Page 1 : 424, , MATHEMATICS, , Chapter, , 10, , VECTOR ALGEBRA,  In most sciences one generation tears down what another has built and what, one has established another undoes. In Mathematics alone each generation, builds a new story to the old structure. – HERMAN HANKEL , , 10.1 Introduction, In our day to day life, we come across many queries such, as – What is your height? How should a football player hit, the ball to give a pass to another player of his team? Observe, that a possible answer to the first query may be 1.6 meters,, a quantity that involves only one value (magnitude) which, is a real number. Such quantities are called scalars., However, an answer to the second query is a quantity (called, force) which involves muscular strength (magnitude) and, direction (in which another player is positioned). Such, quantities are called vectors. In mathematics, physics and, engineering, we frequently come across with both types of, W.R. Hamilton, quantities, namely, scalar quantities such as length, mass,, (1805-1865), time, distance, speed, area, volume, temperature, work,, money, voltage, density, resistance etc. and vector quantities like displacement, velocity,, acceleration, force, weight, momentum, electric field intensity etc., In this chapter, we will study some of the basic concepts about vectors, various, operations on vectors, and their algebraic and geometric properties. These two type of, properties, when considered together give a full realisation to the concept of vectors,, and lead to their vital applicability in various areas as mentioned above., , 10.2 Some Basic Concepts, Let ‘l’ be any straight line in plane or three dimensional space. This line can be given, two directions by means of arrowheads. A line with one of these directions prescribed, is called a directed line (Fig 10.1 (i), (ii))., , 2019-20

Page 2 : VECTOR ALGEBRA, , 425, , Fig 10.1, , Now observe that if we restrict the line l to the line segment AB, then a magnitude, is prescribed on the line l with one of the two directions, so that we obtain a directed, line segment (Fig 10.1(iii)). Thus, a directed line segment has magnitude as well as, direction., Definition 1 A quantity that has magnitude as well as direction is called a vector., Notice that a directed line segment is a vector (Fig 10.1(iii)), denoted as, or, simply as , and read as ‘vector, , ’ or ‘vector ’., , starts is called its initial point, and the, The point A from where the vector, point B where it ends is called its terminal point. The distance between initial and, terminal points of a vector is called the magnitude (or length) of the vector, denoted as, |, , |, or | |, or a. The arrow indicates the direction of the vector., , Note, , Since the length is never negative, the notation | | < 0 has no meaning., , Position Vector, From Class XI, recall the three dimensional right handed rectangular coordinate, system (Fig 10.2(i)). Consider a point P in space, having coordinates (x, y, z) with, respect to the origin O (0, 0, 0). Then, the vector, having O and P as its initial and, terminal points, respectively, is called the position vector of the point P with respect, (or ) is given by, to O. Using distance formula (from Class XI), the magnitude of, |, , |=, , x2 + y 2 + z2, , In practice, the position vectors of points A, B, C, etc., with respect to the origin O, are denoted by , , , etc., respectively (Fig 10.2 (ii))., , 2019-20

Page 3 : 426, , MATHEMATICS, , Fig 10.2, , Direction Cosines, Consider the position vector, of a point P(x, y, z) as in Fig 10.3. The angles α,, β, γ made by the vector with the positive directions of x, y and z-axes respectively,, are called its direction angles. The cosine values of these angles, i.e., cos α, cos β and, cos γ are called direction cosines of the vector , and usually denoted by l, m and n,, respectively., Z, C, , P(x,y,z), , z, r, O, , y, B, , Y, P, , x, A, , O, 90°, , X, , A, , Fig 10.3, , X, , From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we, have, , . Similarly, from the right angled triangles OBP and, , y, z, and cos γ = . Thus, the coordinates of the point P may, r, r, also be expressed as (lr, mr,nr). The numbers lr, mr and nr, proportional to the direction, cosines are called as direction ratios of vector , and denoted as a, b and c, respectively., , OCP, we may write cos β =, , 2019-20

Page 4 : VECTOR ALGEBRA, , Note, , 427, , One may note that l2 + m2 + n2 = 1 but a2 + b2 + c2 ≠ 1, in general., , 10.3 Types of Vectors, Zero Vector A vector whose initial and terminal points coincide, is called a zero vector, (or null vector), and denoted as . Zero vector can not be assigned a definite direction, as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any, represent the zero vector,, direction. The vectors, Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The, unit vector in the direction of a given vector is denoted by â ., Coinitial Vectors Two or more vectors having the same initial point are called coinitial, vectors., Collinear Vectors Two or more vectors are said to be collinear if they are parallel to, the same line, irrespective of their magnitudes and directions., Equal Vectors Two vectors, are said to be equal, if they have the same, magnitude and direction regardless of the positions of their initial points, and written, ., as, Negative of a Vector A vector whose magnitude is the same as that of a given vector, (say,, , ), but direction is opposite to that of it, is called negative of the given vector., , For example, vector, , is negative of the vector, , , and written as, , =–, , ., , Remark The vectors defined above are such that any of them may be subject to its, parallel displacement without changing its magnitude and direction. Such vectors are, called free vectors. Throughout this chapter, we will be dealing with free vectors only., Example 1 Represent graphically a displacement, of 40 km, 30° west of south., represents the required, Solution The vector, displacement (Fig 10.4)., Example 2 Classify the following measures as, scalars and vectors., (i) 5 seconds, (ii) 1000 cm3, , 2019-20, , Fig 10.4

