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 quantities, namely, scalar quantities such as length, mass, 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)). 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. uuurNotice that a directed line segment is a vector (Fig 10.1(iii)), denoted as AB or uuur simply as a r , and read as ‘vector AB ’ or ‘vector a r ’. uuurThe point A from where the vector AB starts is called its initial point, and the 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 uuur |AB|, or | a r |, or a. The arrow indicates the direction of the vector. 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 uuurrespect to the origin O(0, 0, 0). Then, the vector OP having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect uuurto O. Using distance formula (from Class XI), the magnitude of OP (or r r ) is given by uuur2 22|OP| = x + y + z In practice, the position vectors of points A, B, C, etc., with respect to the origin O r rrare denoted by a , bc,, etc., respectively (Fig 10.2 (ii)). Fig 10.2 Direction Cosines uuur rConsider the position vector OP or r of a point P(x, y, z) as in Fig 10.3. The angles α, rβ, γ made by the vector r 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 r r , and usually denoted by l, m and n, respectively. Z From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we x r stands for | | . Similarly, from the right angled triangles OBP andhave cosα= (rr )r zOCP, we may write cos β= y and cos γ= . Thus, the coordinates of the point P may rr 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 r r , and denoted as a, b and c, respectively. 10.3 Types of Vectors Zero Vector A vector whose initial and terminal points coincide, is called a zero rvector (or null vector), and denoted as 0 . Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as uuur uuurhaving any direction. The vectors AA, BB represent the zero vector, Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The runit vector in the direction of a given vector a is denoted by aˆ. 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. rrEqual Vectors Two vectors a and b are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and writtenrr as =.ab Negative of a Vector A vector whose magnitude is the same as that of a given vector uuur(say, AB ), but direction is opposite to that of it, is called negative of the given vector. uuur uuur uuuruuur For example, vector BA is negative of the vector AB , and written as BA =−AB . 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. uuurSolution The vector OP represents the required displacement (Fig 10.4). Example 2 Classify the following measures as scalars and vectors. (i) 5 seconds (ii) 1000 cm3 Fig 10.4 (iii) 10 Newton (iv) 30 km/hr (v) 10 g/cm3 (vi) 20 m/s towards north Solution (i) Time-scalar (ii) Volume-scalar (iii) Force-vector (iv) Speed-scalar (v) Density-scalar (vi) Velocity-vector Example 3 In Fig 10.5, which of the vectors are: (i) Collinear (ii) Equal (iii) Coinitial Solution rrr(i) Collinear vectors : ,dacand . rr(ii) Equal vectors : aand c. rr (iii) Coinitial vectors : , d.bcr and 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 (ii) distance (iii) force (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) a r and −a r are collinear. (ii) Two collinear vectors are always equal in magnitude. Fig 10.6 (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal. 10.4 Addition of Vectors uuurA vector AB 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 uuurpoint A to the point C, is given by the vector AC and Fig 10.7expressed as uuuruuur uuur =AC AB +BC This is known as the triangle law of vector addition. rrIn general, if we have two vectors a and b (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)). Fig 10.8 rFor example, in Fig 10.8 (ii), we have shifted vector b without changing its magnitude and direction, so that it’s initial point coincides with the terminal point of a r . Then, the rrvector ab, represented by the third side AC of the triangle ABC, gives us the sum + rr(or resultant) of the vectors a and b i.e., in triangle ABC (Fig 10.8 (ii)), we have uuur uuuruuur =AB +BC AC uuur uuur Now again, since AC =−CA , from the above equation, we have uuur ruuur uuur uuurAB +BC +CA =AA =0 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)). uuuuruuur Now, construct a vector BC′so that its magnitude is same as the vector BC , but the direction opposite to that of it (Fig 10.8 (iii)), i.e., uuuruuuurBC′ = −BC Then, on applying triangle law from the Fig 10.8 (iii), we have ruuur uuuruuuur uuur uuuurr( BC)abAC′=AB +BC ′= AB +− =− uuuurrrThe 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. rrIf 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 rrsum + is represented in magnitude andab direction by the diagonal of the parallelogram Fig 10.9through their common point. This is known as the parallelogram law of vector addition. Properties of vector addition rrProperty 1 For any two vectors a and b , rrrrab+ = ba(Commutative property)+ Proof Consider the parallelogram ABCD uuur uuur rr(Fig 10.10). Let AB aand BC b, then using the triangle law, from triangle ABC, we have AC uuurabr +r = Now, since the opposite sides of a parallelogram are equal and parallel, from uuur uuur rFig 10.10, we have, AD = BC = b and uuur uuur rDC= AB= a . Again using triangle law, from Fig 10.10triangle ADC, we have uuuuruuur uuur r AC =AD+ DC= bar+r r Hence ab= +r+ bar rrrProperty 2 For any three vectors ,andcab r rr rr r( + )+ = a bc+ ) (Associative property)ab c+ ( r uuuruuur uuurrrabProof Let the vectors ,andcbe represented by PQ, QR and RS , respectively, as shown in Fig 10.11(i) and (ii). Fig 10.11 r uuur uuur uuurrThen ab= PQ+QR=PR + uuur uuuruuurr rand bc= QR+RS=QS + uuuruuuruur r So (ab) + = r + c rPR +RS=PS uuuruuuruurrrrand a +(b +c) = PQ+QS=PS rrrrrr c ++ (Hence (ab++ ) = a bc) Remark The associative property of vector addition enables us to write the sum of rrrrr rthree vectors ab,,c as a +b +c without using brackets. rNote that for any vector a , we have rrrrr a +0 =0 +aa= rHere, the zero vector 0 is called the additive identity for the vector addition. 10.5 Multiplication of a Vector by a Scalar rrLet a be a given vector and λ a scalar. Then the product of the vector a by the scalar rrλ, denoted as λ a , is called the multiplication of vector a by the scalar λ. Note that, rrrλ a is also a vector, collinear to the vector a . The vector λ a has the direction same r(or opposite) to that of vector a according as the value of λ is positive (or negative). rrAlso, the magnitude of vector λ a is |λ | times the magnitude of the vector a , i.e., rλa r|| λa | =|||A geometric visualisation of multiplication of a vector by a scalar is given in Fig 10.12. Fig 10.12 rrWhen λ = –1, then λa =−a, which is a vector having magnitude equal to the rrrmagnitude of a and direction opposite to that of the direction of a . The vector – a is rcalled the negative (or additive inverse) of vector a and we always have rrrrr a (–) (–) a +=+ a = a0 rr1Also, if λ= r, provided a 0, i.e. a is not a null vector, then ||a r1 r a| a r|| λ|| a | r||1λ= = ||a r rSo, λ a represents the unit vector in the direction of a . We write it as 1 r aˆ = r a||a 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 uuuruuur uuur|OA| 1,|OB| = 1and |OC| =1= uuur uuur uuur The vectors OA, OB and OC , each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by ˆˆ,ij and ˆk , respectively (Fig 10.13). Fig 10.13 uuur Now, consider the position vector OP 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. We, thus, see that P1 P is ˆˆˆparallel to z-axis. As ,k are the unit vectors along the x, y and z-axes,ij and uuur uuur respectively, and by the definition of the coordinates of P, we have 1 ˆ .PP = OR = zk uuur uuur uuurSimilarly, QP1 = OS = yjˆ and OQ = xiˆ. uuur ˆˆuuur uuur Therefore, it follows that OP1 = OQ + QP 1 = xiyj + uuur uuuur ˆˆ ˆand =OP+PP = xiyj + uuur + zk OP 11 Hence, the position vector of P with reference to O is given by uuur rˆˆ ˆOP (or r ) = xiyj ++ zk This form of any vector is called its component form. Here, x, y and z are called ˆ, ˆ ˆas the scalar components of r r , and xiyj and zk are called the vector components rof r along the respective axes. Sometimes x, y and z are also termed as rectangular components. r ˆˆThe length of any vector rxiyjˆ + zk , is readily determined by applying the=+ Pythagoras theorem twice. We note that in the right angle triangle OQP1 (Fig 10.14) uuuuruuuur uuur 2 222|OP | = |OQ| +|QP| = x + y ,1 1and in the right angle triangle OP1P, we have uuuruuur uuur2 2 222= |OP | |PP| (xy ) zOP 11 rHence, the length of any vector rxiyj= ˆ+ ˆ+ zk ˆ is given by r 2 22r ˆˆ ˆ|| =| xiyj ++ zk |= x + y + z rrIf a and b are any two vectors given in the component form aiˆ +a ˆj + ak ˆ and123 biˆ + bj ˆ + bk ˆ , respectively, then 12 3 rr(i) the sum (or resultant) of the vectors a and b is given by rrab =(a + bi )ˆ + (a + b )ˆj + (a + b )kˆ+ 112 2 33 r (ii) the difference of the vector a rand b is given by r ˆrab(a − bi ˆ + (a − b )ˆj + (− =11) 22 a3 − b3)k rr(iii) the vectors a and b are equal if and only if a1 = b1, a2 = b2 and a3 = b3 r(iv) the multiplication of vector a by any scalar λ is given by λa r =( ai)ˆ ( a )ˆj ( a )kˆ 12 3 The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws: rrLet a and b be any two vectors, and kand mbe any scalars. Then rr r(i) ma=(k+ )ka+ ma rr(ii) kma () =(kma ) rrrr(iii) ka( b) ka kb Remarks (i) One may observe that whatever be the value of λ, the vector λa r is always rr rcollinear to the vector a. In fact, two vectors a and b are collinear if and only r if there exists a nonzero scalar λ such that b=λa r . If the vectors ar and b are rr r aai ˆ + 2ˆ ˆ = ˆˆˆgiven in the component form, i.e. = 1 aj +ak 3 and bbi +bj +bk ,12 3 then the two vectors are collinear if and only if biˆ +bj ˆ +bk ˆ = λ(aiˆ +aj ˆ +ak ˆ)123 123 ⇔ bi+bj ˆ +bk ˆ(λai ( a )ˆj ( ˆˆ =)ˆ +λ +λ a )k123 123 ⇔ b =λa , b =λa , b =λ a112 23 3 b bb1 23⇔ = = =λ aaa1 23 r r(ii) If aai ˆ aj ˆ ak ˆ , then a1, a2, a are also called direction ratios of a.123 3(iii) In case if it is given that l, m, n are direction cosines of a vector, then liˆ +mj ˆ +nkˆ = (cos )iˆ (cos β)ˆj+(cos γ)ˆα+ k is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with x, yand zaxes respectively. rExample 4 Find the values of x, y and z so that the vectors = ˆ 2ˆj+ ˆaxi + zk and rb=2iˆ +yj ˆ +kˆ are equal. Solution Note that two vectors are equal if and only if their corresponding components rrare equal. Thus, the given vectors a and b will be equal if and only if x= 2, y= 2, z= 1 rrrr rrExample 5 Let ai=+2ˆjand b ij+ . Is ||a=| b| ? Are the vectors a equal? ˆ=2ˆˆ and b r bSolution We have ||ar=12 +22 =5 and || 2212 5 rrSo, ||| =b|a . But, the two vectors are not equal since their corresponding components are distinct. rExample 6 Find unit vector in the direction of vector a=2iˆ 3ˆjk ˆ++ Solution The unit vector in the direction of a vector a r is given by aˆ = 1 r a r .||a rNow || =a 23 1 Therefore aˆ = 1 (2iˆ ++3ˆjk ˆ) = iˆ + ˆj+ kˆ 14 1414 14 Example 7 Find a vector in the direction of vector aiˆ2ˆjthat has magnituder=− 7 units. rSolution The unit vector in the direction of the given vector ais 1 r 1 12 aˆ = ra=(iˆ −2 ) ˆj= iˆ − ˆj||5 55a Therefore, the vector having magnitude equal to 7 and in the direction of a r is ∧⎛1 ∧ 2 ∧⎞ 7 ˆ 14 ˆi− j7 a=7 i− j=⎜ ⎟⎝55 ⎠55 Example 8 Find the unit vector in the direction of the sum of the vectors, rrˆˆˆ 2ˆˆˆi – 5 and b ijk.a=2 +2 jk = ++ 3Solution The sum of the given vectors is rrrab c ( ,say)=4 ˆ 3ˆ 2ˆi jk r22 2and || =4 3 (2) = 29 c + +− Thus, the required unit vector is 1 r 1 432 cˆ = r c= (4ˆ +− 3ˆ 2) ˆ = iˆ + ˆj− kˆi jk ||c 29 2929 29 Example 9 Write the direction ratio’s of the vector ˆˆ ˆ and hence calculater =+−2aij kits direction cosines. rrxˆˆjzˆ are justSolution Note that the direction ratio’s a, b, cof a vector =iy++ k the respective components x, yand z of the vector. So, for the given vector, we have a= 1, b= 1 and c = –2. Further, if l, mand nare the direction cosines of the given vector, then a 1 b 1 c −2 r r = , m== , n= r = r= 6l= r as | | | r|6 ||r 6 ||r 6 ⎛11 2 ⎞Thus, the direction cosines are , , – .⎜ ⎟⎝66 6 ⎠ 10.5.2 Vector joining two points If P1(x1, y1, z1) and P2(x2, y2, z2) are any two points, then the vector joining P1 and P2 uuuuris the vector PP (Fig 10.15).12Joining the points P1 and P2 with the origin O, and applying triangle law, from the triangle OP1P2, we have uuuuruuur uuuurOP +PP =OP. 112 2Using the properties of vector addition, the above equation becomes uuuuruuuuruuur Fig 10.15PP =OP −OP 1221 uuuuri.e. PP = x2iyˆ +2ˆ +2kˆ) −(x1iy1ˆ +1kˆ)( jzˆ + jz12=(x−xi y yjzzˆ +( − ) +( − ˆ) ˆ)k2121 21 uuuurThe magnitude of vector 12 is given byPP uuuur2 22PP = (x−x) +( yy ) +(zz )−−1221 2121 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 uuur and Q is the terminal point. So, the required vector joining P and Q is the vector PQ , given by uuurPQ = −−2) iˆ +−− 3) ˆj+(4 ˆ(1 (2 −− 0) k uuuri.e. PQ = 3iˆ 5ˆj4ˆ−− −k. 10.5.3 Section formula uuur uuur Let P and Q be two points represented by the position vectors OP and OQ , respectively, 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 uuur the position vector OR for the point R with respect to the origin O. We take the two cases one by one. Case I When R divides PQ internally (Fig 10.16). Fig 10.16uuuruuuruuur If R divides PQ such that mRQ = nPR , uuurwhere mand nare positive scalars, we say that the point R divides PQ internally in the ratio of m: n. Now from triangles ORQ and OPR, we have uuuruuur uuur r r RQ = OQ −OR =−br uuur uuurr ruuurand PR =OR −OP ra,=− r rrrTherefore, we have mb r( =( −) (Why?)