UNIT 10 VECTOR ALGEBRA
Structure 10.1 Introduction Objectives 10.2 Basic Concepts 10.3 Components of a Vector 10.4 Operations on vectors 10.4.1 Addition of Vectors 10.4.2 Properties of Vector Addition 10.4.3 Multiplication of Vectors by S&lars 10.5 Product of Two Vectors 10.5.1 Scalar or Dot Product 10.5.2 Vector Product of Two Vectors 10.6 Multiple Products of Vectors 10.6.1 Scalar Triple Product 10.6.2 Vector Triple Product 10.6.3 Quadruple Product of Vectors 10.7 Polar and Axial Vectors 10.8 Coordinate Systems for Space 10.8.1 Cartesian Coordinates 10.8.2 Cylindrical Coordinates 10.8.3 Spherical Coordinates 10.9 Summary
1 0.1 INTKODUCTION
Vectors are used extensively in almost all branches of physics, mathematics and engineering. The usefulness of vectors in engineering mathematics results from the fact that many physical quantities-for example velocity of a body, the forces acting on a body, linear and angular momentum of a body, magnetic and electrostatic forces may be represented by vectors. In several respects, the rules of vector calculations are as simple as the rules governing thesystem of real numbers. It is true that any problem that can be solved by the use of vectors can also be treated by non-vectorial methods, but vectol analysis simplifies many calculations considerably. Further more, it is a way of visualizing physical and geometrical quantities and relations between them. For all these reasons, extensive use is made of vector notation in modern engineering literature. It is a very useful tool in the hands of scientists and engineers. Many of you may have studied vectors at school level while some of you might not have done so. This unit has been developed keeping inview the interest of all the students. We have listed important results of vector algebra. The proofs of some of elementary properties have been omitted. Interested learner may look up for the proofs in the books recommended for your further study, which are available at your study centres. We have begun this unit by giving basic definitions in Sec. 10.2. The operations on vectors, such as addition and subtraction of vectors, multiplication of a vector by a scalar, etc. have been taken up in Sec. 10.3. We have devoted Sec. 10.4 to the representation of the vectors in component forms. Sec. 10.5 has been used to discuss vector products. We have given the geometrical and physical interpretations, wherever possible. In the following units we shall take up differentiation and integration of vectors. vector Calculus Objectives After studying this unit you should be able to * distinguish between scalars and vectors, define a null (or zero) vector, a unit vector, negative of a vector and equality of vectors, identify coinitial vectors, like and unlike vectors, free and localized vectors, coplanar and co-linear vectors, add and subtract vectors, graphically and analytically, multiply a vector by a scalar, define the system of linearly independent and dependent vectors, compute scalar and vector products of two vectors and give their geometrical interpretation , compute the scalar triple products and vector triple products and give their . geometrical interpretation , compute quadruple product of vectors, and solve problems on the application of vector algebra. 10.2 BASIC CONCEPTS
In physics, geometry or engineering mathematics, we come across physical quantities such as mass of a body, the charge of an electron, the spccific heat of water, the resistance of a resistor, the diameter of a circle and the volume of a cube. With a suitable choice of units of measure, each of these quantities is described by a single number. Such a quantity is called a scalar. Length, temperature, time, density, frequency are some other familiar examples of scalars. Thus we may define a scalar as. follons:
A phydical quantity is called a scalar if it can be completely specified by a single number (with a suitable choice of units of measure). On the other hand physical quantities like velocity, acceleration, displacement, momentum, force, electric field intensity, etc. are entities which cannot be specified by a single number. They require for their complete characterization the specifilcation of a direction as well as a magnitude. Velocity u
Trajectory of Earth
fForce
Figure 10.1: Force and Velocity Figure 10.1 shows the force of attraction for the earth's motion around the sun. The instantaneous velocTty of the earth may be represented by an arrow of suitable length and direction. This illustrates that velocity is characterized by a magnitude and a direction. Figure 10.2: Displacement Figure 10.2 shows a displacement (without rotation) of a triangle in the plane. We may represent this displacement graphically by a directed linc segment whose initialpoint is the original position of a point P of the given triangle and where terminalpoint is the new position Q of that point after displacement.
Definition : Quantities which are specified by a magnitude and a directiort are called vector quantities. We know that a directed line segment is a line segment with an arrow-head showing direction (Figure 10.3).
Figure 10.3 Directed line segment AB A directed line segment is characterized by (i) Length : Length of directed line segment AB is the length of line segment AB (ii) Support : The support of a directed line segmentAB is the line P of infinite length of which AB is a portion. It is also called line of action of that directed line segment
(iii) Sense : The sense of a directed line segment ABis from its tail or the initial pointA to its head or the terminalpoint B. Thus we,may also define a vector as follows. A directed line segment is called a vector. Its length is called the length or Euclidean norm or magnitude of the vector and its direction iv called the direction of the vector. In Figure 10.3, the direction of directed lineAB is fromA to B. The two endpoints of a directed line segments are not interchangeable and you must think of directed line segments AB and BA as different. I From the definition of a vector, we see that a vector may be translated (displaced without rotation) or, in other words, its initial point may be chosen in an arbitrary fashion. Once we choose a certain point as the initial point of a given vector, the terminal point of the vector is uniquely determined. For the sake of completeness, we mention that in physics there are situations where we want to impose $he restrictions on the initial point of a vector. For example, in mechanics a force acting on a rigid body may be applied to any point of the body. This suggests the concept of sliding vector, i.e., " a vector where initial point can be Vector Calculus any point on a straight line which isparallel to the vector. " However, a force acting on an elastic body is a vector where initial point cannot be changed at all. In fact, if we choose another point of application of the force, the effect of the force will in general be different. This suggests the notion of a bounded-vector, i.e., 'a vector having a certain fied initial point (orpoint of application). ' When there is no restriction to choose the initial point of a vector, it is called afree vector. When there is restriction on the choice of a certain point as the initial point of a vector, then it is called a localized vector. Throughout the block, we shall denote vectors by bold face letters, e.g., v, a, F, etc. ** You may also denote vectors by drawing arrows above them, e.g., v, a, For by drawing a !ine (straight or curly) below them, e.g., 5 g, F or v,--- a, F. The magnitude of vector v is denoted by I v 1, called modulus of v or by v (a light lettcr in itali~).In diagrams we shall show vectors as straight lines with arrowheads on them (as in Figure 10.3) and denote directed line segment AB as AB. Before taking up the algebra of vectors, we state some definitions related to vcctors. Zero Vector or Null Vector : A vector whose length is zero is called a Null or Zero Vector and is den~tedby 0. Clearly, a null vector has no direction. Thus a = AB is a null vector if and only if [ a I = 0, i.e., if and only if I AB 1 = 0, i.e., if and only ifA and B coincide Any non-zero vector is called a proper vector. Unit Vector :A vector where length (modulus or magnitude) 1s rtr~ityis called a unit vector. Generally, a unit vector is denoted by a single letter with a cap '^' over it. Thus $ denotes a unit vector -''a Againa = Ia(a or a = -. la I Co-initial Vectors :All vectors having the same initial poinl are called co-initial vectors Hence AB, AC, AD are all co~initialvectors. Like and L'nlike. Vectors :Vectors are said to be like $hey have the same directioh and unlike if they have opposite directions. (Figure 10.4)
Figure 10.4(a): Like Vectors Figure 10.qb): Unlike Vectors Figure 10.4(c) : Vectors having dif- ferent directions Both unlike and like vectors have the same line of action or have the lines of action parallel to one another; such vectors are also called Collinear or Parallel Vectors. 1 Coplanar Vectors :Vectors are said to be coplanar if they are parallel to the same plane or they lie in the same plane. Negative of a Vector :A vector whose magnitude is the same as that of the given vector a but has the direction opposite to that of a is called the negative of a and is denoted by - a. I Thus if AB represents the vector a, then BA represents the vector -a. It is evident that ( a I - ( -a I. Equal Vectors :Two vectors are said to be equal if they have Vector Algebra I (i) the same length I (ii) the same or parallel supports (iii) the same sense. I In Figure 10.5 the three vectors a, b, c represented by the directed line Segments
. F~gure10.5: Equal Vectors. AB, CD, EF respectively have different initial and terminal points but the same length, the same or parallel supports and the same sense and hence are equal. We write a = b - corAB = CD =EF. Thus equal vectors may be represented by directed segments of equal length in the same sense along the same or parallel supports. We may remark here that if AB = CD and AB and CD do not lie along the same line, then it is evident that ABCD is a parallelogram.
I Further, two vectors cannot be equal if (i) they have different magnitudes, or
1 I (ii) they have inclined supports, or I (iii) they have different senses. Let us consider an example.
Example 1 : .I 1 ABCDEF is a regular hexagon. If P = AB, Q = BC and R = CD, name the vectors represented by AF, ED and FE.