Page 5 : 428, , MATHEMATICS, , (iii) 10 Newton, (iv) 30 km/hr, (vi) 20 m/s towards north, Solution, (i) Time-scalar, (iv) Speed-scalar, , (ii) Volume-scalar, (v) Density-scalar, , Example 3 In Fig 10.5, which of the vectors are:, (i) Collinear, (ii) Equal, , (v) 10 g/cm3, , (iii) Force-vector, (vi) Velocity-vector, , (iii) Coinitial, , Solution, ., , (i) Collinear vectors :, (ii) Equal vectors :, (iii) Coinitial vectors :, , Fig 10.5, , EXERCISE 10.1, 1. Represent graphically a displacement of 40 km, 30° east of north., 2. Classify the following measures as scalars and vectors., (i) 10 kg, (ii) 2 meters north-west (iii) 40°, (iv) 40 watt, , (v) 10–19 coulomb, , (vi) 20 m/s2, , 3. Classify the following as scalar and vector quantities., (i) time period, (iii) force, , (ii) distance, , (iv) velocity, (v) work done, 4. In Fig 10.6 (a square), identify the following vectors., (i) Coinitial, , (ii) Equal, , (iii) Collinear but not equal, 5. Answer the following as true or false., (i) and – are collinear., (ii) Two collinear vectors are always equal in, magnitude., (iii) Two vectors having same magnitude are collinear., , Fig 10.6, , (iv) Two collinear vectors having the same magnitude are equal., , 2019-20

Page 6 : VECTOR ALGEBRA, , 429, , 10.4 Addition of Vectors, A vector, simply means the displacement from a point, A to the point B. Now consider a situation that a girl, moves from A to B and then from B to C, (Fig 10.7). The net displacement made by the girl from, point A to the point C, is given by the vector, and, expressed as, , Fig 10.7, , =, This is known as the triangle law of vector addition., In general, if we have two vectors and (Fig 10.8 (i)), then to add them, they are, positioned so that the initial point of one coincides with the terminal point of the other, (Fig 10.8(ii))., C, , C, , b, , a, , +b, , A, a, (i), , b, , b, B, , a, , A, , a, (ii), , a –, b, (iii), , B, –b, , C’, , Fig 10.8, For example, in Fig 10.8 (ii), we have shifted vector without changing its magnitude, and direction, so that it’s initial point coincides with the terminal point of . Then, the, vector + , represented by the third side AC of the triangle ABC, gives us the sum (or, resultant) of the vectors and i.e., in triangle ABC (Fig 10.8 (ii)), we have, Now again, since, , =, , from the above equation, we have, , This means that when the sides of a triangle are taken in order, it leads to zero, resultant as the initial and terminal points get coincided (Fig 10.8(iii))., , 2019-20

Page 7 : 430, , MATHEMATICS, , , Now, construct a vector, so that its magnitude is same as the vector BC , but, the direction opposite to that of it (Fig 10.8 (iii)), i.e.,, , , BC′ = − BC, , Then, on applying triangle law from the Fig 10.8 (iii), we have, , ,   ,  , AC′ = AB + BC′ = AB + (− BC) = a − b, , , , The vector AC′ is said to represent the difference of a and b ., Now, consider a boat in a river going from one bank of the river to the other in a, direction perpendicular to the flow of the river. Then, it is acted upon by two velocity, vectors–one is the velocity imparted to the boat by its engine and other one is the, velocity of the flow of river water. Under the simultaneous influence of these two, velocities, the boat in actual starts travelling with a different velocity. To have a precise, idea about the effective speed and direction, (i.e., the resultant velocity) of the boat, we have, the following law of vector addition., , , If we have two vectors a and b represented, by the two adjacent sides of a parallelogram in, magnitude and direction (Fig 10.9), then their,  , sum a + b is represented in magnitude and, direction by the diagonal of the parallelogram, through their common point. This is known as, the parallelogram law of vector addition., , Note, or, , Fig 10.9, , From Fig 10.9, using the triangle law, one may note that,  , , OA + AC = OC,  ,  , , (since AC = OB ), OA + OB = OC, , which is parallelogram law. Thus, we may say that the two laws of vector, addition are equivalent to each other., Properties of vector addition, , , , Property 1 For any two vectors a and b ,,  ,  , a +b = b + a, , 2019-20, , (Commutative property)

Page 8 : VECTOR ALGEBRA, , 431, , Proof Consider the parallelogram ABCD, (Fig 10.10). Let, , then using, , the triangle law, from triangle ABC, we have, , Now, since the opposite sides of a, parallelogram are equal and parallel, from, Fig 10.10,, , we have,, , and, , . Again using triangle law, from, , Fig 10.10, , triangle ADC, we have, , Hence, =,  , , Property 2 For any three vectors a , b and c, =, Proof Let the vectors, shown in Fig 10.11(i) and (ii)., , (Associative property), , be represented by, , Fig 10.11, Then, , =, , and, , =, , So, , =, , 2019-20, , , respectively, as

