Chapter 6 Vectors And Scalars
Chapter 6135Vectors and ScalarsChapter 6Vectors and Scalars6.1Introduction:In this chapter we shall use the ideas of the plane to develop a newmathematical concept, vector. If you have studied physics, you haveencountered this concept in that part of physics concerned with forces andequilibrium.Physicists were responsible for first conceiving the idea of avector, but the mathematical concept of vectors has become important inits own right and has extremely wide application, not only in the sciencesbut in mathematics as well.6.2Scalars and Vectors:A quantity which is completely specified by a certain numberassociated with a suitable unit without any mention of direction in space isknown as scalar. Examples of scalar are time, mass, length, volume,density, temperature, energy, distance, speed etc. The number describingthe quantity of a particular scalar is known as its magnitude. The scalarsare added subtracted, multiplied and divided by the usual arithmeticallaws.A quantity which is completely described only when both theirmagnitude and direction are specified is known as vector. Examples ofvector are force, velocity, acceleration, displacement, torque, momentum,gravitational force, electric and magnetic intensities etc. A vector isrepresented by a Roman letter in bold face and its magnitude, by the sameletter in italics. Thus V means vector and V is magnitude.6.3Vector Representations:A vector quantity is represented by a straight line segment, say. The arrow head indicate the direction from P to Q. The length of theVector represents its magnitude. Sometimes the vectors are represented bysingle letter such as V oror by just V, where . The magnitude of a vector is denoted by V means modulus ofwhich is a positive valueFig. 1
Chapter 66.4136Vectors and ScalarsTypes of Vectors:1.Unit Vector:A vector whose magnitude is unity i.e., 1 and direction along thegiven vector is called a unit Vector. If a is a vector then a unit vector inthe direction of a , denoted by a (read as a cap), is given as,a a a ora a a2.Free Vector:A vector whose position is not fixed in space. Thus, the line ofaction of a free vector can be shifted parallel to itself. Displacement is anexample of a free vector as shown in figure 1:3.Localized or Bounded Vectors:A vector which cannot be shifted parallel toitself, i.e., whose line of action is fixed is called alocalized or bounded vector. Force and momentumare examples of localized vectors.4.Coplanar Vectors:The vectors which lies in the same planeare called coplanar vectors, as shown in Fig. 2.5.Concurrent Vectors:The vectors which pass through thecommon point are called concurrent vectors. In thefigure no.3 vectors a, b and c are calledconcurrent as they pass through the same point.6.Negative of a Vector:The vector which has the same magnitudeas the vector a but opposite in direction to a iscalled the negative to a . It is represented by a . Thus of AB a then BA aFig. 4
Chapter 6137Vectors and Scalars7.Null or Zero Vector:It is a vector whose magnitude is zero. We denote the null vectorby O. The direction of a zero vector is arbitrary.The vectors other than zero vectors are proper vectors or non-zero vectors.8.Equal Vectors:Two vectors a and b are said to be equal if they have the samemagnitude and direction. If a and b are equal vectors then a b9.Parallel and Collinear Vectors:The vectors a and b are parallel if for any real number n,a n b . If(i)n 0 then the vectors a and b have the same direction.(ii)n 0 then a and b have opposite directions.Now, we can also define collinear vectors which lie along the samestraight line or having their directions parallel to one another.10.Like and Unlike Vectors:The vectors having same direction are called like vectors and thosehaving opposite directions are called unlike vectors.11.Position Vectors (PV):If vector OA is used to specify the position of a point A relativeto another point O. This OA is called the position vector of A referred toO as origin. In the figure 4 a OA and OB b are the position vector(P.V) of A and B respectively. The vector AB is determined as follows:By the head and tail rules,OA AB OBAB OB OA b aOrYBAOXFig. 5
Chapter 6138Vectors and Scalars6.5Addition and Subtraction of Vectors:1. Addition of Vectors:Suppose a and b are any twovectors. Choose point A so thata OA and choose point C so thatb AC . The sum, a b of a andb is the vector is the vector OC . Thusthe sum of two vectors a and b isperformed by the Triangle Law ofaddition.2.Subtraction of Vectors:If a vectoris to be subtracted from a vectorvectorcan be obtained by adding vectorsThe vector , the differenceand .is a vector which is equal and parallel to that ofvector but its arrow-head points in opposite direction. Now the vectorsand can be added by the head-to-tail rule. Thus the lineAC represents, in magnitude and direction, the vector. b BCba ( b)a bAaFig . 7Properties of Vector Addition:i.Vector addition is commutativei.e., a b a b where and b are any two vectors.