. Solution : , Since ABCDEF is a regular hexagon,
Figure 10.6: A regular hexagon.
Also AB (1 ED, BC 11 FE, and CD IIAF. Vector Calculus Further sense of AB-isthe same as that of ED, sense of BC is the same as that of FE and sense of CD is the same as that of AF. :. AB = ED, BC - FE and CD = AF Thus ED = P, FA!? = QandAF = R. So far We have introduced vectors geometrically (using the notion of a directed line segment). We have defined vectors without referring them to any coordinate system. A point in three-dimensional space is a geometric object, hut if we introduce a coordinate system, we may describe it by (or even regard it as) an ordered triple of numbers (called its coordinates). Similarly if we use a coordinate system, we may describe vectors in algebraic terms. This alternative way to represent the vectors is also termed as analytic approach. In thc next section we shall discuss how to express a vector analytically, i.e., in terms of its components. We shall also give a new and precise way of defining vectors which will be of practical use to you in the study of your other courses also.
10.3 COMPONENTS OF A VECTOR
Let us introduce a coordinate system in space whosc axes are three mutually perpendicular straight lines. On all the three axes, we choose thc s;imc scale. Then the three unit points on the axes, whose coordinates are (I,(),(I), (0, 1.0) and (0,0, -1Lhive the same distance from the origin, the point of intersection ol' the axes. The rectangular coordinate system thus obtained is called a Cartesian coordinate system in space (refer Figure 10.7(a)).
I z ------./.Q -- (O,O,l) rJ"
1...- p.. , , ,
ia2- I b - ,,'" y (0,l.O) Y i ; ,' ... I .,' ... .I.. I' . .. : ,*' .._ ~ , . ----.>:...>,. ',,:'
X 0 X ' i1
Figure 10.7(a): Cartesian Coordinate System. Figure 10.7(b): Componenrsof a vector. - 1 We now introduce a vector a obtained by directing a line segment PQ such that Pis the initial point and Q is the terminal point (Figure 10.7(b)). Let ( XI, yl, zl ) and ( x2, yi, 22 ) be the coordinates of the point P and Q respectively. Then the numbers
are called the components of the vector a with respect to that coordinate system. By deifinition, the length or magnitude ( a I of the vector a is the distance and from equation (10.1) and the theorem of Pythagoras, it follows that
(a(- ,/- al+a2+a3 (10.2)
For hstance, the vector a with initial point R(3, 1,4) and terminal point Q:(l,-2,4)hasthecomponentsaL= 1-3 = -2, a2 = -2-1 = -3, a3 - 4 - 4 - 0 and the length Conversely, if ahas the components - 2, - 3 and 0 and if we choose the initial point of aas ( - 1,5,8 ) then the corresponding terminal point is ( - 1 - 2,5 - 3,8 + 0 ), i.e., ( - 3, 2, 8 ). You may observe from equation (10.1) that if we choose the initial point of a vector to be the origin, then its components are equal to the coordinates of the terminal L point and the vector is then called the position vector of the terminal point (with respect to our coordinate system) and is usually denoted by r (refer Figure 10.8).
Figure 10.8: Position Vecror rof a pointA (.r,y, 2).
We see that r=OA
= OM+mo+~ooK r A = xr + yJ+ zt AAA where i,j, k are unit vectors parallel to the axes of x, y, zrespcctively. AfiA The vectors i,J, k are mutually perpendicular. From relations (10.1), we can also see that the components al, a2, a3 of the vector a are independent of the choice of initial point of a. This is because if we translate (displace without rotation) athen corresponding coordinates of P and Q are altered by the same amount. Hence, given a fixed Cartesian coordinate system, each vector is uniquely I. determined by the ordered triple of its components w.r. to that coordinate system. 1 I We may introduce the null vector or zero vector 0 as the vector with components 0, 0,o. Further two vectors aand b are equal if and only if corresponding components of I these vectors are equal. Consequently, a vector equation. a-b
is equivalent to three equations for the components of aand 4, j.e.,
, al = bl, a2 = h, a3 = b3, where the components a~,a2, a3 of aand bl, b2, b3 of b refer to the same Cartesian I I coordinate system.
From equation (10.1), you may note that al, a2, a; are tlie projections of aas i A A t L a = ali^+a2j+a3k I Vector Calculus In Figure 10.7(b), if a,p, y are ,the angles which PQ makes with the three axes, then
a1 = 1 a ( cos a, a2 = ( a ( cos p, a3 = I a I cos y The three angles a, p, y which any vector makes with the three coordinate axes are called direction angles and the cosines of these angles are called direction cosines. Let us now consider an example.
Exampler 2 : If cos a, cos p, cos y are the direction cosines of r, show that
r 9 A A r = - = 1.1 cosar+cospj+cosyk.
Solution : Let the position vector of a point with coordinates (x, y, z) be r, so that x, y, z are the projections of r 01; the coordinate axes.
AAA ..' Wehave r = xi+yj+tk
Also x = I r 1 cos a,y = 1 r 1 cos P, z = I r ( cos y A A A ... r = IrIcosai+IrJcospj+IrIcosyk
r h A A => - = r" = cosai+cospj+cosyk. 1.1 You may now try the following exercises
E 1. If P (3. - 2, 1) and Q (1,2, - 4) are the initial and terminal points respectively of a vector a, find the components of a and 1 a I.
13 If -, 1, - are the components of a vector a and P 1, is a particular 2 2 (- i, $1 initial point of a, find the corresponding terminal point and the length of a.
We now introduce algebraic operations for vectors in the next section. These operations help us to do various calculations with the vectors. 10.4 OPERATIONS ON VECTORS
By 'algebraic operations on vectors', we mean various ways of combining vectors and scalars, satisfying different laws, called laws of calculations. Let us take this one by one. 10.4.1 Addition of Vectors Vector Algebra The motivation for addition of two vectors is provided by displacements. Also, experiments show that the resultant of two forces can be determined by the familiar parallelogram law or by the triangle law. We now define addition of vectors and their properties. Let a and b be two vectors. Let the vector a be the directed line segment AB and the vector b be the directed segment BC (so that the terminal point B of a is the initial point of b) (see Figure 10.9).Then the directed line segment AC (i.e, AC) represents the sum (or resultant) of a and b and is written as a + b.
Figure 10.9: Addillon of two Vectors.
ThusAC = AB+BC = a+b. The method of drawing a triangle in order to define the vector sum ( a + b ) is balled triangle law of addition oftwo vectors, which states as follows
taken in the reverse
Since any side of a triangle is less than the sum of the other two sides of the triangle; hence modulus of AC is less than the sum of moduli of AB and BC. It may be noted that the vector sum does not depend upon the choice of initial point of the vector as can be seen from Figure 10.10.
Figure 10.10: Vwtor sum is independent of initial point. If in some fixed coordinate system, a has the components al, a2, a3 and bhas the compolhents bl, h,b3, then the components cl, c2, c3 of the sum vector c - a + b are obtained by the addition of corresponding components of a and b ;thus
This fact is represented in Figure 10.11 in the case of plane. In space the situation is similar.
I - - ?. ------I I I I T I I " 2 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I jt-I- b1 -; - 0 xb 4 1 b
Figure 10.11: Vector addition in terms of components in a plane.
10.4.2 Properties of Vector Addition From the definition of vector addition and using relation (10.3), it can be shown that vector addition has the following properties. (i) Vector addition is commutative: [faand bare any two vectors, then a+b = b+a
C a B From equations (10.4) and (10.5), we have
a+b = b+a
(ii) Vector addition is associative: If a, b, c are any three vectors,'then
a+(b+c) = (a+b)+c
The above equation can be easily verified from Figure 10.13. Note that the sum of three vectors a, b, c is independent of the order in which they are added and is written as
Figure 10.13: Associative Law of Vector Addition. (iii) Existence ofAdditive Identity: For any vector a,
a+O = a = O+a,
where 0 is a null (or zero) vector. I i . Thus 0 is called additive identity of vector addition. (iv) Existence ofAdditive Inverse: For any vector a, there exists another . vector4 such that f a+(-a) = 0, where -a denotes the vector having the length I a 1 and the direction opposite to that of a. In view of the above property, the vector ( -a ) is called the additive inverse of vector h. t Let us take an example from geometry to illustrate the use of vector addition. 1 Example 3 :
Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero. I Solution : k In A ABC, AD, BE, and CF are the median. Vector Calculus
Figure 10.14: Medians of a triangle Now AD+BE+CF = (AB+BD)+(BC+CE)+(CA+AF)
By triangle law of addition, AB + BC = AC
From equations (10.6)and (l0.7), we get AD+BE+CF
You may now attempt the following exercise
Show that the sum of vectors represented by the sides AB and DC of any quadrilateral ABCD is equal to the sum if the vectors represented by the diagonals AC and DB. ' It may be mentioned that associative law of vector addition is true even if PQRS,as shown in Figure 10.13, is a skew quadrilateral (i.e., when a, b, c are non-coplanar , vectors). Further, instead of a + a, we also write 2a. This and the notioq-a suggests that we define the second algebraic operation for vectors viz, the multiplication of a vector by an arbitrary real number (called a scalar). I 10.4.3 Multiplication of Vectors by Scalars Definition : Let a be any vector and rn be any given scalar. Then the vector m a (product of vector a and scalar m> is a vector whose (i) magnitude 1 ma 1 = I m 1 . 1 a 1
L (ii) support is the same or parallel to that of support of a if a z 0 and m > 0 L and (iii) ma has the direction of vector a if a z 0 and m > 0 and ma has the direction opposite to vector a if a z 0 and m < 0. Further, if a = 0 or rn = 0 ( or both ), then ma = 0. Geometrically, we can represent ma as follows: Let AB = a, then AC = ma if m > 0. Here we choose the point C on AB on the same side of A as B (see Figure 10.15(a)).