Page 9 : 432, , MATHEMATICS, , and, Hence, , =, =, , Remark The associative property of vector addition enables us to write the sum of, without using brackets., three vectors, , Note that for any vector a , we have, Here, the zero vector, , =, is called the additive identity for the vector addition., , 10.5 Multiplication of a Vector by a Scalar, Let be a given vector and λ a scalar. Then the product of the vector by the scalar, λ, denoted as λ , is called the multiplication of vector by the scalar λ. Note that, λ, is also a vector, collinear to the vector . The vector λ has the direction same (or, opposite) to that of vector according as the value of λ is positive (or negative). Also,, the magnitude of vector λ is | λ | times the magnitude of the vector , i.e.,, , –2, 1, 2 a, , 1, 2 a, , a, , 2a, , a, , |λ | = |λ | | |, A geometric visualisation of multiplication of a vector by a scalar is given, in Fig 10.12., , Fig 10.12, , When λ = – 1, then λ = – , which is a vector having magnitude equal to the, magnitude of and direction opposite to that of the direction of . The vector – is, called the negative (or additive inverse) of vector and we always have, + (– ) = (– ) +, , =, , 1, Also, if λ =  , provided ≠ 0 i.e. is not a null vector, then, |a |, , | λ | =| λ | | | =, , 2019-20

Page 10 : VECTOR ALGEBRA, , So, λ, , 433, , represents the unit vector in the direction of . We write it as, 1 , â =  a, |a |,  , Note For any scalar k, k 0 = 0., , , , 10.5.1 Components of a vector, Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and, z-axis, respectively. Then, clearly, |, , | = 1, |, , | = 1 and |, , |= 1, , , each having magnitude 1,, The vectors, are called unit vectors along the axes OX, OY and OZ,, respectively, and denoted by iˆ, ˆj and kˆ , respectively, (Fig 10.13)., Fig 10.13, , Now, consider the position vector, of a point P (x, y, z), as in Fig 10.14. Let P1 be the foot of the perpendicular from P on the plane XOY., , Fig 10.14, , We, thus, see that P1 P is parallel to z-axis. As iˆ, ˆj and kˆ are the unit vectors along the, x, y and z-axes, respectively, and by the definition of the coordinates of P, we have, . Similarly,, , and, , 2019-20, , .

Page 13 : 436, , MATHEMATICS, , Example 5 Let, , and, , Solution We have, , . Is, , ? Are the vectors, , equal?, , and, , So,, . But, the two vectors are not equal since their corresponding components, are distinct., Example 6 Find unit vector in the direction of vector, Solution The unit vector in the direction of a vector is given by, Now, , Therefore, , =, , 2 2 + 32 + 12 = 14, , 2 ˆ, 3 ˆ, 1 ˆ, i+, j+, k, 14, 14, 14, , 1, (2iˆ + 3 ˆj + kˆ) =, 14, , aˆ =, , ., , Example 7 Find a vector in the direction of vector, 7 units., , that has magnitude, , Solution The unit vector in the direction of the given vector is, =, , 1 ˆ, 1 ˆ 2 ˆ, (i − 2 ˆj ) =, i−, j, 5, 5, 5, , , Therefore, the vector having magnitude equal to 7 and in the direction of a is, , 7 ˆ 14 ˆ, ∧,  1 ∧ 2 ∧, i−, j, 7a = 7, i−, j =, 5, 5, 5, 5, , , Example 8 Find the unit vector in the direction of the sum of the vectors,, and, , ., , Solution The sum of the given vectors is, , and, , =, , 42 + 32 + ( −2)2 = 29, , 2019-20

Page 15 : 438, , MATHEMATICS, , Example 10 Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed, from P to Q., Solution Since the vector is to be directed from P to Q, clearly P is the initial point, , and Q is the terminal point. So, the required vector joining P and Q is the vector PQ ,, given by, = (−1 − 2)iˆ + (−2 − 3) ˆj + ( − 4 − 0) kˆ, = −3iˆ − 5 ˆj − 4kˆ., , i.e., , 10.5.3 Section formula, Let P and Q be two points represented by the position vectors, with respect to the origin O. Then the line segment, joining the points P and Q may be divided by a third, point, say R, in two ways – internally (Fig 10.16), and externally (Fig 10.17). Here, we intend to find, for the point R with respect, the position vector, to the origin O. We take the two cases one by one., , , respectively,, , Case I When R divides PQ internally (Fig 10.16)., If R divides, , such that, , =, , Fig 10.16, , ,, , where m and n are positive scalars, we say that the point R divides, ratio of m : n. Now from triangles ORQ and OPR, we have, , internally in the, , =, and, , =, , Therefore, we have, , =, , or, , =, , ,, (Why?), (on simplification), , Hence, the position vector of the point R which divides P and Q internally in the, ratio of m : n is given by, =, , 2019-20