Chapter 6(ii)139Vectors and ScalarsVectors Addition is Associative:i.e. a b c a b c where a , b and c are any three vectors.(iii)O is the identity in vectors addition:Fig.9For every vector aa O aWhere O is the zero vector.Remarks: Non-parallel vectors are not added or subtracted by theordinary algebraic Laws because their resultant depends upon theirdirections as well.6.6Multiplication of a Vector by a Scalar:If a is any vectors and K is a scalar, then K a a K is a vectorwith magnitude K . a i.e., K times the magnitude of a and whosedirection is that of vector a or opposite to vector a according as K ispositive or negative resp. In particular a and a are opposite vectors.Properties of Multiplication of Vectors by Scalars:1.The scalar multiplication of a vectors satisfiesm(n a ) (mn) a n(m a )2.The scalar multiplication of a vector satisfies the distributive laws(m n) a m a n ai.e.,
Chapter 6and140Vectors and Scalarsm( a b ) m a m bWhere m and n are scalars and a and b are vectors.The Unit Vectors i, j, k (orthogonal system of unit Vectors):Let us consider three mutually perpendicular straight lines OX, OYand OZ. These three mutually perpendicular lines determine uniquely theposition of a point. Hence these lines may be taken as the co-ordinatesaxes with O as the origin.ZWe shall use i, j and k to denote thekUnit Vectors along OX, OY andOX respectively.j6.7iXOFig. 9a6.8 Representation of a Vector in theForm of Unit Vectors i, j and k.Let us consider a vector r OP as shown in fig. 11. Then x i, y jand z k are vectors directed along the axes,becauseandOQ xi yiBecauseQP zkOP OQ QPr OP xi yj zkandHere the real numbers x, y and z are the components of Vectorr or the co-ordinates of point P in the direction of OX, OY and OZrespectively. The vectors xi, yj and zk are called the resolved parts of thevector r in the direction of the Unit vectors i, j and k respectively.Y
Chapter 6141Vectors and ScalarsZCzkP(x,y,zryjO)zkYxiBxi y jFig. 10Q(x,y,o)AX06.9Components of a Vector when the Tail is not at the Origin:Consider a vector r PQ whose tail is at the point P(x1, y1, z1)and the head at the point Q (x2, y2, z2). Draw perpendiculars PP and QQ on x-aixs.P Q x2 – x1 x-component of rNow draw perpendiculars PPo and QQo on y-axis.Then PoQo y2 – y1 y-component of rSimilarlyz2 – z1 z-component of rHence the vector r can be written as,r PQ (x2 – x1)i (y2 – y1)j (z2 – z1)kOr, r PQ (x2 – x1, y2 – y1, z2 – z1)
Chapter 6142Vectors and Scalars6.10Magnitude or Modulus of a Vector:Suppose x, y and z are the magnitude of the vectors OA , OB andOC as shown in fig. 10.In the right triangle OAQ, by Pythagorean TheoremOQ2 x2 y2Also in the right triangle OQP, we haveOP2 OQ2 QP2OP2 x2 y2 z2Orr OP x 2 y2 z 2r PQ xi yj zkThus ifThen , its magnitude isrIfThen x 2 y2 z 2r (x2 – x1)i (y2 – y1)j (z2 – z1)kr (x 2 x1 ) 2 (y 2 y1 ) 2 (z 2 z1 ) 2Example 1:If P1 P(7, 4, 1) and P2 P(3, 5, 4), what are the componentsof P1P2? Express P1 P2 in terms of i, j and k.Solution:x-component of P1 P2 x2 – x1 3 – 7 –4y-component of P1 P2 y2 – y1 –5 –4 –9andz-component of P1 P2 z2 – z1 4 – (–1) 5alsoP1 P2 (x2 – x1)i (y2 – y1)j (z2 – z1)kP1 P2 –4i –9j 5kExample 2:Find the magnitude of the vectoru 3 22 3i j k5 55Solution:22 2 3 3 2 u 5 5 5 u 194 1225 25 25 25252