r - A a B C Figure 10.15 (a) Now if m < 0 and AB = a, then AC = ma where we have choosen the point C on AB on the side ofA opposite to that of B. (see Figure 10.15 (b)),
Figure 10.15 (b) Further, if a has the component al, aa as, then ma has the components mal, maz, ma3, (w.r. to the same coordinate system). L Multiplication of a vector by a scalar helps us to define linearly dependent and independent vectors, which we take up next. Linearly Dependent and Independent Vectors
Two non-zero vectors a and b are said to be linearly dependent if there exists a scalar t ( z 0 ) such that a - tb Thus two vectors can be linearly dependent if and only if the vectors a and b are parallel. Further, if a = AB and b = BC, then a and b are linearly dependent if and only if A,B,C lie in a straight line. If the vectors a and b are not linearly dependent, they are said to be linearly independeni and in this case a and b are not parallel vectors. I From the definition, we have the following properties of multiplication of a vector by a scalar. vector Calculus (I) m(a+b) = ma+mb
Figure 10.16 Mere OA' = ma and A'B' = mh ... OB' = ma + mb
Also OB' = m(a+b) (11) (m + n) a = ma + na (Distributive law) (111) m ( n a ) = ( mn ) a = mna ( Associative Law ) (IV) la = a (Existence of multiplicative identity) (V) Oa = 0
(VI) (-l)a = -a
A Also if a is the unit vector having the same direction as a,then
A a a = la(a and a = -. la I Let us take up some examples to illustrate the above properties.
Example 4 :
If a is a non-zero vector, find a scalar h such that 1 h a ( = 1.
Solution : We have to determine h such that
(ha1 = 1
3 (h((a1= 1
1 3 1 h I - ( ... a 1s non-zero, :. 1 aI z 0 ) - 1.1
1 h = - the + signu is to be taken when h r 0 and la I the -sign is to be taken when h < 0
Example 5 :
A Ah A A A Shpw that the vectors a = i + 2j + k, b = 3k and c = 2i + 4j constitute a linearly dependent set. Solution : Veclor Algebra
A AA A = 2(i+2j+k)-2k
A A 2 A = 2(i+2j+k)--.3k 3 2 = 2a--b 3
Hence the vectors a, b, c , as given here, constitute a linearly dependent set. We can now define the difference of two vectors Difference of Two Vectors
The difference a - b of two vectors is defined as the sum a + ( - b ), where ( - b ) lhe negative of b. We can geometrically represent the difference a - b as in Figure
I I I\ cqident that
a-a = 0
and a+O = a.
Figure 10.17: D~fferenceof two Vectors If two vectors are given in their component forms then to obtain their difference, subtract the vectors component wise.
A A A A A A For example, if a = a1 i + a2j +f3 k, and b =A bl i + b2j + bz k, then C a-b = (a1-b1)i+(a2-b2)j+(a~-b~)k Let us take a few examples.
If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is 6.
Solution : Let OA and AB be two unit vectors and b^. Then by triangle law of addition,
A A a+b = OB Vector Calculus
We are given . . OA = AB = OB - 1 A A A Let AC = -b.ThenAC = IACI = I-bl = (bl = 1 Since OA - AB = AC, then by geometry A BOC is a right-anglcd triangle, with LBOC = rc/2 AA A Nowa- b = a+(-;) = OA+AC = OC
Now kc2 = O@
Example 7 :
What is the eeometrical significance of the relation I a + b I = I a - b 1 ?
Solution : Let a = AB We complete the parallelogram ABCD having AB and AD as adjacent sides. Let us draw the two diagonals AC and BD also. DB = DA +AB
= -.AD+AB - -b+a - a-b :. la-b1 - lDBl We are given
+ diagonals of the parallelogram are equal
=+ parallelogram ABCD is a rectangle and hence a is perpendicular to b.
You may now attempt the following exercises:
E 4. Prove that ci> la+bl s laI+Jbl (ii) 1a1-lbl s la-bl
E5. Show that the vectors
a - 3&- 3i b = 6ic 3&+ 3j form the sides of an equilateral triangle. E6. Show that the three points with position vectors c A A A 2f+ 3~3i+ and 5i + 0.75j are collinear. 4
E7. Show that the set of vectors 5a + 66 + 7c, 7a - 86 + 9c, 3a + 20 b + 5c are coplanar.
So far we have defined addition and subtraction of vectors as well as multiplication of vectors by scalars. We shall now introduce multiplication of vectors by vectors. 10.5 PRODUCT OF TWO VECTORS
When ane vector is multiplied with another vector, result can be a scalar or a vector. There are in general two differentways in which vectors can be multiplied. These are the $culaF or dot or inner product which is a mere number (or scalar) having magnitude alone and the other is ealled vector or cross product, which is a vector having a definite direction. We shall now take up these two products one by one.
I 10.5.1 Scalar or Dot Product The scalar or dot or inner product of two vectors a and b in the three-dimensional space is written as a .b (read as 'a dot b') and is defined as
when a - 0 or b - 0
/
-b b > 0 a.6 = 0 Vector Algebra where y ( 0 s y 5 rc ) is the angle between a and b (computed when the vectors have their initial points coinciding) (Refer Figure 10.20). The value of the dot product is a scalar (a real number), and this motivates the term "scalar product". Since the cosine in equation (10.8) may be positive, zero or negative, the same is true for the dot product. Angle y in equation (10.8) lies between 0 and rc and we know that cos y = 0, if and only if y = rc/2, we thus have the following important result: Two non-zero vectors are orthogonal (perpendicular) if and only if their dot product is zero. If we put b = a in equation (10.8), we have a .a = 1 a* ( and this shows that the length (or Euclinear norm or modulus or magnitude) of a vector can be written in terms of scalar product as (a1 ==(t 0). (10.9) From relations (10.8) and (10.9), we obtain the angle y between two non-zero vectors as a.b a.b cos y 5 -- 1alI.bl - fim The scalar product has the following properties: P(1) ( qla + q2b ) . c - qp . c + q2b . c (linearity) P(2) a . b = b . a (symmetry or commutative law) P(3) a.az0 ( positive definiteness ) Also a.a = 0 if and only if a = 0
In property P(l), with ql = 1 and 92 = 1, we have P(4) (a + b ) . c = a .c + b . c (distributivity) Hence scalar product is commutative and distributive with respect to vector addition. From the definition of scalar product, we get
P(5) 1 a . b 1 5 1 a 1 ( b 1 ( -.. I cos y 1 5 1 )(Schwarz Inequality) Also using the definition and simplifying, we get
P(6) 1 a + b l2 + ( a - b l2 = 2 ( 1 a l2 + I b \2 ) (parallelogram equality) AAA Further, if i, j, k are unit vectors forming an orthogonal triad, then, from definition of scalar product, we have
If vectors a and b are represented in terms of components, say,
A A A AAA a = ali+a2j+a3k and b = bli+b;?j+b3k, then their scalar product is given by the formula
/ a . b = alb~+ a2b2 + a3b3 (using W)
( ah+a2b2 + a363 ) andcosy = (a.b/(a(IbI) = ), where y is the (dal+a2+as b:+bi+b: angle between a and b. Before we take up some applications of scalar products, we give below the geometrical interpretation of scalar product of two vectors. Geometrical Interpretation of Dot Product The scalar product of two vectors is the product of the modulus of either vector and the resolution,(projection) of the other in its direction.
Figure 10.21(a) :Projection of a in the direction of b Figure 10.21(b) : Projection of b in the direction of a Let OA = a, OB = b and LBOA = y,
From the point B draw BC perpendicular on OA(Figure 10.21(b)) and from the point A, draw AD perpendicular on OB (Figure 10.21(a)) :. OC = Projection of OB on OA = OB cos y = ) b I cos y and OD = Projection of OA on OB = OA cos y = I a I cos y Thusa.b - IaIIb(cosy - IaI()bIcosy) = (aI(projectionofbinthe direction of a).