Page 16 : VECTOR ALGEBRA, , 439, , Case II When R divides PQ externally (Fig 10.17)., We leave it to the reader as an exercise to verify, that the position vector of the point R which divides, the line segment PQ externally in the ratio, m : n i.e., , PR m, =, QR n, , is given by, , =, , Fig 10.17, , Remark If R is the midpoint of PQ , then m = n. And therefore, from Case I, the, , midpoint R of PQ , will have its position vector as, =, Example 11 Consider two points P and Q with position vectors, and, . Find the position vector of a point R which divides the line joining P and Q, in the ratio 2:1, (i) internally, and (ii) externally., Solution, (i) The position vector of the point R dividing the join of P and Q internally in the, ratio 2:1 is, =, (ii) The position vector of the point R dividing the join of P and Q externally in the, ratio 2:1 is, =, Example 12 Show that the points A(2iˆ − ˆj + kˆ ), B(iˆ − 3 ˆj − 5kˆ), C(3iˆ − 4 j − 4kˆ) are, the vertices of a right angled triangle., Solution We have, = (1 − 2)iˆ + (−3 + 1) ˆj + (−5 − 1)kˆ = −iˆ − 2 ˆj − 6kˆ, = (3 − 1)iˆ + (−4 + 3) ˆj + (−4 + 5) kˆ = 2iˆ − ˆj + kˆ, and, , = (2 − 3)iˆ + ( −1 + 4) ˆj + (1 + 4)kˆ = −iˆ + 3 ˆj + 5kˆ, , 2019-20

Page 17 : 440, , MATHEMATICS, , Further, note that, =, Hence, the triangle is a right angled triangle., , EXERCISE 10.2, 1. Compute the magnitude of the following vectors:, 1 ˆ 1 ˆ 1 ˆ, i+, j−, k, 3, 3, 3, 2. Write two different vectors having same magnitude., 3. Write two different vectors having same direction., , = iˆ + ˆj + k ;, , = 2iˆ − 7 ˆj − 3kˆ;, , =, , 4. Find the values of x and y so that the vectors 2iˆ + 3 ˆj and xiˆ + yjˆ are equal., 5. Find the scalar and vector components of the vector with initial point (2, 1) and, terminal point (– 5, 7)., , 6. Find the sum of the vectors = iˆ − 2 ˆj + kˆ, = −2iˆ + 4 ˆj + 5kˆ and c = iˆ − 6 ˆj – 7 kˆ ., , 7. Find the unit vector in the direction of the vector a = iˆ + ˆj + 2kˆ ., , 8. Find the unit vector in the direction of vector PQ , where P and Q are the points, (1, 2, 3) and (4, 5, 6), respectively., 9. For given vectors, = 2iˆ − ˆj + 2kˆ and = −iˆ + ˆj − kˆ , find the unit vector in the, direction of the vector, , ., , 10. Find a vector in the direction of vector 5iˆ − ˆj + 2kˆ which has magnitude 8 units., 11. Show that the vectors 2iˆ − 3 ˆj + 4kˆ and − 4iˆ + 6 ˆj − 8kˆ are collinear., 12. Find the direction cosines of the vector iˆ + 2 ˆj + 3kˆ ., 13. Find the direction cosines of the vector joining the points A (1, 2, –3) and, B(–1, –2, 1), directed from A to B., 14. Show that the vector iˆ + ˆj + kˆ is equally inclined to the axes OX, OY and OZ., 15. Find the position vector of a point R which divides the line joining two points P, and Q whose position vectors are iˆ + 2 ˆj − kˆ and – iˆ + ˆj + kˆ respectively, in the, ratio 2 : 1, (i) internally, , (ii) externally, , 2019-20

Page 18 : VECTOR ALGEBRA, , 441, , 16. Find the position vector of the mid point of the vector joining the points P(2, 3, 4), and Q(4, 1, –2)., 17. Show that the points A, B and C with position vectors,, , = 3iˆ − 4 ˆj − 4kˆ,, , = 2iˆ − ˆj + kˆ and = iˆ − 3 ˆj − 5kˆ , respectively form the vertices of a right angled, triangle., 18. In triangle ABC (Fig 10.18), which of the following is not true:, (A), (B), (C), (D), 19. If, , Fig 10.18, , are two collinear vectors, then which of the following are incorrect:, , (A), (B), (C) the respective components of, (D) both the vectors, , are not proportional, , have same direction, but different magnitudes., , 10.6 Product of Two Vectors, So far we have studied about addition and subtraction of vectors. An other algebraic, operation which we intend to discuss regarding vectors is their product. We may, recall that product of two numbers is a number, product of two matrices is again a, matrix. But in case of functions, we may multiply them in two ways, namely,, multiplication of two functions pointwise and composition of two functions. Similarly,, multiplication of two vectors is also defined in two ways, namely, scalar (or dot), product where the result is a scalar, and vector (or cross) product where the, result is a vector. Based upon these two types of products for vectors, they have, found various applications in geometry, mechanics and engineering. In this section,, we will discuss these two types of products., 10.6.1 Scalar (or dot) product of two vectors, Definition 2 The scalar product of two nonzero vectors, , 2019-20, , , denoted by, , , is

