoted by O, called the origin of the coordinates system. Both axis are scaled so that each point of the plane corresponds to a unique (ordered) pair of real numbers, called the coordinates of this point. In the Figure 14 below we depict the point P with coordinates (3, 2) so that we may write P = (3, 2). 7 Figure 14. When we have u = −→OP with P = (3, 2), we also call (3, 2) the coordinated of u and we write u = (3, 2); see Figure 14 above. But now we have a poblem: when we write (3, 2), does it mean the point P or the vector −→OP ? Isn’t there some conflict with our notation? Our answer is, whether (3, 2) should be P or should be −→OP is only our subjective opinion, that is, we can interpret it in either way. All we need is to put a word in front we indicate our choice of interpretation, such as “point (3, 2)” or “vector (3, 2)”. The importance of using coordiantes for vectors lies in the fact that the two basic vector operations, addition and scalar multiplication, are carried out in the “coordinatewise manner”: If v1 =(x1, y1) and v2 =(x2, y2), then v1 + v2 =(x1 + x2, y1 + y2) and av1 =(ax1, ay1). (Here a is any scalar.) Instead of giving a routine but pretty boring proof os this basic statement, we let you check its validity for some special cases to convince you the validity of it in the following easy exercises. Exercise 7. In each of the following cases, use graphic paper to draw the vector −→AB and check the identity −→AB = −−→OB −→OA: (a) A = (1, 2), B = (3, 5); (b) A = ( 2, 5), − − B = (3, 2); (c) A =( 1, 2), B = (1, 2); (d) A =( 3, 2), B =( 1, 4). − − − − − − Exercise 8. (a) Find the midpoint M between P ( 5, 4) and Q(9, 6). (b) If M(2, 3) − is the midpoint of P (1, 1), what is Q? Use graphic paper to check your answer in both parts. 8 Exercise 9.. Let A B and C be arbitrary points and let M be the midpoint of AB and N be the midpoint of BC. Express the vector −−→MN as a linear combination of u = −→AB and v = −−→BC.