−) nr ar r rmb na or = + (on simplification)r mn+ Hence, the position vector of the point R which divides P and Q internally in the ratio of m: nis given by r uuurmb na r+ =OR mn+ 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 PR m m: ni.e. is given byQR nr uuurmbna − r OR = Fig 10.17mn− Remark If R is the midpoint of PQ , then m= n. And therefore, from Case I, the uuur midpoint R of PQ , will have its position vector as rr +uuurab=OR 2 uuur rrExample 11 Consider two points P and Q with position vectors OP =3a−2band uuur rrOQ ab. 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 rrrr r uuur2(ab) +(3 −2) 5a+ ab ==OR 21 3+ (ii) The position vector of the point R dividing the join of P and Q externally in the ratio 2:1 is rrrr ruuur2(ab) −(3 − b+ a2) r = =4 −baOR 21− ˆˆˆˆˆˆ ˆ ˆ), B( i 5 ), C(3 i 4) Example 12 Show that the points A(2i jk 3 jk 4 jkare the vertices of a right angled triangle. Solution We have uuurˆ(3 ˆ(5 kˆ iˆ 2ˆjk ˆAB =(1−2) i+−+ 1) j+−− 1) 6 uuurBC = − i(4 3) j(4 5) ˆ =2i jkˆˆ +(3 1) ˆ +−+ ˆ +−+ k − ˆ uuurand CA =(2 −3) iˆ (1 4)ˆj+(1 +4)kˆ iˆ 3ˆ + ˆ+−+ 5=− + jkFurther, note that uuuruuur uuur |AB|2 = 41 =+35 = 2 +| CA | 26 |BC| Hence, the triangle is a right angled triangle. 1. Compute the magnitude of the following vectors: rr r 111ˆˆ ˆˆˆ ˆˆ kˆ=++ ; bi 2 −− 7 3; ai jk = j kc= i+ j− 333 2. Write two different vectors having same magnitude. 3. Write two different vectors having same direction. 4. Find the values of xand yso that the vectors 2i+3ˆj and xˆ +ˆjare equal.ˆ iy5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7). rr ˆˆ ˆ6. Find the sum of the vectors ai=ˆ −+ ˆ, −+ + iˆ 4ˆ 5 and ciˆ6 –7k.2 jkb =2 jk r =−ˆj 7. Find the unit vector in the direction of the vector ˆˆ ˆ . r =++ 2aij kuuur8. 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. rr ˆˆˆ ˆˆˆ9. For given vectors, a= −+2b=i jk , find the unit vector in the2ij k and −+− rrdirection of the vector ab.+ 10. Find a vector in the direction of vector 5ˆˆ−+2ˆ which has magnitude 8 units.ij k11. Show that the vectors ˆ −ˆ +ˆ ˆˆjk ˆ are collinear. 2i3 jk 4 and −+ −4i68 12. Find the direction cosines of the vector iˆ +2ˆjk+3ˆ. 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. ˆˆˆ14. Show that the vector ++ is equally inclined to the axes OX, OY and OZ.i jk15. Find the position vector of a point R which divides the line joining two points P ˆ ˆˆ ˆˆˆand Q whose position vectors are i2jk and – i jk ++ respectively, in the +− ratio 2 : 1 (i) internally (ii) externally 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, a= 3iˆ − 4ˆj− 4 , ˆr k rrb 2ˆˆˆ ˆˆ− and == i jk + ci− 3ˆj− 5k, respectively form the vertices of a right angled triangle. 18. In triangle ABC (Fig 10.18), which of the following is not true: uuur uuuur uuur r(A) AB+BC+CA=0 uuur uuuruuur r (B) AB + BC − AC = 0 uuur uuuruuur r (C) AB + BC −CA = 0 uuur uuuruuur r Fig 10.18(D) AB − CB + CA = 0 rr19. If aand bare two collinear vectors, then which of the following are incorrect: r r(A) b=λa, for some scalar λ rr(B) a=±b rr(C) the respective components of aand bare not proportional rr(D) both the vectors aand bhave 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 r rr rDefinition 2 The scalar product of two nonzero vectors aand b, denoted by ab⋅ , is rrrrdefined as ab= ab| cos ⋅ ||| θ, rrwhere, θ is the angle between aand b, 0 (Fig 10.19). r rrrIf either a=0or b=0, then θ is not defined, and in this case, Fig 10.19 rr we define ab 0 Observations rr1. ab is a real number. ⋅ r rrr2. Let aand bbe two nonzero vectors, then ⋅= if and only if aand br ab 0 r are perpendicular to each other. i.e. rrrrab 0 ab⋅=⇔ ⊥ rrrr3. If θ = 0, then ⋅=||| |ab ab rr rIn particular, aa ||, as θ in this case is 0.⋅=a2 rrrr4. If θ = π, then ⋅= ab |ab−||| rr rIn particular, a( a)| a|2, as θ in this case is π. 5. In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆˆˆ, and kwe haveij , ˆˆˆ ˆˆii jjˆ⋅=⋅=⋅ = kk 1, ˆˆˆˆ ˆˆijjkki 0⋅=⋅ = r 6. The angle between two nonzero vectors arand bis given by rrr rab –1 ⎛ab.