A1soa.b = )a)(b)cosy= )b)(IaIcosy)= IbI(pro-jectionofainthe direction of b). Hence the result. , ,/ ' a.b Remember that ifa and b ar& vectors and a * 0, then p = 1 b cos y = - I la1 called the component of b in the direction of a, or the projection of b in the direction of a, where y is the angle between a and b. If a = 0, then y is undefined and we setp = 0. It follows that Ip ) is the length of the orthogonal projection of b on a straight line 1 in the direction of a. Here p may be positive, zero or negative. See Figure 10.22 below.
Figure 10.22: Components of b in the direction of a
In particular, if a is a unit vector, then we simply have p = a.b The following examples illustrate the applications of dot product.
Give a representation of work done by a force in terms of scalar product.
Solution : Consider a particle P on which a constant force a acts. Let the particle be given a displacement d by the application of this force. Then the work done W by a in this displacement is defined as the product of ( d 1 and the component of a in the direction of d, i.e., W = (aI(Id(cosa)= a.d, where a is the angle between a and d.
P d Figure. 10.23: work done by a force
Find a representation of the straight line ll through the point Pin the xy-plane and perpendicular to the line 12 represented by x - 2y + 2 = 0.
Solution : Any straight line 11 in the xy-plane can be represented in the form alx + bly - c. If c = 0, then 11 passes through the origin. If c + 0, then alx + bly = 0 represents a line 11' through the orign and parallel to 11. The fi position vector of a point on the line 11' is r = x i + y J. A A If we introduce the vector a = a1 i + a2j, then, by the definition of dot product, we can give a representation of 11' as a.r - 0
Certainly a 0 and vector a is perpendicular to rand, therefore, perpendicular to the line 1'1. It is called normal vector to 1'1.
b X Figure 10.24: Normal lines Since 11 and 1'1 are parallel lines, thus a is also normal to line 11.
Hence two lines 11 and 12 are perpendicular OF orthogonal if and only if their normal vectors say a and b are orthogonal, i.e., a .b - 0 (see Figure 10.24). You may note that this implies that the slopes of the lines are negative reciprocals. e A Vector Calculus For the give: line l2 the normal vector is b = a - 3 and a vector perpendicular to b is a - 2i +j. Hence the representation of line 11 must be of the form 2r + y = c. The value of c can be obtained by substituting the coordinate of P in this repre$entation.
Example 10 : Find the projection of bon the line of a if a = ?+;+ f and b - 2E+ 4;+ 51 Solution : Here a. b - 1.2 + 1.4 + 1.5 - 2 + 4 + 5 - 11
:. Projection of bon the line of a a.b 11 116
Example 11 : Show by vector method that the diagonals of a rhombus are at right angles.
Solution : Let ABCD be a rhombus. Let A be taken on the origin. Let band dbe the position vectors of vertices B and D respectively referred to the origin A (ref. Figure 10.25) Then the position vector of C (using triangle or parallelogram law of addition) is b+ d. Also BD * d-b Now ABCD is a rhombus.
A (Origin) b
Figure 10.25: NowAC - b+d. ... AC.BD - (b+d).(d-b) - (d+b).(d-b) - 8-b2 Hence AC is perpendicular to BD Vector Algebra Thus diagonals AC and BD of the rhombus ABCD are at right angles. You may try the following exercises
E 8. Prove by vector method that the angle in a semi-circle is a right angle
E 9. AAA Afi A A particle is acted on by constant forces - 3i + 2j + 5k and 2i + J - 3k and is displaced from the point ( 2, - 1, - 3 ) to the point ( 4, - 3,7 ). Find the total work done by the forces.
E 10. AAA AAA Vectors a and b are given by a = 3i - 8 + k and b = - 2i + 2j + 4k. Find (i) magnitudes of a and b and the angle between them (ii) projection of the vector ( a + -i) b onto a (iii) Which of the following vectors are perpendicular to a ? AAA AA AAA c = -i-4j+2k, d = -3i+k, e = 2i+2j-2.k.
Dot multiplication of two vectors gives the product as a scalar. Various applications suggest another kind of multiplication of vectors such that the product is again a vector. We next take up such vector product or cross product between two vectors. Cross products play an important role in the study of eledricity and magnetism. Vector Calculus 10.5.2 Vector Product of Two Vectors The vectorproduct or the crossproduct of two vectors a and b, denoted by a x b, is &fined as axb = IaIIblsin€~;, (10.10)
Figure 10.26: Vector product where I a ( and I b I are the magnitudes of the vectors a and b respectively, 8 k the angle between wctors a and b and ;; is a unit wctorpeipndicular to both a and b and is such that a, b, i,in tltk order, form a right-Iunded triple or right- Iund hiod (see Figure 10.26). The term right-handed comes from the, fact that the vectors a, b, i,in this order, assume the same sort of orientation as the thumb, index finger, and middle finger of right hand when these are held as shown in Figure 10.27(a). Let us now look at the screw shown in Figure 10.27(b). You may observe that if a is rotated in the
Right* handed screw
Figure. 10.27(a): Right-handed triple of vectors a, h, A Figure. 10.27(b): Rhght-handed screw.
directian of b through an angle 8 ( < n ), then ;; advances in a direction pointing towards the reader or away from the reader according as the screw is right-handed or left-handed. On the same account an ordered vector triad a, b, is right-handed or left-handed. .You may note here that (i) OS~SJC (ii) is perpendicular to the plane which contains a and b both (iii) vector product of two vectors is always a vector quantity (iv) The sine of the angle between two vectors aand bis given by Vector Algebra sin0 = (axb)/()al.Ibl)
Vector product a x b, is also called the cross product because of the notation used and is read as across b.
From the definition of vector product, you know that bx ais a vector of magnitude ( a I 1 bI sin 0 and is normal to aand band in a direction such that b, aand b x a from a right-handed system (see Figure 10.28). This is possible only if b x a is opposite to the direction of a x b. Since bx a and a x bhave equal magnitudes,
bxa = -axb,
a xb
i b Xa
F~gure10.28 which shows that the cross-product of vectors is not commutative. Hence the order of the factors in a vector product is of great importance and must be carefully observed. It may be observed that the unit vector normal to both aand b, namely, is given by
nn-A axb la4 If the vectors a and bare parallel (or collinear) then 0 - 0 or 180" sin 0 - 0 Hence a x b = 0 is the condition for the two vectors aand bto be parallel.
In particular a x a = 0 C
AAA You know that i,j, krepresent unit vectors along the axes of a artesian coordinate system. Also since [ig in this order, form a right-handed system of mutually perpendicular vector:, therefore, fx jA is a vector having modulus as unity and direction parallel to k.
c"' c' A A Similarly jxk-r--kxj
AA A A kxi-J =-ixk
fir AA 1x1 = JXJ = kxk = 0. From the definition of cross product, it follows that for any constant A (Aa)xb= A{axb) = ax(hb) Vector Calculus Further, cross-multiplication is distributive w.r.to vector addition, i.e.,
ax(b+c) = axb+axc
(a+b)xc = axc+bxc. Cross multiplication has a very unusual and important property, namely, Cross mubiplication is not associative, i.e., in general
ax(bxc) t (axb)xc. Ce AA A For instance, rx(ixj) = ixk = -j, Cc' c' A whereas (1xr)xj = Oxj = 0 From the definition of cross-product and dot product of two vectors, we have
)axbI2= IaI2)b(2sin28
= (a)2)b)2(1-cos28)
= 1~~(b~-[)~11b)~~~8;~J
= (a.a)(b.b)-(a.b)2
From the above identity, we obtain a useful formula for the modulus of a vector product as
laxbl = \/(a.a)(b.b)-(a.b)'
Befare taking up geometrical interpretation of vector product we list all the properties of vector product discussed above for the ready reference.
PC1 : axb = -bxa, i. e., vector product is not commutative.
PC2 : a x b = ( 1 a I ( b I sin 0 ) ;I = 0, if a is Parallel to b. Vectpr product of parallel vectors is zero. In particular, a x a = 0 and a x 0 = 0
PC3 : Ifmandnarescalarsthen(rnaxnb) = rnn(axb)
A A A PC4 : For a right handed triad of unit vectors i, j, k fifi fifi AA 1x1 = JXJ = k~k= 0 C fi A3 A fiA C rxj = k,jxk = r,kxr = j AA p 3 C c'C A c' JXl = -k,kxJ = -1,lXk = -J
PC5 : Vector product is distributive w.r.t vector addition, i.e., a.x(b+c) = axb+axc
PC6 : ax(bxc) t (axb)xc i. e..., vector product is not associative. We next take up geometrical interpretation of vector product. Geometrical Interpretation of Vector Product Consider a parallelogram OACB with a and b as adjacent sides.