Chapter 6143Vectors and ScalarsNote: Two vectors are equal if and only if the corresponding componentsof these vectors are equal relative to the same co-ordinate system.Example 3:Find real numbers x, y and z such thatxi 2yj – zk 3i – j 4i 3kSolution:Since (x 3)i (2y – 1)j (–z)k 4i 3kComparing both sides,we getx 3 4, 2y – 1 0,–z 3x 1, y Note 2:,z –3r1 x1i y1 j z1kIfr2 x 2i y2 j z2kThen the sum vector r1 r2 (x1 x 2 )i (y1 y2 )j (z1 z2 )kr1 r2 (x1 x2 , y1 y2 , z1 z2 )OrExample 4:a 3i 2 j 5k and b 2i j k .Find 2 a 3 b and also its unit vector.Solution:2 a 3 b 2(3i – 2j 5k) – 3(–2i – j k) 6i – 4j 10k 6i 3j – 3k 12i – j 7kIf we denote 2 a 3 b c , then c 12i j 7kand Therefore,c c12i j 7k c1941211i j k194194194Note 3: Two vectors r1 x1i y1 j z1k and r2 x 2i y2 j z2k arex1 y1 z1parallel if and only if .x 2 y2 z2c
Chapter 6144Vectors and Scalars6.11Direction Cosines:Let us consider that the vector r OP which makes angles , and with the coordinate axes OX, OY and OZ respectively. ThenCos , Cos and Cos are called the direction cosines of the vectorOP . They are usually denoted by l, m and n respectively.If OP r xi yj zk,then x, y and z are defined asthe direction ratios of thevectorandrr x 2 y2 z 2 .Sincethe angles A, B and C are rightangles (by the fig. 11), so in theright triangles.OAP, OBP and OCPthe direction cosines of r canbe written as,l cos m cos andn cos x ry rz rxx 2 y2 z2yx 2 y2 z2zx 2 y2 z2rxi yj zk rrNote 1: Since the unit vector r̂ xyzi j krrrr̂ Cos i Cos j Cos kr̂ li mj nkr̂ OrTherefore the co-efficient of i, j and k in the unit vector are thedirection cosines of a vector.Note 2:l 2 m2 n 2 x2r2 y2r2 z2r2
Chapter 6145 x 2 y2 z 2r2Vectors and Scalars x 2 y2 z 2 1x 2 y2 z 2Example 5:Find the magnitude and direction cosines of the vectors3i 7j – 4k, i – 5j – 8k and 6i – 2j 12k.Solution:a 3i 7j 4kLetb i 5 j 8kc 6i 2 j 12kNowâ a 3i 7j 4k a7437i j 74743So the direction cosines of a are:,744k7474, 7474158Similarly the direction cosines of b are:, , 9090906212and the direction cosines of c are:, ,184184 184 Exercise 6.1Q.1Q.2Q.3Q.4Q.5If a 3i j 4k, b 2i 4j 3k and c i 2j k .Find unit vector parallel to 3 a 2 b 4 c .Find the vector whose magnitude is 5 and which is in the directionof the vector 4i – 3j k.For what value of m, the vector 4i 2j – 3k and mi – j 3 khave same magnitude?Given the points A (1, 2, 1), B ( 3, 1, 2) and C (0, 4, 3)(i) find AB, BC, AC(ii) Show that AB BC ACFind the lengths of the sides of a triangle, whose vertices areA (2, 4, –1), B (4, 5, 1), C (3, 6, –3) and show that thetriangle is right angled.