⎞ cos rr, or θ=cos ⎜rr⎟||| | ⎝||| | ⎠ab ab 7. The scalar product is commutative. i.e. rrab⋅= ⋅rr ba (Why?) Two important properties of scalar product rr rProperty 1 (Distributivity of scalar product over addition) Let , beaband c any three vectors, then rrrrr ⋅+ = r⋅ r +⋅ab( c) ab ac rrProperty 2 Let a and b be any two vectors, and λ be any scalar. Then rr rrrrrr()⋅ () b (ab) a (λab = ab) rrIf two vectors a and b are given in component form as aiˆ + a ˆj +ak ˆ and12 3 biˆ + bj ˆ + bk ˆ , then their scalar product is given as12 3 rr⋅ ˆ ˆ ˆ)( ˆˆˆab =(ai + aj + ak ⋅ bi + bj + bk )12 3 123 = ai ˆ ⋅ (biˆ + bj ˆ + bk ˆ) + a ˆj ⋅ (biˆ + bj ˆ +bk ˆ) + akˆ ⋅(bi ˆ + bj ˆ + bk ˆ)1123 2123 3123 ˆˆˆˆˆˆ ˆˆ ˆˆ ˆˆ=( ⋅ ) + ab i ⋅ j) + ab i ⋅ k) + ab ( ji ) + ab ( jj ) + ab ( jk )ab i i (( ⋅⋅ ⋅1112 13 21 22 23 ˆˆ ˆˆ+ ab ki ( ⋅ ˆ) + ab (k ⋅ ˆj) + ab (kk ) (Using the above Properties 1 and 2)⋅3132 33 = ab + ab + ab(Using Observation 5)112233 rr⋅ ab + ab Thus ab = + ab 1122 33 10.6.2 Projection of a vector on a line uuurSuppose a vector AB makes an angle θ with a given directed line l (say), in the uuurr anticlockwise direction (Fig 10.20). Then the projection of AB on l is a vector p uuur r(say) with magnitude |AB | cos θ , and the direction of p being the same (or opposite) rto that of the line l, depending upon whether cos θ is positive or negative. The vector p B B AC A p C p 00 00(0 < θ <90) (90< θ <180) (i) (ii) θθp pC l C θ l A B B (180 < < 270 ) 0 0θ (270 < < 360 ) 0 0θ (iii) Fig 10.20 (iv) ris called the projection vector, and its magnitude | p | is simply called as the projection uuurof the vector AB on the directed line l. For example, in each of the following figures (Fig 10.20(i) to (iv)), projection vector uuur uuurof AB along the line l is vector AC . Observations 1. If pˆ is the unit vector along a line l, then the projection of a vector a r on the line rl is given by apˆ. rr2. Projection of a vector a on other vector b , is given by r rr ⎛b ⎞ 1 rr ⋅ˆab, or a ⋅⎜ r ⎟, or r (a ⋅b) ⎝||b ⎠ ||b uuuruuur3. If θ = 0, then the projection vector of AB will be AB itself and if θ = π, then the uuuruuur projection vector of AB will be BA . uuurπ 3π4. If θ= or θ= , then the projection vector of AB will be zero vector. 22 rRemark If α, β and γ are the direction angles of vector aai = ˆ +a ˆj +ak ˆ , then its12 3 direction cosines may be given as raiˆ a1 a2 a3cos rr , cos r , and cos r|||| aiˆ| a | ||a ||a rr rAlso, note that | a | cos α, | |cos β aa and | |cos γare respectively the projections ofrr a along OX, OY and OZ. i.e., the scalar components a1, a2 and a3 of the vector a , rare precisely the projections of a along x-axis, y-axis and z-axis, respectively. Further, rif a is a unit vector, then it may be expressed in terms of its direction cosines as ˆˆ ˆa r =cos i cos β+ α+ j cos γk rrExample 13 Find the angle between two vectors a and b with magnitudes 1 and 2 rrrespectively and when ⋅=1ab. r rrrSolution Given ab 1, | a | 1and |b | 2. We have rr 1 ab 11 cos r cos rab | 23||| rr ˆˆˆ ˆˆˆExample 14 Find angle ‘θ’ between the vectors ai jk=+− and =−+ bi jk . rrSolution The angle θ between two vectors aand bis given by rr⋅ab cosθ = rr ||| |ab rr ˆˆˆ ˆˆˆNow ab=(ijk ijk)( )111 −1.⋅ +− ⋅ −+ =−−=−1 Therefore, we have cosθ = 3 11 hence the required angle is θ = cos 3 rr =5 −− and bi 5 ˆ , then show that the vectorsExample 15 If aij k ˆˆ3ˆ =+ − ˆ 3ˆjk rrrr−ab+ and ab are perpendicular. Solution We know that two nonzero vectors are perpendicular if their scalar product is zero. rrˆˆˆ ˆˆˆ ˆˆˆHere ab= (5 −−3) ( i3 jk =6 +− 8+ ij k ++− 5) i2 jk ˆˆ ˆand ab= (5ij k −+− (ˆ jk 5) =4ˆ jk r− r −−3 ) ˆ i3ˆ i4ˆ ˆ−+ 2 rrr rSo (ab ab ) = iˆ 2ˆjk ˆ ⋅iˆ 4ˆjk ˆ =24 −− 16 )( (6 +− 8 ) (4 −+ 2) 8 =0. +⋅− rrrr−Hence ab+ and ab are perpendicular vectors. rExample 16 Find the projection of the vector =2iˆ ˆ 2 ˆ on the vectora ++ 3 jk r biˆ2ˆjkˆ.=+ + rrSolution The projection of vector aon the vector bis given by rr (2×1+×+× 3221) 10 51(ab) = == 6r⋅ 222 3b|| (1) +(2) +(1) 6 rr rrr rExample 17 Find | −| , if two vectors and b a 2,| 3ab a are such that || b| rr and ab 4.⋅= Solution We have r rrrrr =( −⋅− | ab|2 ab)( ab) rr rr rrr r =.bbabbaaa−⋅−⋅+⋅ rrrr =||a2 2( a)| 2bb|− ⋅+ = (2) 2 −2(4) +(3) 2 Therefore | ab=5r− r| rrr rr rExample 18 If ais a unit vector and (xa xa ) 8, then find ||.