Let a = OA, b = OB Vector Algebra
and let BN be perpendicular to OA from B Figure 10.29 Let LBON = 8 :. BN - OB sin 8
Now Iaxb( = (a((b(sin8 = OA . BN - Area of the parallclogram OACB.
Thus, the magnitude of a x b is equal to the area of the parallelogram whose adjacent sides are the vectors a and b. The sign may also be assigned to the area. When a person travels along the boundary and the area lies to his left side, the area is positive and if the area lies to his right side, then the area is negative. Thus a x b = vector area of the parallelogram OACB and b x a = vector area of the parallelogram OBCA We shall now represent a vector product in terms of the components of its factors with respect to a Cartesian coordinate system. In this connection it is important to ' note that there are two types of such systems, depending on the orientation of the axes, namely, right-handed and left-handed. Vector Product in Terms of Components We shall first define right-handed and left-handed triples; yz call a Cartesian coordinate system to be right-handed if the unit vectors i, j,.k in the directions of positive x, y, z-axes fo~~aiight-handedtriple (Figure 10.30(a) ;and it is called lefr-handed, if vectors i, j, k form a left-handed triple. In applications, we usually consider a right-handed system. Let us consider 3 right-handed Cartesian coordinate system and let al, a2, a3 and bl, b2, b3 be the components of two vectors a and b respectively, so that A $ A A A A a = ali+a2j+a3kandb = bli+b2j+b3k A :. axb = (ali+azf+a3i)x(bl?+b2f+b3i)
= a1bl(?x?)+a1b2(~xf)+al~(ix~) + a2 bl (3~i^) t a2 b? (ixp) + a2 b3 (3~i ) +a3b1(Ex~)+a3h(PxT)+a3b3(~~P)
A A = o+al bzk-alb3j A A 5 e - a2blk+~+azb3i+a3bl~-a3b21+O(UsingPC4) A A vector Calculus a oxb = (a2&-a3b2)i+(a3b1-alb3)j^+(alb2-a2bl)k,orinkmsof second-order determinants,
Figure. 10.3qa): Right- handed system Figure. 10.30@):Left-handed system This enables us to obtain the cross product by expanding the third order determinant viz.
by its first row. Remember it is not an ordinary determinant as the elements of the first row are vectors and it must be expanded only by the first row. i A A In a left-handed Cartesian coordinate system, i x j = - i (Figure 10.30@)), and other similar expression lead to i
Linear dependence or independence of two vectors can also be tested by their cross product. We say that 'Two vectorsform a linearly dependent set if and only iftheir vector product is zero'.
Cross product has many applications. We shall now take up a few examples to I illustrate them.
Example 12 : Express Velocity of a rotating body as a vector product.
Solution : A roaion of a rigid body B in space can be simply and uniquely described by a vector o.The direction of o is that of axis of rotation of the body and such that A the rotation appears clockwise, if one looks from the initial point of o to its terminal point. The magnitude of o is equal to the angular speed o ( > 0 ) of the rotation. 4 The angular speed rotation of a body is the linear sped of a joint of the body divided by its distance from the axes of rotation.
Figure 10.31: Rotation of a rigid body Now let P be any point of body B and let d be its distance from the axis. Then P has the speed w d . utr be the position vector of P referred to a coordinate system with origin 0 on the axis of rotation. Then d = I r I sin y, where y is the angle between o and r. Therefore,
From the above result and the definition of vector product, wt see that the velocity V of P can be represented in the form
This formula is useful for determining velocity Vat any point P of the body B.
Example 13 : Find the moment of a force about a point in terms of vector product.
Solution : In mechanics the momentum of a forcep about a point Q is defined as the product m = 1 p ( d, where d is the perpendicular distance between Q and the line of action 1 of p (Refer Figure 10.32). 1 k Q I
If r is the vector from Q to any point A on 1, then
and
where Y is the angle between r and Po Therefore, m = \p\\r\~in~ = Irxpl Vector Calculus The vector m - r x p, is called the moment vecur or vector moment of p about Q.Its magnitude is m and its direation is that of the axis of the rotation about Q whichp has the tendancy to produce.
Example 14 : Express force on a charged particle and an element of current carrying conductor placed in a magnetic field in terms of cross product.
Solution : The magnitude of the force on a point charge q moving with velocity V in a magnetic field B is proportional to I V ( times the perpendicular component of B. Thus it can be expressed as a vector product as F =qVxB If I is the current through the conductor, then the force on an element dl of a current carrying conductor in a magnetic field B is
F = dZxB
Example 15 : If a, b, c are the position vectors ofA, B, C in a AABC, show that the vector area of the triangle is
Hence deduce the condition that the three points A, B, C may be collinear.
Solution : Let 0 be the origin of reference with respect to which a, b, c are the position vectors ofA, B, C in A ABC (See Figure 10.33)
0 Figure 10.33
- Now BC= OC- OB = c-b - 'i r -- BA=OA-~D,, r Vector Algebra . Vector area of the triangle ABC I ,
Now if the points A, B, C are collinear, then the vector area of the A ABC must be zero, i.e.,
=+ axb+bxc+cxa - 0, which is the required condition
You may now attempt the following exercises.
AA A Consider 2 f%rce F = ( - 3i +j + 5k ) newtons acting at a point P ( 7i + 3j + k ) m. What is the torque about the origin? (Hint: Torque t is a measure of the ability of an applied force to produce a twist or to rotate a body => t = r x F ).
E 12. Show that for any two vectors a and b, (a-b)x(a+b)- 2(axb), and give its geometrical interpretation. Vector Calculus Find the values of a for with the vectors 3i^+ 2;+ 9$ and i^+ a;+ 3; are (i) perpendicular (ii) parallel.
E 14. A A AA A A force represented by 5j+ k is acting at a point 9i -j + 2k. Find its moment about the point 3i + 2j + k.
Tn physics, repeated products of more than two vectors occur very often. For example, the electromotive force dE induced in an element of a conducting wire dI moving with velocity V through a magnetic field B is represented by dE = (VxB).dZ It is a real economy in thinking to represent the result in the compact vector form that removes the necessity of carrying factors such as sinc ol the angle between 9 and V and the cosine of the angle between the normal to their plane and the vector dl. These are taken into account by the given product of the three vectors. Let us now discuss the product of three or more vectors in the next section. 10.6 MULTIPLE PRODUCTS OF VECTORS
Some of the cbmmonly occuring multiple products in physical and engineering applications are the scalar and vector products of three vectors. We 'mow that if b and c are two vectors, then b x c is a vector perpendicular to the plane of b and c. If a is a third vector, then scalar (or dot) product of a with ( b x c ) is a scalar. This is called scalar triple product or mixed triple product. The cross (or vector) product of a with ( b x c ) yields a vector and is called the vector triple product. Let us discuss the two types of products one by one. 10.6.1 Scalar Triple Product The scalar triple product of three vectors a, b and c, denoted by a . ( b x c ), is defined as
a.(bxc) = IaIIbxcIcosp, where p is the angle between a and the vector ( b x c ). Here a . ( b x c ) is evidently a scalar. The scalar triple product a . ( b x c ) can be interpreted as the component of a along the vector ( b x c ). Note that ( a .b ) c merely represents a vector in the direction of c, whose modulus is (a ,b ) times the modulus of c, whereas ( a .b ) x c is meaningless since a . b is not a vector but a scalar. The different ways the scalar triple product of vectors a, b, c is written are as follows: Vector Algebra a.(bxc)or[abc]or[a,b,c]or(abc) The absolute value of the scalar triple product has a geometrical meaning too. Let us see what it is?
Geometrical Interpretation of the Scalar Triple Product Let a, b, c be any three vectors. Consider a parallelopiped with its three coterminus edges having the magnitude and directions as of a, b and c respectively (see Figure 10.34). Let 0 be the angle between b and c and P be the angle between a and ;, the unit
Figure 10.34: Geometrical interpretation of a scalar triple product. vector normal to b and c. Then, we have
Now ( I b ( I c I sin 9 ) is the area of the parallelogram formed by the vectors b and c, i.e., the base of the parallelopiped. Also ( I a I cos P ) is the height h of the parallelepiped gives the volume of the parallelopiped. Hence, the absolute value of the scalar triple product a . ( b x c ) is equal to the volume of theparallelopiped with a, b, c as adjacent edges. From this geometrical consideration it follows that the value of the scalar triple product is a real number which is independent of the choice of Cartesian coordinates in space. In the parallelopiped in Figure 10.34,if we take the area of the face formed by a and b and multiply it by the height perpendicular (projection of c on a x b), we get the volume of the same parallelopiped. Thus
Thus thepositions of dot and cross in a scalar tripleproduct are interchangable provided the cyclic order of the factors is maintained.