Chapter 6146Vectors and ScalarsQ.6If vectors 3i j – k and i – 4j 4k are parallel, find the valueof .Q.7Show that the vectors 4i – 6j 9k and –6i 9j –Q.8Find real numbers x, y and z such that(a)7xi (y – 3)j 6k 10i 8j – 3zk(b)(x 4)i (y – 5)j (z – 1)k 0Given the vectors a 3i 2j 4k and b 2i j 3k find themagnitude and direction cosines of(i )(ii)3Q.927k are parallel.2Q.10 If the position vector of A and B 5i – 2j 4k and i 3j 7krespectively, find the magnitude and direction cosines of AB .Answers 6.115Q.2(17i 3j 10k)(4i 3j k)39826Q.4( 4, 1, 3),(3, 5, 1),( 1, 6,4)Q.3 5Q.5AB AC 3, BC 3 2Q.6 1210x y 11 z - 2Q.8(a)(b)x -4 , y 5 , z 171 3 11 86Q.9(a)(b) 5 5;11;11 11 115 5 5 5 5 453Q.1050;,,505050Q.16.12 Product of Vectors:1.Scalar Product of two Vectors:If a and b are non-zero vectors,and θ is the angle between them, then thescalar product of a and b is denoted bya . b and read as a dot b . It is defined bythe relationa . b a b Cos θ .Fig.14(1)If either a or b is the zero vector, then a . b 0
Chapter 6147Vectors and ScalarsRemarks:i. The scalar product of two vectors is also called the dot productbecause the “.” used to indicate this kind of multiplication.Sometimes it is also called the inner product.ii. The scalar product of two non-zero vectors is zero if and only ifthey are at right angles to each other. For a . b 0 implies thatCos θ 0, which is the condition of perpendicularity of twovectors.Deductions:From the definition (1) we deduct the following:i. If a and b have the same direction, thenθ 0o Cos 0o 1a . b a bIf a and b have opposite directions, then cos 1ii. Iiia . b a ba . b will be positive if 0 and negative if,iv The dot product of a and b isequal to the product of magnitude ofa and the projection of b on a .This illustrate the geometricalmeaning of a . b . In the fig.15 b Cos θ is the projection ofb on a .vFrom the equation (1)b . a Cos θ a b Cos θb . a a . bHence the dot product is commutative.
Chapter 6148Vectors and ScalarsCorollary 1:If a be a vector, then the scalar product a . a can be expressedwith the help of equation (1) as follows:a . a aOra a Cos 0o a2 . (2)This relation gives us the magnitude of a vector in terms of dotproduct.Corollary 2:If i, j and k are the unit vectors in thedirections of X , Y and Z axes, thenfrom eq. (2)i2 i.i i i Cos0oi2 1soi2 j2 k2 1andi . j j . i 0 Because Cos90o 0i.k k.i 0k.i i.k 0Corollary 3:( Analytical expression of)Scalar product of two vectors in terms of their rectangularcomponents.For the two vectorsa a1i a2j a3kb b1i b2j b3kandthe dot product is given as,a . b (a1i a2j a3k) . (b1i b2j b3k) a1b1 a2b2 a3b3 as i2 j2 k2 1and i . j j . k k . i 0Also a and b are perpendicular if and only if a1b1 a2b2 a3b3 0Example 6:If a 3i 4j – k, b –2i 3j k find a . bSolution:a . b (3i 4j – k) . (–2i 3j k) –6 12 –1 5