Page 19 : 442, , MATHEMATICS, , defined as, , =, , where, θ is the angle between, , (Fig 10.19)., , then θ is not defined, and in this case, we, , If either, , Fig 10.19, , define, Observations, 1., , is a real number., , 2. Let, , be two nonzero vectors, then, , if and only if, , are, , perpendicular to each other. i.e., , 3. If θ = 0, then, as θ in this case is 0., , ,  , 4. If θ = π, then a ⋅ b = − | a | | b |, In particular,, , , as θ in this case is π., , In particular,, , 5. In view of the Observations 2 and 3, for mutually perpendicular unit vectors, , iˆ, ˆj and kˆ, we have, iˆ ⋅ iˆ = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1,, iˆ ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ iˆ = 0, 6. The angle between two nonzero vectors, , is given by, , or, 7. The scalar product is commutative. i.e., (Why?), Two important properties of scalar product, Property 1 (Distributivity of scalar product over addition) Let, any three vectors, then, , 2019-20, , be

Page 21 : 444, , MATHEMATICS, , is called the projection vector, and its magnitude | | is simply called as the projection, on the directed line l., of the vector, For example, in each of the following figures (Fig 10.20 (i) to (iv)), projection vector, of, along the line l is vector, ., Observations, 1. If p̂ is the unit vector along a line l, then the projection of a vector on the line, , l is given by a ⋅ pˆ ., , 2. Projection of a vector on other vector b , is given by, , , a ⋅ bˆ,, , or, , 3. If θ = 0, then the projection vector of, projection vector of, 4. If θ =, , will be, , will be, , itself and if θ = π, then the, , ., , π, 3π, or θ = , then the projection vector of, 2, 2, , Remark If α, β and γ are the direction angles of vector, direction cosines may be given as, , will be zero vector., = a1iˆ + a2 ˆj + a3 kˆ , then its, , Also, note that, are respectively the projections of, along OX, OY and OZ. i.e., the scalar components a1, a2 and a3 of the vector , are, precisely the projections of along x-axis, y-axis and z-axis, respectively. Further, if, is a unit vector, then it may be expressed in terms of its direction cosines as, = cos αiˆ + cos βˆj + cos γkˆ, Example 13 Find the angle between two vectors, respectively and when, Solution Given, , =1., . We have, , 2019-20, , with magnitudes 1 and 2

Page 23 : 446, , MATHEMATICS, , =, = (2)2 − 2(4) + (3) 2, Therefore, , =, , Example 18 If, , is a unit vector and, , Solution Since, , is a unit vector,, , or, , 5, , then find, , ., , . Also,, =8, , or, Therefore, , = 3 (as magnitude of a vector is non negative)., , Example 19 For any two vectors, Schwartz inequality)., , , we always have, , (Cauchy-, ,  ,  , Solution The inequality holds trivially when either a = 0 or b = 0 . Actually, in such a, , . So, let us assume that, , situation we have, Then, we have, , ., , = | cos θ | ≤ 1, Therefore, C, , Example 20 For any two vectors, have, , b, a +, , , we always, , (triangle inequality)., , b, , A, , Solution The inequality holds trivially in case either, (How?). So, let, , a, , B, , . Then,, Fig 10.21, , =, =, , (scalar product is commutative), , ≤, , (since x ≤ | x | ∀ x ∈ R ), , ≤, , (from Example 19), , =, , 2019-20

Page 25 : 448, , MATHEMATICS, , 6. Find, , , if, , ., , 7. Evaluate the product, , ., , 8. Find the magnitude of two vectors, , , having the same magnitude and, , such that the angle between them is 60o and their scalar product is, 9. Find, , , if for a unit vector ,, , ., , 10. If, perpendicular to , then find the value of λ., 11. Show that, , 1, ., 2, , are such that, , is perpendicular to, , is, , , for any two nonzero, , ., , vectors, 12. If, , , then what can be concluded about the vector ?, , 13. If, , are unit vectors such that, , , find the value of, , ., 14. If either vector, , . But the converse need not be, , true. Justify your answer with an example., 15. If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2),, and, respectively, then find ∠ABC. [∠ABC is the angle between the vectors, ]., 16. Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear., 17. Show that the vectors 2iˆ − ˆj + kˆ, iˆ − 3 ˆj − 5kˆ and 3iˆ − 4 ˆj − 4kˆ form the vertices, of a right angled triangle., 18. If is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ is unit, vector if, (A) λ = 1, , (B) λ = – 1, , (C) a = | λ |, , (D) a = 1/| λ |, , 10.6.3 Vector (or cross) product of two vectors, In Section 10.2, we have discussed on the three dimensional right handed rectangular, coordinate system. In this system, when the positive x-axis is rotated counterclockwise, , 2019-20