The change of cyclic order of factors brings about a change of sign in the value of the scalar triple product. A A A The scalar triple product of the orthonormal right handed vectors i, j, k is equal to unity because . Vector Calculus Next consider three coplanar vectors a, b, c. Now bx c is a vector perpendicular to the plane of band c and is, therefore, perpendicular to aalso. :. Scalar product of aand bx cmust be zero
Thus, ifthree vectors are coplanar, their scalar triple product is zero. * Further if we consider a, a, band form a scalar triple product, then
a.(axb)= (axa).b
If two vectors in a scalar triple product are equal or parallel, tlten scalar triple product vanishes. With respect to any right-handed artesian coordinate system, let
Then
Hence
Since interchanging of two rows reverses'the sign of the determinant, we have
[a,b,c] = -[b,a,cI
Let us consider some examples giving applications of scalar triple product.
Example 16 : Show that the force acting on a charged particle q which moves with velocity V in the magnetic field B does not bring about any change in the energy of the charge.
Solution : The force acting on a charge particle q moving with velocity V in the magnetic field B is given by F=qVxB We know that the change in energy is equal to the work done Now Work done on the charge fol' an infinitesimal displacement dr is
If the displacement ,drtakes place in time dt, then dr = V dt :. W = q(VxB). Vdt
= sdV,B,"I = qdt .O - 0 Since the work done is zero, hence the force F does not bring about any change in the energy of the charge.
Example 17 : Find the volume of the tetrahedron whose three sides are given by
CC A 2:- 3& 4i, f+ 2;- i and 31 -j + 2k.
Solution: We know that'volume of a tetrahedron is one-sixth the volume of the parallelepiped with three sides of tetrahedron as its three adjacent edges. 1 1 :. :. Regdvolume = -a.(bxc) = -(axb).c 6 6 i
AAA where a = 2i-3j+4k
AA Now axb = - -5i+6j+7k
7 Neglecting negative sign, the volume of tetrahedron - - 6'
Example 18 : Find the constant A so that the vectors AAA AAA AAA a = 2i-j+k, b = i+g-3k, c = 3i+hj+5karecoplanar.
Solution : If a, b, c are coplanar, then their scalar triple product is zero, i.e., a.(bxc) - 0 2 -1 1 => 12-3 =o 3A 5.
=> 2(10+3A)+(5+9)+(A-6) = 0
You may now attempt a few exercises to test your knowledge. Vector Calculus
A A IftheAvolumeof the parallelepiped whose edges are - 12f+ d , 3j- k and 2i + j - 15k in 546, determine a.
A AhA A Tie pzsiti2n vect2rs 01theApoints A,B,C,D are 3i - 2j- k, 2i + 3j- 4k, - i +j + 2k and 4i + 5j+ hk respectively. If the pointsA,B,C,D lie in a plane, find the value of h .
Let us now discuss the second type of vector product, that is, the vector triple product. 10.6.2 Vector Triple Product We first give the definition of vector triple product.
Definition : The vector product of two vectors, one of which is itself the vector product of two vectors, is a vector quantity called the vector triple product
Thus if a, b, c be three vectors, then a x ( b x c ) and ( a x b ) x c are vector triple products. We next give geometrical interpretation for vector triple product Vector Algebra Geometrical Interpretation of Vector Triple Product a x ( b x c ) You know that b x c, by definition, is perpendicular to both b and c, that is, b x c is perpendicular to the plane containing b and c. Also a x ( b x c ), being the vector product of vector a and vector ( b x c ), is perpendicular to both a and ( b x c ). Hence a x ( b x c ) must be parallel to the plane determined by b and c and is perpendicular to a. From the above geometrical interpretation of vector triple product, you may note that where as a x ( b x c ) is a vector parallel to the plane of b and c and is perpendicular to the vector a, the vector triple product ( a x b ) x c is a vector parallel to the plane of a and h is perpendicular to the vector c. Hence, in general,
ax(bxc)z (axb)xc
Thus cross (or vector) product is not associative. As a word of caution we may mention that always keep track of the brackets for correct interpretation and expansion of vector triple product. The geometrical interpretation of vector triple product can also be used tofind its expansion formula. Expansion Formula for Vector triple product a x ( b x c ) : Since a x ( h x c ) lie in the plane containing b and c, we can, theretore, resolve a x ( b x c ) into components parallel to b and c, i.e, it is possible to find scalars m and n such that
ax(bxc)= mb+nc (10.11) -
Multiplying both sides by a, we get
a.ax(bxc) = m(a.b)+n(a.c) (10.12) - The left-hand side of relation (10.12) is a scalar triple product of a, a and ( b x c ). Since in this scalar triple product, two of the vectors are equal, therefore it is zero.
m(a.b)+~z(a.c)= 0
--m - n (a.c) -(a.b) = P, say Putting the values of m and n in relation (10.11), we get
ax(bxc)= p(a.c)b-p(a.b)c (10.13)
We note that both the sides of relation (10.13) are equally balanced in a, b and c. Hence p must be some numerical constant independent of a, band c. Also relation (10.13) is true for any three vectors a, b, c. Thus to determine p, we take the special case where e e a=b=~andc=j.
e ee eec cee 'x('xJ) = P(l.~)l-~(1.1)~
A A A A - ixk =p(O)i-p(1)j e => -J = -pi -p=l Vector Calculus Substituting forp = 1in equation (10.13), we get the required expansion formula as
Let us consider a physical example of the vector triple product from electromagnetism.
Example 19 : Show that parallel wires carrying current in the same direction attract each other.
Solution : Consider two wires wl and w2 carrying currents il and iz respectively. Let dI1 and dI2 be the infinitesimal elements of the wire in the direction of current flow. Then the force experienced by the infinitesimal element & due to dI1 is given by p dJzx(dI1xr) F = -il i2 9 43~ r3 where p is the permeability of the free space and r is the position vector of dI2 w. r. to dI1. If we take dl1 and r in the plane of this paper, then dIl x r is a vector perpendicular to the plane of this paper and points to it. Thus dl2 x ( dI1 x r ) is a vector in the plane of this paper and perpendicular to d12. This means that parallel wires carrying current in the same direction attract each other. In case the direction of any one of the currents is reversed, the wires will repulse each other. Let us take another example to illustrate the use of vector triple products.
Example 20 : Prove that ax(bxc)+bx(cxa)+cx(axb) = 0.
Solution : From the expansion of vector triple products, we have ax(bxc) - (a.c)b - (a.b)c
bx(cxa) = (b.a)c - (b.c)a
cx(axb) = (c.b)a - (c.a)b
Since scalar product is commutative, hence adding equations (10.14), (10.15) and (10.16) we see that the terms in the right cancel and we get
ax(bxc)+bx(cxa)+cx(~xb)= 0
You may now attempt the following exercises. E 18.
AAA A A A A AA Ifa P i-23-3k, b = 2i+j-k, c = i+3j-k, find(i) ax(bxc),aqd (ii) (axb)xc Vector Algcbrr
L Prove that k for any arbitrary vector a
1 If a, 6, c are three unit vectors, such that a x ( b x c ) = 2 h,
find the angle which a makes with b and c, given that b and c are non-parallel.
Show that ( a x b ) x c - a x ( b x c ) if and only if a and c are collineer vectors.
Using the results of the Secs 10.6.1 and 10.6.2, the product of four (or more) vectors can also be evaluated. It will be seen that other repeated products occurs in applications may be expressed in terms of dot, vector or scalar triple products. We now discuss briefly the product of four vectors. 10.63 Quadruple Product of Vectors Quadruple product means the product of four vectors. Some of the relevant quadrupk products are Vector Calculus (i) (axb).(cxd) (ii) (axb) x (cxd) (iii) ax[bx(cxd)]. Let us now find their simplified expressions.
(i) Letp I c x d :.- (axb).(cxd) = (axb).p = a . ( b x p ), since dot and cross product are interchangable in scalar triple product.
Hence, (axb).(cxd) = (a.c)(b.d) - (b.c)(a.d), (10.17)
which is known as Identity of Lagrange. Next consider
(ii) (axb)x(cxd) = (axb)xp, wherep = cxd = -px(axb) - -[(p.b)a - (p.a)bl = (p.a)b - (p.b)a
Hence,
l(axb) x (cxd) = [a,c,d]b - [b,c,d]al (10.18)
In case we start-with a x b = q and expand vector triple product ( a x b ) x ( c x d ) as q x ( c x d ), we get another expression for this quadruple product as
Lastly, we consider (iii) ax[bx(cxd)] = ax[(b.d)c - (b.c)d] - (b.d)(axc) - (b.c)(axd) Hence ax[bx(cxd)] - (b.d)(axc) - (b.c)(axd) (10.20) Let us consider some example illustrating the use of quadruple products
Example 21 : Prove that
Solution : Using result (10.18), we have (bxc)x(cxa) - [b,c,a]c - [b,c,c]a Vector Algebra = [b,c,a]c (': [b,c,c]= 0) flence (axb).[(bxc)x(cxa)]
= (axb).{[b,c,a]c/
= [b,c,a]{(axb).~}
= [a,b,c][a,b,c] cyclic change in scalar triple product
= [a.(bxc)12 Hence from the above result, it immediately follows that ifa, b, care three non-koplanar vectors, then a x b, b x c and c xu are also non-coplanar You may now try this exercise
Prove that
(axb)x(axc),d= (a.d)[a,b,c]
On the basis of the above knowledge, we shall like to discuss in brief in the next section the two categories of vectors into which all vector quantities of mechanics /physics can be grouped. These are termed as proper (or polar) vectors and pseudo vectors (or axial vectors).