Chapter 6149Vectors and ScalarsExample 7:For what values of , the vectors 2i – j 2k and 3i 2 j areperpendicular?Solution:a 2i – j 2k and b 3i 2 jLetSince a and b are perpendicular,a . b 0So(2i – j 2k) . (3i 2 j) 06 – 2 0Or 3Example 8:Find the angle between the vectors a and b , wherea i 2j – k and b –i j –2k.Solution:a . b a b Cos θAsThereforeCos θ a.b, a . b 1 2 2 3a . ba 1 4 1 6,b 1 1 4 636 63 1Cos θ 6 21θ Cos 1 60o2Cos θ Example 9:Consider the points A, B, C, D where coordinates are respectively(1, 1, 0), ( 1, 1, 0), (1, 1, 0), (0, 1, 1). Find the directioncosines of AC and BD and calculate the angle between them.Solution:Now we have A(1, 1, 0), B( 1, 1, 0), C(1, 1, 0), D(0, 1, 1)a AC (1 1)i ( 1 1)j (0 0)k 2jAC 2 j jUnit vector along AC AC2 The direction cosines of AC are 0, 1, 0Now
Chapter 6150Vectors and Scalarsb BD (0 1)i ( 1 1)j (1 0)k i 2j kBD i 2 j k i 2 j kUnit vector along BD BD1 4 16 The direction cosines of BD are:1 2 1,,6 6 6Let, θ be the angle between AC and BD then:Cos θ ( 2j).(i 2j k)(2) 6( 2)( 2)2 (2) 66 2 θ Cos 1 6 Example 10:Show that if a b a – b then a and b are perpendicular.Solution:We have a b a – b a b 2 a – b 2 taking square.a2 b2 2a.b a2 b2 – 2a.b4a.b 0 or a.b 0Hence a and b are perpendicular.2. Vector Product:If a and b are non-zero vectors and θ is the angle betweena and b , then the vector product of a and b , denoted by a x b ,is the vector c which is perpendicular to the plane determined by a andb . It is defined by the relation,c a x b a b Sin θ n
Chapter 6Where151Vectors and Scalarsa b Sin θ is the magnitude of c and n is the UnitVector in the direction of c . The direction of c is determined by theright hand rule.The vector product is also called the „cross product‟ or „Outerproduct‟ of the vectors.Remarks:If we consider b x a , then b x a would be a vector which isopposite in the direction to a x b .Hence a x b b x aWhich gives that a x b b x ain generalHence the vector product is not commutative.Deductions:The following results may be derived from the definition.i. The vector product of two non-zero vectors is zero if a andb are parallel, the angle between a and b is zero. Sin 0o 0,Hence 0.ii.For a x b 0 implies that Sin θ 0 which is the condition ofparallelism of two vectors. In particular a x a 0 . Hence for theunit vectors i, j and k,ixi jxj kxk 0If a and b are perpendicular vectors,then a x b is a vector whose magnitudeis a b and whose direction is such thatthe vectors a, b, a x b form a right-handedsystem of three mutually perpendicular
Chapter 6iii.152Vectors and Scalarsvectors. In particular i x j (1) (1) Sin 90o k (k being perpendicularto i and j) kSimilarly j x i k, i x k j, k x j iHence the cross product of two consecutive unit vectors is the thirdunit vector with the plus or minus sign according as the order ofthe product is anti-clockwise or clockwise respectively.Since a x b a b Sin θ (2)Which is the area of the parallelogram whose two adjacent sidesare a and b .Hence, area of parallelogram OABC a x band area of triangle OAB 1a x b2If th vertices of a parallelogram are given, thenarea of parallelogram OABC OA x OBand, area of triangle OAB iv.1OA x OB2If n is the unit vector in the directions of c a x b thenorn ca x b caxbn a x ba b Sin θfrom equation (2) we also find.Sin θ 6.13a x ba bRectangular form of a x b(Analytical expression of)If a a1i a2j a3kand b b1i b2j b3kthen a x b (a1i a2j a3k) x (b1i b2j b3k) (a1b2k – a1b3j – a2b1k a2b3j a3b1j – a3b2j) (a2b3 – a3b2)i – (a1b3 – a3b1)j (a1b2 – a2b1)kThis result can be expressed in determinant form as