Page 26 : VECTOR ALGEBRA, , 449, , into the positive y-axis, a right handed (standard) screw would advance in the direction, of the positive z-axis (Fig 10.22(i))., In a right handed coordinate system, the thumb of the right hand points in the, direction of the positive z-axis when the fingers are curled in the direction away from, the positive x-axis toward the positive y-axis (Fig 10.22(ii))., , Fig 10.22 (i), (ii), , Definition 3 The vector product of two nonzero vectors, , , is denoted by, , and defined as, =, , ,, , 0 ≤ θ ≤ π and n̂ is a, , where, θ is the angle between, , unit vector perpendicular to both, , , such that, , form a right handed system (Fig 10.23). i.e., the, moves in the direction, right handed system rotated from, , Fig 10.23, , of n̂ ., If either, Observations, , , then θ is not defined and in this case, we define, , ., , 1., , is a vector., , , 2. Let a and b be two nonzero vectors. Then, parallel (or collinear) to each other, i.e.,, =, , 2019-20, , if and only if, , are

Page 27 : 450, , MATHEMATICS, , In particular,, and, , since in the first situation, θ = 0 and, in the second one, θ = π, making the value of sin θ to be 0., 3. If θ =, , π, , then, ., 2, 4. In view of the Observations 2 and 3, for mutually perpendicular, unit vectors iˆ, ˆj and kˆ (Fig 10.24), we have, iˆ × iˆ =, , iˆ × ˆj = kˆ, ˆj × kˆ = iˆ, kˆ × iˆ = ˆj, , Fig 10.24, , 5. In terms of vector product, the angle between two vectors, given as, , may be, , sin θ =, 6. It is always true that the vector product is not commutative, as, Indeed,, , , where, , i.e., θ is traversed from, , , Fig 10.25 (i). While,, , =, , ., , form a right handed system,, , form a right handed system i.e. θ is traversed from, Fig 10.25(ii)., , , where, ,, , Fig 10.25 (i), (ii), , , Thus, if we assume a and b to lie in the plane of the paper, then nˆ and nˆ1 both, will be perpendicular to the plane of the paper. But, n̂ being directed above the, paper while n̂1 directed below the paper. i.e. nˆ1 = − nˆ ., , 2019-20

Page 28 : VECTOR ALGEBRA, , Hence, , 451, , =, , =, 7. In view of the Observations 4 and 6, we have, ˆj × iˆ = − kˆ, kˆ × ˆj = − iˆ and iˆ × kˆ = − ˆj., 8. If, , represent the adjacent sides of a triangle then its area is given as, ., , By definition of the area of a triangle, we have from, Fig 10.26,, Area of triangle ABC =, But, , 1, AB ⋅ CD., 2, , (as given), and CD =, , Fig 10.26, , sin θ., , Thus, Area of triangle ABC =, 9. If, by, , represent the adjacent sides of a parallelogram, then its area is given, ., , From Fig 10.27, we have, Area of parallelogram ABCD = AB. DE., But, , (as given), and, ., Fig 10.27, , Thus,, Area of parallelogram ABCD =, , We now state two important properties of vector product., Property 3 (Distributivity of vector product over addition): If, are any three vectors and λ be a scalar, then, (i), (ii), , 2019-20

Page 30 : VECTOR ALGEBRA, , Example 23 Find a unit vector perpendicular to each of the vectors, where, , 453, , and, , ., , Solution We have,  ,  , A vector which is perpendicular to both a + b and a − b is given by, , =, , Now, = 4 + 16 + 4 = 24 = 2 6, Therefore, the required unit vector is, , c, −1 ˆ 2 ˆ 1 ˆ, i+, j−, k,  =, |c |, 6, 6, 6, , Note, , There are two perpendicular directions to any plane. Thus, another unit, , vector perpendicular to, , will be, , be a consequence of, , 1 ˆ 2 ˆ 1 ˆ, i−, j+, k . But that will, 6, 6, 6, , ., , Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3), and C(2, 3, 1) as its vertices., Solution We have, is, , . The area of the given triangle, , ., , iˆ, , ˆj kˆ, , Now,, , = 0 1 2 = − 4iˆ + 2 ˆj − kˆ, 1 2 0, , Therefore, , =, , Thus, the required area is, , 1, 2, , 16 + 4 + 1 = 21, , 21, , 2019-20

Page 31 : 454, , MATHEMATICS, , Example 25 Find the area of a parallelogram whose adjacent sides are given, by the vectors, Solution The area of a parallelogram with, by, , as its adjacent sides is given, , ., , iˆ, , ˆj, , kˆ, , Now, , = 3 1 4 = 5iˆ + ˆj − 4kˆ, 1 −1 1, , Therefore, , =, , and hence, the required area is, , 25 + 1 + 16 = 42, , 42 ., , EXERCISE 10.4, 1. Find, , ., , 2. Find a unit vector perpendicular to each of the vector, , , where, , ., 3. If a unit vector, , makes angles, , π, π, with iˆ, with ˆj and an acute angle θ with, 3, 4, , k̂ , then find θ and hence, the components of ., 4. Show that, , 5. Find λ and µ if, 6. Given that, , ., and, , . What can you conclude about the vectors, , ?, 7. Let the vectors, , be given as a1iˆ + a2 ˆj + a3kˆ, b1iˆ + b2 ˆj + b3kˆ,, , c1iˆ + c2 ˆj + c3 kˆ . Then show that, , ., , 8. If either, then, . Is the converse true? Justify your, answer with an example., 9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)., , 2019-20