−)( ⋅+ = x rrSolution Since a is a unit vector, ||=a 1 . Also, rr rr(xa xa ) ( ) =8−⋅+rrrrrrror xr⋅xx+⋅−⋅−⋅xa =8aaa r 1xor ||2 =8 i.e. | xr|2 = 9 rxTherefore ||= 3 (as magnitude of a vector is non negative). rrrr rrab |||| Example 19 For any two vectors aand b, we always have | ⋅≤| ab (Cauchy-Schwartz inequality). r rrrSolution The inequality holds trivially when either a=0 or b=0 . Actually, in such a rr rrr rsituation we have | ⋅ ==|| ab | . So, let us assume that ||a|.ab0 || ≠≠0 b| Then, we have r | ab|r⋅ rr= |cos θ≤|1||| |ab rr rr⋅| abTherefore | ab≤ ||| | rrExample 20 For any two vectors and b, we always a rr have | ab|| a| +| b| (triangle inequality).r+≤ r ASolution The inequality holds trivially in case either rr rr rrr a=0or b=0 (How?). So, let ||a 0|b| . Then, r rrrr2 r 2 rr Fig 10.21| ab|(ab) =(ab ab )+ = + +⋅+ )( rr rrrrr r = aaabbabb⋅+⋅+⋅+⋅ rr rr2 bb|+⋅+ =||a2 a| 2 (scalar product is commutative) rrrr a2 ⋅+ 2 x≤||≤ || +2| ab|| b| (since x∀∈x R ) rrrr≤ ||2 +2| ab | +| b|2 (from Example 19)a || rr = (| a|| b|) 2 r rrr| ≤ || b|Hence | ab a | Remark If the equality holds in triangle inequality (in the above Example 20), i.e. r rrr| ab| || +| b|+ = a , uuuruuur uuur then |AC| = |AB| +|BC | showing that the points A, B and C are collinear. Example 21 Show that the points A( 2 3 jk 5 ), B( i++jk 3 ) C(7ikˆ −−+ + iˆ ˆ ˆ ˆ2ˆ ˆ and ˆ) are collinear. Solution We have uuurAB =(1 2) iˆ (2 3) ˆj(3 5) ˆ3ˆ ˆ2ˆ,k ij kuuurBC =(7 1) iˆ (0 2) ˆj(1 3) ˆ 6ˆ 2ˆ 4ˆki jk, uuurAC = ˆ ˆ ˆ9ˆˆ6ˆ(7 2) i(0 3) j(1 5) ki 3 jk uuuruuur uuur |AB| = 14, |BC | 2 14 and |AC | uuuruuur uuur AC = |AB| +|BC | Therefore Hence the points A, B and C are collinear. rr1. Find the angle between two vectors aand bwith magnitudes 3 and 2 , rrespectively having abr ⋅=2. Find the angle between the vectors and 3. Find the projection of the vector ˆˆ+.− on the vector ˆˆijij4. Find the projection of the vector ˆ 3 ˆjkˆ on the vector 7ˆˆ 8ˆi++7 ij k−+ . 5. Show that each of the given three vectors is a unit vector: 11 1ˆˆˆ ˆˆˆ ˆˆˆ+− 3) (2 i++ 3 jk 6 ), (3i−+ 6 jk 2 ), (6i2 jk 77 7 Also, show that they are mutually perpendicular to each other. r rr rr rrr6. Find || and | ab|, if (ab)( ab ) =8 and | a|=8| b|+⋅− . rrrr7. Evaluate the product (3 b⋅ 7) b.a−5 )(2 a+ rr8. Find the magnitude of two vectors aand b, having the same magnitude and 1such that the angle between them is 60o and their scalar product is . 2 r rr rr9. Find ||, if for a unit vector a, (xa)( xa ) =12 .x r −⋅+ rr r rˆ r ˆ++ ˆ =+ a10. If a=2iˆ 2ˆj 3, kb =−+ + iˆ2 jk and c 3iˆˆjare such that +λb is rperpendicular to c, then find the value of λ. rr rrrr r11. Show that || + is perpendicular to || − |r , for any two nonzeroab ba|| ab ba| r vectors ar and b. r rrr r⋅= , then what can be concluded about the vector b12. If aa 0 and ab ⋅= 0? r rrr rrr13. If ,, are unit vectors such that ++= 0 , find the value ofabc abcrrr rrrab bc ca.⋅+⋅+⋅ r rr rr r14. If either vector a=0 or b 0, then ⋅= 0 . But the converse need not be= ab 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), uuur respectively, then find ∠ABC. [∠ABC is the angle between the vectors BA uuur and BC ]. 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 ˆˆ ˆ ˆ ˆˆˆ ˆ −−ˆ form the vertices2i−+,3 j 5ki 4j ki −− and 3 j 4k of a right angled triangle. r18. If ais a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ a r 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 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) r rrrDefinition 3 The vector product of two nonzero vectors aand b, is denoted by ab and defined as rr rab× = ||| |sin θˆ, rab n rrwhere, θ is the angle between aand b, 0 ≤θ≤πand nˆ is rra unit vector perpendicular to both a and b, such that rr, and nˆ form a right handed system (Fig 10.23). i.e., theab rrright handed system rotated from to moves in theab Fig 10.23 direction of nˆ. rrr rrrrIf either a=0or b=0 , then θ is not defined and in this case, we define ×=ab 0. Observations rr×1. ab is a vector. r rr rrrr2. Let a be two nonzero vectors. Then ×= if and only if a and and b ab 0 b are parallel (or collinear) to each other, i.e., rr rr× 0 rab= ⇔ab� rrrrrr ×=a()−=In particular, aa 0 and × a0 , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0. rrrr3. If then ab ab ||| |. 