10.7 POLAR AND AXIAL VECTORS
The direction of some of the vector quantities is clearly indicated by the direction of motion of a system. Examples of such vectors are displacements, velocity, acceleration, etc. Such vectors are calledpolar vectors. Vectors, namely, angular velocity, angular acceleration, angular momentum, etc. associated with the rotation motion are such that their direction does not indicate the direction of rotation of the body. Their direction is taken to be along the axis of rotation. Such vectors are called axial vectors. / The transformation of coordinates I
x' I - x, yl=-y, zl=-z
is called the parity hrmsformation. Thc difference between polar and axial vectors can be expressed in terms of parity transformation as follows: If a vector changes sign under the parity transformation, it is called a proper or a polar vector. The vectors which donot change sign under a parity transformation are called axial vectors. It may be observed that Vector Calculus The cr&spro8uct of bvo polar vectors is an axial vector. Before we move on to the next units, wherein we shall discuss vector differential calculus and vector integral calculus, it will be worthwhile to look briefly to the three coordinate systems in space and give their relations. We take up these coordinate systems briefly in the next section. 4
10.8 COORDINATE SYSTEMS FOR SPACE 1
Let us look briefly at three coordinate systems for space. The first, cartesian coordinates, is the system we have been using the most in our discussion so far. Cylindrical and spherical polar coordinates will come handy when we study integration of vectors in Units 12 and 13 because surfaces that have complicated repre&ntation in cartesian coordinates sometimes have simpler equations in one of these other systems. 10.8.1 Cartesian Coordinates
Consider' a sysitem of mutually orthogonal cot Figure 1(3.35). The cartesian coordinates of a from the coordinate axes by passing planes through P perpendicular to each axis. The three coordinate planes x - 0, y - 0, z = 0 divide the space into eight cells, callcd octants,
Figure 10.35: Cattesian Coordinales.
The octant in which all the threecoordinates are positive is called the first octant. We next take up cyIinJrical'eoordinaEe&. 10.8.2 Cylindrkal Coordinates It is frequently convenient to use coordinates ( r, 8, z ) to locate a point in space. There are first the polar ( r, 8 ) used instead of (x, y) in the plane z = 0 coupled with the z-coor&ns~,(rcf. F@ 10.36) Equations relating Cartesian and Cylindrical Coordinates are
2 22 x = rcos8 , r = x +y y = r sin 8 tan 8 = y/x 'IC z=z Here r = 0 is the equation for -axis and r = constant describes a circular cylinder of radius r whose axis is the z-axis.
The equation 8 = constant describes a plane containing the z-axis and making an I angle 0 with the ~ositivex-axis.
I Figure 10.36: Cylindrical Coordinates Cylindrical coordinates are convenient when there is an axis of symmetry in a physical problem. i 10.8.3 Spherical Coordinates I Spherical coordinates are useful when thereis a centre of symmetry. The spherical . koordinates ( R, 8, $ ) of a point in space are shown in Figure 10.37 below.
Figure 10.37: Spherical Coordinates. The first coordinate R = ( OP I is the distance from the origin to the point P. It is , never negative. The second coordinate 0 is the same as in cylindrical coordinates, namely, the angle from xz-plane to the plane through P and the z-axis. The third spherical coordinate @ is the angle measured down fiom the z-axis to the line OP. Every point in space can be represented in terms of spherical coordinates restricted to the ranges Vector Cablus Equations relating Cartesian and Cylindrical Coordinates to Spherical Polar Coordinates are
r = R sin ip, x = r cos 0, x = R sin $ cos & z = R cos $, y = r sin 0, y = R sin $ sin 0 0 = 0, z = z, z = R cos $
The equation R = constant describes the surhce of a sphere of radious R with centre at 0. The equation 0 = constant in spherical polar coordinates defines a half-plane (-; R r 0 and 0 s @ s n). The equation $ = constant describes a cone with vertex at 0, axis OZ, and generating angle ip provided we broaden our interpretation of the word 'cone' to include the xy-plane for which $ = n/2 and comes with generating angles greater than m/2. Let us now summarize the contents of this unit.
10.9 SUMMARY
In this unit, you have learnt * A physical quantity completely specified by a single number (with a suitable choice of units of measure) is called a scalar. * Quantities specified by a magnitude and a direction are called rSecfor quantities. * Length, support and sense characterize a directed line segment. Length of a directed line segment is called rnagllitude or modulus or norm of the vector it represents. Direction of a vector is from its initial to terminal point. * A vector with a certain fixed initial point is called a borrnded r~ectorand when there is no restriction to choose the initial point, it is called afree vector. * A vector whose length is zero is called a null vecfor * A vector whose length is unity is called unit vector * All vectors having the same initial point are called coirzitial vectors * Vectors having the same or parallel line of action are called like or parallel or collinear vectors and vectors are called uizlike if they have opposite directions. * Vectors parallel to the same plane or lying in the same plane are called coplanar vectors. Three vectors are coplanar if their scalar triple product is 7er0. * Two vectors are said to he equal if they have thc same length, same or parallel supports and the same sense. * Projections of a vector on the axes of an orth2gonaJ cart5sian coordinate system are called its components. If a = a1 i + a: j + a k in component form, then
a1 a2 a3 and direction cosines of the vector are lal'larm If initial point of a vector is chosen to be the origin of a Cartesian Vector Algebra coordinate system, then components of the vector are the coordinates of its terminal point and the vector is calledposition vector of its terminal point. Two vectors a and b can be added graphically using triangle law or parallelogram law of vector addition . Vector addition is commutative and associative. There exist additive identity (0) and additive increase (negative of the vector). The difference a - b of two vectors a and b is obtained by adding (-b) to a. CC" n n n Givena = a1i+a2j+a3k, b = bli+b2j+b3k, n C A a +. b = (a1 + bl)i+(az + b2)j+(a3 + b3)k. If m is a scalar and a a vector ,then m a is a vector whose magnitude = I m 1 I a I ,support is same or parallel to a and direction of vector m a is same as a if m > 0 and opposite to a if m < 0. Two vectors are linearly dependent if and only if they are parallel, otherwise they are linearly independent. The condition of linear dependence is that their vector product is zero The scalarproduct of two vectors a and b is defined as a.b = IaIIbIcosy,wherey(~s y s rc)istheanglebetweenaandb. In component form, a . b = albl+ azh + a363 Two non-zero vectors are perpendicular to each other if and only if their dot product is zero. a.b (a1 = fiandcosy = -for angle y between a and b. lallbl Geometrically, scalar product of two vectors is the product of the modulus of either vector and projection of the other in its direction. The vector product of two vectors a and b is defined as A axb = Ia)Ib)sinyn, (0s ysrc), n where y is the angle between a and b and n is :unit vector perpendicular to both a and b in the direction such that a, b, n form a right-handed trial.
a x b = 0 is the condition for vectors a and b to be parallel. Geometrically, a x b represents the vector area of the parallelogram having adjacent sides represented by a and b .