Page 32 : VECTOR ALGEBRA, , 455, , 10. Find the area of the parallelogram whose adjacent sides are determined by the, vectors, and, ., 11. Let the vectors, , be such that, , , then, , is a unit, , vector, if the angle between, is, (A) π/6, (B) π/4, (C) π/3, (D) π/2, 12. Area of a rectangle having vertices A, B, C and D with position vectors, 1, 1, 1, 1, – iˆ + ˆj + 4kˆ, iˆ + ˆj + 4kˆ , iˆ − ˆj + 4kˆ and – iˆ − ˆj + 4kˆ , respectively is, 2, 2, 2, 2, 1, (B) 1, (A), 2, (C) 2, (D) 4, , Miscellaneous Examples, Example 26 Write all the unit vectors in XY-plane., be a unit vector in XY-plane (Fig 10.28). Then, from the, Solution Let, figure, we have x = cos θ and y = sin θ (since | | = 1). So, we may write the vector r as, = cos θ iˆ + sin θ ˆj, ... (1), Clearly,, , | |=, , cos 2 θ + sin 2 θ = 1, , Fig 10.28, , Also, as θ varies from 0 to 2π, the point P (Fig 10.28) traces the circle x2 + y2 = 1, counterclockwise, and this covers all possible directions. So, (1) gives every unit vector, in the XY-plane., , 2019-20

Page 33 : 456, , MATHEMATICS, , Example 27 If iˆ + ˆj + kˆ, 2iˆ + 5 ˆj, 3iˆ + 2 ˆj − 3kˆ and iˆ − 6 ˆj − kˆ are the position, vectors of points A, B, C and D respectively, then find the angle between, and, are collinear., Deduce that, , and, , ., , Solution Note that if θ is the angle between AB and CD, then θ is also the angle, between, and, ., Now, , Therefore, Similarly, Thus, , = Position vector of B – Position vector of A, = (2iˆ + 5 ˆj ) − (iˆ + ˆj + kˆ) = iˆ + 4 ˆj − kˆ, |, , (1) 2 + (4) 2 + ( −1) 2 = 3 2, , = − 2iˆ − 8 ˆj + 2kˆ and |CD |= 6 2, , |=, , cos θ =, =, , 1(−2) + 4(−8) + ( −1)(2) −36, =, = −1, 36, (3 2)(6 2), , Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that, Alternatively,, Example 28 Let, , which implies that, , and, , and, , are collinear., , are collinear vectors., , be three vectors such that, , each one of them being perpendicular to the sum of the other two, find, Solution Given, Now, =, +, =, = 9 + 16 + 25 = 50, Therefore, , =, , 50 = 5 2, , 2019-20, , and, .

Page 34 : VECTOR ALGEBRA, , Example 29 Three vectors, , satisfy the condition, , 457, , . Evaluate, ., , the quantity, Solution Since, , , we have, =0, =0, , or, Therefore, , ... (1), , Again,, , =0, , or, , ... (2), , Similarly, , = – 4., , ... (3), , Adding (1), (2) and (3), we have, = – 29, 2µ = – 29, i.e., µ =, , or, , −29, 2, , Example 30 If with reference to the right handed system of mutually perpendicular, unit vectors, , , then express, is parallel to, , Solution Let, Now, , is perpendicular to, , is a scalar, i.e.,, , ., , = (2 − 3λ)iˆ + (1 + λ ) ˆj − 3kˆ ., , , , , , Now, since β 2 is to be perpendicular to α , we should have, , 3(2 − 3λ ) − (1 + λ ) = 0, or, Therefore, , in the form, , λ=, , 1, , =, , 1, 2, 3ˆ 1 ˆ, i − j and, 2, 2, , 2019-20, , . i.e.,

Page 35 : 458, , MATHEMATICS, , Miscellaneous Exercise on Chapter 10, 1. Write down a unit vector in XY-plane, making an angle of 30° with the positive, direction of x-axis., 2. Find the scalar components and magnitude of the vector joining the points, P(x1, y1, z1) and Q (x2, y2, z2)., 3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of, north and stops. Determine the girl’s displacement from her initial point of, departure., 4. If, , , then is it true that, , ? Justify your answer., , 5. Find the value of x for which x(iˆ + ˆj + kˆ) is a unit vector., 6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors, ., , find a unit vector parallel, , 7. If, , to the vector, ., 8. Show that the points A (1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and, find the ratio in which B divides AC., 9. Find the position vector of a point R which divides the line joining two points, P and Q whose position vectors are, externally in the ratio, 1 : 2. Also, show that P is the mid point of the line segment RQ., 10. The two adjacent sides of a parallelogram are 2iˆ − 4 ˆj + 5kˆ and iˆ − 2 ˆj − 3kˆ ., Find the unit vector parallel to its diagonal. Also, find its area., 11. Show that the direction cosines of a vector equally inclined to the axes OX, OY,  1 1 1 , ,, ,, and OZ are ± , .,  3 3 3 , , . Find a vector, , 12. Let, is perpendicular to both and , and, , which, , ., , 13. The scalar product of the vector iˆ + ˆj + kˆ with a unit vector along the sum of, vectors 2iˆ + 4 ˆj − 5kˆ and λiˆ + 2 ˆj + 3kˆ is equal to one. Find the value of λ., 14. If, vector, , are mutually perpendicular vectors of equal magnitudes, show that the, is equally inclined to, , 2019-20, , .