24. In view of the Observations 2 and 3, for mutually perpendicular ˆˆˆunit vectors ij, and k(Fig 10.24), we have ˆˆˆˆˆˆ r ii= ×=×= ×jj kk 0 ˆˆ Fig 10.24ijˆ ˆˆˆˆˆ ˆ×= k jkiki j, ×=,×= rr5. In terms of vector product, the angle between two vectors aand bmay be given as rr | ab|×sin θ = rr ||| |ab r rr r× ba.6. It is always true that the vector product is not commutative, as ab= −× rr rrr rIndeed, ab ab ×=||| | sin θnˆ , where ab form a right handed system,, and nˆ rr r r i.e., θ is traversed from rtor , Fig 10.25 (i). While, ×=||| | sin θn1, wherea b baab ˆ rrr rba,and nˆ1 form a right handed system i.e. θ is traversed from bto a, Fig 10.25(ii). Fig 10.25 (i), (ii) rrThus, if we assume aand b to lie in the plane of the paper, then nˆ and nˆ1 both will be perpendicular to the plane of the paper. But, nˆ being directed above the paper while nˆ1 directed below the paper. i.e. nˆ1 =−nˆ. rr rHence r =ˆab× ||| ab| sin n rr = −||| | sin θˆ1ab n 7. In view of the Observations 4 and 6, we have r bar=− × ˆˆ ˆˆˆˆ ˆˆˆkkj ik =−j.ji× =− ,× =−i and × rr8. If aand brepresent the adjacent sides of a triangle then its area is given as 1 rr | ab|.2 By definition of the area of a triangle, we have from Fig 10.26, 1 ⋅Area of triangle ABC = AB CD. 2 Fig 10.26rrBut AB =| b| (as given), and CD = ||sin θ.a 1 rr 1 rr ba θ× |. Thus, Area of triangle ABC = ||| | sin = | ab 22 rr9. If a and brepresent the adjacent sides of a parallelogram, then its area is rrab. From Fig 10.27, we have Area of parallelogram ABCD = AB. DE. given by | × | rBut AB =| b| (as given), and rDE =| a| sin θ. Thus, Fig 10.27 r rrrθ× |. Area of parallelogram ABCD = ||| ba| sin =| abWe 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 rrr rr rr(i) a bc( + ) ab ac × = r rrrrr(ii) (ab) = λaba ×λ (λ× () ×= b) rrr,andcab rrLet aand b be two vectors given in component form as aiˆ +aj ˆ +ak ˆ and12 3 biˆ +bj ˆ +bk ˆ , respectively. Then their cross product may be given by 12 3 iˆˆjkˆ rraaaab =× 123 bb b123 Explanation We have rrab =(aiˆ +aj ˆ +ak ˆ) ×bi ˆ +bj ˆ +bk ˆ)× (123 123 = abi i (ˆˆ) ab i ˆˆj) ab i ˆ ×+ kˆ) ab j i (ˆ ×ˆ)×+ ( ×+ (11 12 13 21 ˆˆ ˆˆ+ abj j () ab jk )×+ ( ×22 23 + abk i ( ×+ ( j ( ×ˆ ˆ) ab k ˆ ×+ ˆ) abk k ˆ ˆ) (by Property 1)3132 33 ˆˆ ˆˆˆˆ= ab i ( ×j) −abk i ) ab i ×j)( ×− (1213 21 ˆˆˆˆ ˆˆ+ ab jk () abk i ) ab jk )×+ ( ×− ( ×23 31 32 ˆˆ×=×=×= ˆˆˆˆ ˆ ×=ˆ −× ˆ ˆ,ˆ ×ˆ ˆˆ kˆ ×=ˆj −× ˆjk ˆ)(as ii jj kk 0 and ik kiji =−i×j and = abk ˆ −ab j ˆ −abk ˆ +abi ˆ +abj ˆ −abi ˆ12 13 21 2331 32 ˆˆ ˆˆˆˆ ˆˆˆ(as ijk,jk i and ki =j)×=×= ×)ˆ )ˆ ab −ab k ˆ=(ab −ab i −(ab −ab j +()2332 13 31 1221 iˆˆjkˆ aaa= 123 b1 b2 b3 rr Example 22 Find | ab×|, if a=iˆˆj 3 ˆ and b 3iˆ 5ˆj 2ˆrr 2 ++ k =+ − k Solution We have ab× rr = ˆˆ 21 i j ˆ 3 k 35 2− = ˆ(2 15) −−− 4 9) ˆj+(10 –3) kˆ =−17iˆ +13 ˆj+7kˆi −− ( rr2 22Hence | ab| = −+(13) +(7) ( 17) =507 rExample 23 Find a unit vector perpendicular to each of the vectors ( r+ andab) r rr r ˆˆˆ ˆˆˆ(ab), where ai jkbi 3.− =++, =+ + 2 jk rrrrSolution We have ab 2ˆ ++ 3ˆjk ˆ and −=ˆjk ˆ+=i 4 ab −− 2 A vector which is perpendicular to both rrrr(ab)( ab )+× − = rcNow || = Therefore, the required unit vector is r c r = ||c rr aband ab is given byr + r− iˆˆjk ˆ 23 4 0 −− 12 r=−+ − 2iˆ 4ˆjk ˆ(2 =c, say) 416 4 =24 =26++ −12 1ˆˆˆi+ j− k 666 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 ˆˆAB 2j k =+ uuur and AC uuur ˆ ˆ2i j=+ . The area of the given triangle is 1 2 |AB AC | × uuur uuur . ˆˆˆi jk uuur uuur012 4iˆ 2ˆjkˆ=−+ − Now, AB ×AC = 120 uuur uuurTherefore |AB ×AC| = 1 Thus, the required area is 2 Example 25 Find the area of a parallelogram whose adjacent sides are given rr ˆˆˆ ˆˆˆby the vectors ij and =a=3 ++4k bi jk −+ rrSolution The area of a parallelogram with aand bas its adjacent sides is given rrby | ab|.× Now ab× rr = ˆˆ ˆ ˆˆˆ31 4 5 4 i j k i j k = +− 1 1− 1 rrTherefore | ab| =× and hence, the required area is r rrr ˆˆˆ ˆˆˆ×=7 jk =− + 1. Find | ab|, if ai −+ 7 and b 3i 2 j 2k. rrrr −2. Find a unit vector perpendicular to each of the vector ab+ and ab , where rr 3iˆ 2ˆjk ˆ and bi ˆ +− 2ˆ 2ˆ.a=+ + 2 =jk 3. If a unit vector a r makes angles πwith iˆ, πwith ˆjand an acute angle θ with 34 rˆk, then find θ and hence, the components of a. 4. Show that rr rrrr(ab)( ab ) ×)−× + = 2( abr 5. Find λ and μ if (2iˆ +6ˆj+27 ) ˆ ×+(iˆ λˆj+μ = kˆ 0.k ) r rr 6. Given that r and r ×=ab 0 ab 0 . What can you conclude about the vectors rr aand b? rrrˆ ˆ ˆˆˆˆ+ kb+k,7. Let the vectors ,, be given as ai a+ ja, i bjb+abc 12 3123 rrr rr rrˆˆ+ . Then show that ×+ .ci cˆjc+k a bc abac()=×+× 12 3 r rr rrr rab 0 . Is the converse true? Justify your answer with an example. 8. If either a=0 or b=0, then ×= 9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). 