Iaxbl = \l(a.a)(b.b)-(a.b)2 In the component form
The scalar triple product a . ( b x c ) is given by
where flis the angle between a and ( b x c ) . In the component form, * Geometrically, a . ( bx c ) is the volume of the parallclopipcdwith a, b, c as adjacent sides. * The positions of dot and cross products in a scalar triple product are interchangeable provided the cyclic order of the factors is maintained. * The vector triple product of three vectors a, 6, cis given by ax(bxc)= (a.c)b-(a.b)c
* Geometrically, a x ( bx c) is parallel to the plane of hand cand is perpendicular to a. * Some quadruple products of a, h, cand darc (axb).(cxd)= (a.c)(b.d)-(a.d)(h.c)
* The component of a polar vector change their sign under a parity transformation, while those of an axial vector do not. * Equations relating Cartesian and Cylindrical Coordinates to Spherical Polar Coordinates are
r = R sin $ x = r cos 0 x = R sin $ cos 8 z = Rcos$ = rsin9 y = R sin @ sin 0 9=0 Z=Z z = R cos $
10.10 SOLUTIONS / ANSWERS
El If al, a2, a3 are the components of a vector awith initial point P ( 3, - 2, 1 ) and terminal point Q ( 1,2, - 4 ) then a1=1-3=-2,a2=2-(-2)=4,a3=-4-1 =-5
and [a[=,/ al+az+ai2 .7
E2 13 Sinw-, 1, -are components of a vector aand 2 2 P [ - -, 1, -:) is the initial point of aand if Q ( x, y, z ) is the terminal point
i.e., terminal point is (0,2,2) Vector Algebra
--I& 2
In A ABC, by triangle law of addition, AB = AC + CB In A BCD, by triangle law of addtion,
DC = DB + BC Addiing, we get AB + DC = AC + CB + DB + BC
= AC + DB + CB - CB (-a*; + (- a) = 0)
= AC + DB
Hence the result
(i) Let A, B, C,be three non-linear point. Draw a AABC such that
AB = a and BC = b
Then, by addition law As sum of two sides of a triangle is greater than the third side, :. AC CAB + BC :. IAC)c)ABI+ IBC(
When A, B, C are collinear, then
a = AB,b = BC
:. a + b = AC
.:AC = AB + BC :. IACI + JAB1+ )BC(
==>(a+ bl = la1 + lbl
Combining the results for collinear and non-collinear points A, B and C, we get Vector Calculus
Herea = 3fii^- 31 b =6j"andc = 36f+3.; 1 Now
Also
Hence a, b, c form an equilateral triangle. 1 Q I Let us denote the three points by A, B, C. With reference to some origin of regerence 0,we have i
A gA A A Now AB = OB - OA = (3i + ~il- (2i + 3il
AC = OC- OA = (sin+ 4;) - (2?+ 3;) 1
:. AC and AB are parallel, but AC and AB have one point A common. Hence A, B, and C are collinear.
If the three vectors are coplanar, they should be linearly dependeh. Thus one of them can be expressed as linear combination of the remaining two, i.e, there should exist two scalars x and y such that Vector Algebra Vector Calculus
Here a = 3; - 2; + R and b = - 2; + 2; + 4i
If a is the angle between a and b, then
a.b -6-4+4 3 - 1 3 msa - =-- => a = cos - IaI.Ibl - -6 d8T (- 3)
(ii) Here a + Ib = (3;- 2j"+ i)+ 2 L
1 Now projection of ( a + - b ) on a 2
(iii) Vector aisperpendicular to b if a . b = 0 A AA fi $ A Here a.c = (3i - 2j + k).(- a - 4j + 2k) = - 3 + 8 + 2 = 7* 0 :. a is not perpendicular to c. A AA A A a.d = (3i - 2j + k).(- 3i + k) - - 9+0+1=-8#0 :. a is not perpendicular to d Now,a. e = (3:- 2]+ $).(2ih+ 2:- s) -6-4-2 = 0 :. :. a is perpendicular to e Ell
Torque = r x F
= (71+ 3;+ x (- 3in+jn+ 5E) = 2(a x b)
Interpretation :
Let the diagonals AC and BC of the parallelogramABCD intersect at 0.
~etA0= a and OD = b
:. OB = - b
NOW AB = A0 + OB = a - b
and AD = A0 + OD = a + b
Now ( a - ib ) x ( a + b ) = AB x AD = area of parallelogram ABCD.
Again a x b = area of parallelogram whose sides are a and b.
= area of parallelogram with sides as semi-diagonals of parallelogram ABCD.
Hence the area of the parallelogram ABCD is equal to twice the area of the parallelogram whose adjacent sides are semi-diagonals of the first parallelogram. E 13 (i) If the given vectors are perpendicular, then
(ii) If the given vectors are parallel, then
LetA be the point ( 9: - + 2; ) and B the point ( 3; + 2; + ;) AA A A AA A AA ThenBA = r = (9i - j+ 2k) - (3i + 2j + k) = 6i - 3j+ k AA Also the force = F = 5i + k Vector Calculus ':. The required moment = r x F
A 'A = (6;- 3j^+ i)x (5i + k)
E 15
L.H.S = (a + b).[(b+ c)x (c x a)]
= a.(b x c)+ a.(b x'a) + a.(c x a)
= [%he1 + [b,c,al ( .: a scalar triple product is zero when the vectors are equal. )
= [a,b,cl + [a,b,cl
= 2[a,b,c]
= R. H. S.
E 16
A A Let a = - 12i + ak
:. Volume = a.( b x c )
But given volume = 546
:. 528 - 6 a = 546
=> - 6 a = 546 - 528 = + 18 Vector Algebra
Here AB = OB - OA = (2?+ 3;- 4i) - (3;- 2;- i)
AAA =-i+5j-3k
AC - OC -
Since A, B, C, D are coplanar, therefore AB, AC and AD lie in the same plane.
ThusAB.AC x AD = 0
A A Here a = i - 2j - (i) Now b x c = /I2 :1 -I1-1 = ?(- 1 + 3)-;(- 2 + I) + E(6 -
:. :. (axb) x c= Vector Calculus
e e A Let a = a11 + a2J + a3k
E20 We have
1 => a.c - - = 0 (since a.b = 0) 2
( Since band care non-parallel, hence co-efficients of band cmust vanish separately ). If 91 and 92 are the angles between a and c &a and brespectivaly, then
1 cos g1 = - => el = x/3 ( : a, b, c, are unit vectors ) 2
E21
GiVen(a x b) x c = a x (bx c), we have to prove that aand care collinear
Now (a x b) x c = a x (b x c)
=> (a.c)b- (b,c)a= (a.c)b- (a.b)c Vector Algehra => (a.b)c - (b.c)a = 0
=>(axc)xb=O
=> Either a x c = 0 or b = 0
But b # 0
:. :. a x c = 0
=> a and c are collinear
( :. Cross product of collinear vectors is zero )
Conversely, given a and care collinear, we have to prove that
Since aand care collinear,
let c = t a, where t is some scalar
Now (a x b) x c = (a x b)x td
= t [(a.a)b- (a.b)a] ... (1)
and a x (bx c)= a x (bx ta)
= t [(a.a)b- (a.b)a] ... (2)
From (1)& (2),
= [(ax b).c]a - [(a x b).a]c
:. [(ax b) x (a x c)].d = ([a,b,c]a}.d
= [a,b, c](a. d)
= (a.d)[a,b,cl
Hence the result. vector Calculw LTNIT 11 VECTOR DIFFERENTIAL CALCLTLUS
Structure 11.1 Introduction Objectives 11.2 Scalar and Vector Fields 1 1.3 Vector Calculus 11.3.1 Limit and Continuity 11.3.2 Differentiability 11.3.3 Applications of Derivatives 11.4 Directional Derivative and Gradient Operator 11.5 Divergence of a Vector Field 11.5.1 Physical Interpretation 11.5.2 Formulae on Divergence of Vector Functions 11.B Curl of a Vector Field 11.6.1 Physical Interpretation 11.6.2 Rotation of a Rigid Body 11.6.3 Formulae on Curl, Divergence and Gradient 11.7 Summary
11.1 INTRODUCTION
In physical problems we often come across quantities such as temperature of a liquid, distance between two points, density of a gas, velocity and acceleration of a particle or a body, tangent to a curve and normal to a surface. In Unit 10, you have learnt that physical quantities can be categorised either as a scalar or as a vector. You might have noticed that some physical quantities, whether scalars or vectors, are variable. That is, their values are not constant or static but change with the change in variable. For example, the density of a gas, which is a scalar quantity, changes from place to place and is different at different times. Similarly, tangent to a curve may have different directions at different points of the curve. This variable character of scalars and vectors give rise to scalar functions and vector functions. Further, you may notice that at different positions the temperature or velocity of a body does not remain same. The distribution of temperature or velocity is therefore defined at each point of a given domain in space which leads to the idea of scalar fields and vector fields. We shall discuss about scalar functions and scalar fields, vector functions and vector fields in Section 11.2. In Section 11.3, we shall extend, in a very simple and natural way, the basic concepts of differential calculus to vector-valued functions. We shall also discuss about . * physically and geometrically important concepts related to scalar and vector fields namely, directional derivatives, gradient, divergence and curl in Section 11.4 and give their applications. We shall introduce the vector operator V in Sectioi~11.5 and give the physical interpretation of the divergence of a vector field and some basic formulas involving it. Finally, the concept of curl of a vector. field and its invariance is discussed in Section 11.6. Formulas involving curl, divergence, curl and gradient, divergence and curl and Laplacian operator, v', are also developed here. Objectives After studying this unit you should be able to * define a scalar function, a scalar field, a vector function and a vector field, * state conditions for the convergence of a sequence of vectors, * define limit, continuity and differeritiability of vector functions, * differentiate sum, difference and products i~ivolviiigvectors,