Page 36 : VECTOR ALGEBRA, , 15. Prove that, , , if and only if, , 459, , are perpendicular, given, , ., Choose the correct answer in Exercises 16 to 19., 16. If θ is the angle between two vectors, , π, 2, (C) 0 < θ < π, , , then, , only when, , π, 2, (D) 0 ≤ θ ≤ π, , (A) 0 < θ <, , (B) 0 ≤ θ ≤, , be two unit vectors and θ is the angle between them. Then, , 17. Let, , is, , a unit vector if, (A) θ =, , π, 4, , (B) θ =, , π, 3, , (C) θ =, , π, 2, , 18. The value of iˆ.( ˆj × kˆ) + ˆj ⋅ (iˆ × kˆ) + kˆ ⋅ (iˆ × ˆj ) is, (A) 0, (B) –1, (C) 1, 19. If θ is the angle between any two vectors, , (D) θ =, , 2π, 3, , (D) 3, , then, , when θ, , is equal to, (A) 0, , (B), , π, 4, , (C), , π, 2, , (D) π, , Summary, , , , Position vector of a point P(x, y, z) is given as, magnitude by, , , , , , and its, , x2 + y 2 + z2 ., , The scalar components of a vector are its direction ratios, and represent its, projections along the respective axes., The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of, any vector are related as:, , a, l= ,, r, , b, c, m= , n=, r, r, , 2019-20

Page 37 : 460, , , , , , MATHEMATICS, , The vector sum of the three sides of a triangle taken in order is ., The vector sum of two coinitial vectors is given by the diagonal of the, parallelogram whose adjacent sides are the given vectors., The multiplication of a given vector by a scalar λ, changes the magnitude of, the vector by the multiple | λ |, and keeps the direction same (or makes it, opposite) according as the value of λ is positive (or negative)., , , , For a given vector , the vector, , , , of ., The position vector of a point R dividing a line segment joining the points, , gives the unit vector in the direction, , P and Q whose position vectors are, (i), , internally, is given by, , (ii) externally, is given by, , , , respectively, in the ratio m : n, ., ., having angle θ between, , The scalar product of two given vectors, them is defined as, ., , Also, when, is given, the angle ‘θ’ between the vectors, determined by, , may be, , cos θ =, , , , If θ is the angle between two vectors, given as, , , then their cross product is, , where n̂ is a unit vector perpendicular to the plane containing, that, , , , . Such, , form right handed system of coordinate axes., , If we have two vectors, , , given in component form as, and λ any scalar,, , 2019-20

Page 38 : VECTOR ALGEBRA, , then, , 461, , = (a1 + b1 ) iˆ + (a2 + b2 ) ˆj + ( a3 + b3 ) kˆ ;, λ = (λa1 )iˆ + (λa2 ) ˆj + (λa3 ) kˆ ;, , = a1b1 + a2b2 + a3b3 ;, , and, , iˆ, = a1, a2, , ˆj, b1, b2, , kˆ, c1 ., c2, , Historical Note, The word vector has been derived from a Latin word vectus, which means, “to carry”. The germinal ideas of modern vector theory date from around 1800, when Caspar Wessel (1745-1818) and Jean Robert Argand (1768-1822) described, that how a complex number a + ib could be given a geometric interpretation with, the help of a directed line segment in a coordinate plane. William Rowen Hamilton, (1805-1865) an Irish mathematician was the first to use the term vector for a, directed line segment in his book Lectures on Quaternions (1853). Hamilton’s, method of quaternions (an ordered set of four real numbers given as:, a + biˆ + cjˆ + dkˆ, iˆ, ˆj , kˆ following certain algebraic rules) was a solution to the, problem of multiplying vectors in three dimensional space. Though, we must, mention here that in practice, the idea of vector concept and their addition was, known much earlier ever since the time of Aristotle (384-322 B.C.), a Greek, philosopher, and pupil of Plato (427-348 B.C.). That time it was supposed to be, known that the combined action of two or more forces could be seen by adding, them according to parallelogram law. The correct law for the composition of, forces, that forces add vectorially, had been discovered in the case of perpendicular, forces by Stevin-Simon (1548-1620). In 1586 A.D., he analysed the principle of, geometric addition of forces in his treatise DeBeghinselen der Weeghconst, (“Principles of the Art of Weighing”), which caused a major breakthrough in the, development of mechanics. But it took another 200 years for the general concept, of vectors to form., In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist and, mathematician, and Oliver Heaviside (1850-1925), an English engineer, created, what we now know as vector analysis, essentially by separating the real (scalar), , 2019-20

Page 39 : 462, , MATHEMATICS, , part of quaternion from its imaginary (vector) part. In 1881 and 1884, Gibbs, printed a treatise entitled Element of Vector Analysis. This book gave a systematic, and concise account of vectors. However, much of the credit for demonstrating, the applications of vectors is due to the D. Heaviside and P.G. Tait (1831-1901), who contributed significantly to this subject., , ——, , 2019-20