10. Find the area of the parallelogram whose adjacent sides are determined by the rr ˆˆ=− +3=− jk vectors aij kˆˆ and bi2ˆ 7ˆ + . rr rrr 2 r11. Let the vectors aand bbe such that a=3 and | b|= , then ab|| × is a 3 rrunit vector, if the angle between aand bis (A) π/6 (B) π/4 (C) π/3 (D) π/212. Area of a rectangle having vertices A, B, C and D with position vectors 111 1ˆˆˆˆj ki ˆˆjk ˆˆjk 4ˆ –iˆ − ˆ 4ˆ– i++4, ++4, i−+ and + , respectively isjk 222 2 1(A) (B) 1 2 (C)2 (D)4 Miscellaneous Examples Example 26 Write all the unit vectors in XY-plane. r∧∧Solution Let r=xiy be a unit vector in XY-plane (Fig 10.28). Then, from the + j rfigure, we have x= cos θ and y= sin θ (since | r r | = 1). So, we may write the vector ras uuurr r(=OP)= cos iˆ sin ˆj ... (1) rrClearly, || = 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 vectorin the XY-plane. Example 27 If ijk i ˆˆ ˆ,2ˆ ˆji ˆ 2ˆjk ˆ and ˆ jkˆ are the position5,3 3i 6ˆ uuurvectors of points A, B, C and D respectively, then find the angle between AB and uuuruuur uuurCD . Deduce that AB and CD are collinear. Solution Note that if θ is the angle between AB and CD, then θ is also the angle uuur uuur between AB and CD . uuur Now AB = Position vector of B – Position vector of A = (2ˆ +5 ) ˆj (ijk i ˆˆ ˆ)ˆ4ˆjk ˆi −++=+ − uuur22 2Therefore |AB | = (1) +(4) +( 1) =3− 2 uuur uuurSimilarly CD −− + 2iˆ 8ˆ 2ˆ =62= jk and |CD| uuur uuur AB CD uuur uuur Thus cos θ = |AB||CD| 1(2) 4(8) (1)(2) − 36−+−+− = = =−1 (3 2)(6 2) 36 uuuruuurSince 0 ≤ θ ≤ π, it follows that θ = π. This shows that AB and CD are collinear. uuur uuuruuur uuur1Alternatively, AB CD which implies that AB and CD are collinear vectors. 2 r rrrrraband be three vectors such that || 3, | b =Example 28 Let ,ca= |=4,| | c 5 and rrreach one of them being perpendicular to the sum of the other two, find | ++|abc. r rrrrrrrrSolution Given ⋅+ ( = 0, bca) =0, cab ( =0. abc) ⋅( + ⋅+ ) r r rrrrrrrrrrNow | abc2 = abc ++)2 )( ++|( = ++⋅++ (abc abc ) r rrrrrrr rr = aaabc bbbac() (⋅+⋅ + +⋅+⋅+ ) rrr rr cab c+ .( ++.c) rrr222=|| +| b| +| c|a = 9 + 16 + 25 = 50 r Therefore | abc| = r ++r50 =52 rrrrrrrExample 29 Three vectors ,csatisfy the condition abc 0 . Evaluateaband ++= rr rr rrrr rthe quantity μ= ⋅+⋅+⋅ ab bc ca , if | a|=1, | b|=4 and | c|=2. rrrrSolution Since abc 0, we have++= rr rr( ) =0aab crr r r rr or aa abac=0⋅+ ⋅+ ⋅ rrrr r⋅+⋅ −a 2 =−1 ... (1)Therefore ab ac= rrrAgain, babcr=0⋅(++) rrr2r⋅+⋅r or ab bc= −b =−16 ... (2) rrrrSimilarly ac bc =– 4. ⋅+⋅... (3) Adding (1), (2) and (3), we have rrrrrr2( ab b⋅+⋅+⋅ c ac) =–21 −21 or 2μ = – 21, i.e., μ = 2 Example 30 If with reference to the right handed system of mutually perpendicular r rrˆˆˆ ˆˆ ˆˆˆα= − i , then express βin the formunit vectors ij, and k,3ij, β=2 +j–3 k rrr r r rr,where is parallel to and is perpendicular to α.121 2 r rrSolution Let 1 , is a scalar, i.e., 1 3 iˆ λβ=λ−ˆj. r rrNow β2 =β−β1 = (2 −λ+ +3 ) ˆ (1 )ˆj 3kˆ.i λ− rrrrNow, since βis to be perpendicular to α, we should have α⋅β= 0 . i.e.,22 3(2 −λ− +3) (1λ) =0 1 or λ = 2 rr31 13Therefore = iˆ − ˆj and β= iˆ + ˆj–3kˆβ1222 22 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. r rr rr4. =+r , then is it true that || | = ? Justify your answer. If abc ab | +| c| (ˆˆ ˆ5. Find the value of xfor which ijk) is a unit vector. x++6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a r =2iˆ +3ˆjk −ˆ and bi r ˆ 2ˆjk +ˆ=− . rr r7. If =++ ˆ, =2ˆ −+ 3 and ci ˆ2 +ˆ , find a unit vector parallelˆˆ ˆˆ =−jkaijkb ij k ˆ rrrto the vector 2–ab+3c. 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 rrrrP and Q whose position vectors are (2ab) and ( ab – 3 ) 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 iˆˆ +5ˆ ˆ2ˆjk ˆ.2 −4 jk and i−− 3 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 11and OZ are r rr rˆˆˆ ˆˆˆ ˆˆˆ12. Let ai 4 +2, =3 −2 +7 and c=2ij k =+jkbi jk −+ 4 . Find a vector d r r which is perpendicular to both and b, and cd15 .a rr ⋅= ˆˆ+ + with a unit vector along the sum of13. The scalar product of the vector i jkˆ vectors ˆ ˆ −5ˆ and ˆ 2ˆjk ˆ is equal to one. Find the value of λ.2i+4 jkλi++3 rrr14. If ab,, c are mutually perpendicular vectors of equal magnitudes, show that r rr rrthe vector abc++r is equally inclined to aband c,. rrr rrrr r15. Prove that +⋅ )| 2 +| | 2 , if and only if , are perpendicular, (ab ab )( += a| b ab rr rrgiven a≠0, b≠0. Choose the correct answer in Exercises 16 to 19. rr rr16. If θ is the angle between two vectors a and b, then ⋅≥ only whenab 0 π π(A) 0 <θ< (B) 0 ≤θ≤ 2 2 (C) 0 < θ < π (D) 0 ≤θ≤π r r aand be two unit vectors and θ is the angle between them. Then abis a unit vector if 17. Let rb r + π ππ 2π(A) θ= (B) θ= (C) θ= (D) θ= 4 323 ˆˆˆ ˆˆˆ ˆˆˆijk jik kijis18. The value of .( ) ( ) ( ) (A)0 (B)–1 (C)1 (D)3 rrrr rr19. If θ is the angle between any two vectors aand b, then | ab⋅= ab× || | when θ is equal to ππ(A) 0 (B) (C) (D) π42 —•—