Chapter 6153ija x b a1 a 2b1 b2Vectors and Scalarska3b3Example 11:If a 2i 3j 4k b I – j k, Finda x b(i)(ii)Sine of the angle between these vectors.(iii)Unit vector perpendicular to each vector.Solution:(i)i j ka x b 2 3 41 -1 1a x b i(3 4) – j(2 – 4) k(–2 – 3) 7i 2j – 5k(ii)Sin θ Sin θ (iii)a x ba b 7 2 22 ( 5) 222 32 42 . 12 ( 1) 2 127829 32629If n̂ is the unit vector perpendicular to a and b thenn̂ a x b 7i 2j 5k 78ax bExample 12:a 3i 2kb 4i 4j – 2k,c i – 2j 3kd 2i – j 5k,Compute ( d x c ).( a b )Solution:d x ci j k 2 -1 51 -2 3 i(–3 10) –j(6 – 5) k(–4 1) 7i – j – 3k
Chapter 6154Vectors and Scalarsa b –i –4j 4kAlsoHence ( d x c ).( a b ) (7i – j – 3k).( –i –4j 4k) –7 4 –12 –15Example 13:Find the area of the parallelogram with adjacent sides,a i j k, and b 2j 3kSolution:a x bi j k 1 -1 10 2 -3 i(3 – 2) –j(–3 –0) k(2 0) i 3j 2kArea of parallelogram a x b 1 9 414 square unit.Example 14:Find the area of the triangle whose vertices areA(0, 0, 0), B(1, 1, 1) and C(0, 2, 3)Solution:Since AB (1 – 0, 1 -0, 1 – 0) (1, 1, 1)AC (0 – 0, 2 – 0, 3 – 0)andAC (0, 2, 3)AB x ACi j k 1 1 10 2 3 i(3 – 2) –j (3 – 0) k(2 – 0) i – 3j 2kArea of the triangle ABC 1 AB x AC 2 1 21 ( 3)2 22214square unit2Example 15:Prove by the use of cross-product that the pointsA(5, 2, 3), B(6, 1, 4), C( 2, 3, 6) and D( 3, 2, 1) are thevertices of a parallelogram.
Chapter 6155Vectors and ScalarsSolution:Since AB (1, 1, 7)DC ( 1, 1, 7)BC ( 8, 4, 2)andAD ( 8, 4, 2)AB x DCAB x DCAlsoBC x ADi j k 1 -1 71 -1 7 i( 7 7) j( 7 7) k(1 1) 0, so, AB and DC are parallel.ij k -8 -4 2-8 -4 2 i(0) j(0) k(0) 0, so, BC and AD are parallel.BC x ADHence the given points are the vertices of a parallelogram.Exercise 6.2Q.1Find a . b and a x b(i)a 2i 3j 4k(ii)Q.2Q.3a i j kb i j kb 5i 2j 3k(iii)a i j kb 2i jShow that the vectors 3i – j 7k and –6i 3j 3k are at rightangle to each other.Find the cosine of the angle between the vectors:(i)a 2i 8j 3kb 4j 3k(ii)a i 2j kb j 2ka 4i 2j kb 2i 4j kIf a 3i j k, b 2i j k and c 5i 3k , find(iii)Q.4(2 a b ). c .Q.5What is the cosine of the angle between P1P2 and P3P4If P1(2,1,3) , P2( 4, 4, 5) , P3(0, 7, 0) and P4( 3, 4, 2)?
Chapter 6156Vectors and ScalarsQ.6If a [a1,a 2 ,a 3 ] and b [b1,b2 ,b3 ] , prove that:Q.72 b Find ( a b ).( a b ) if a i 2j 3k and b 2i j k .Q.8Prove that for every pair of vectors a and ba.b 1 a b2 2 a( a b ).( a b ) aQ.9 b2Find x so that a and b are perpendicular,(i)anda 2i 4j 7kb 2i 6j xka xi 2j 5k(ii)Q.1022b 2i j 3kandIf a 2i 3j 4k and b 2j 4kFind the component or projection of a along b .Q.11Q.12Q.13Under what condition does the relation ( a . b )2 a22b holdfor two vectors a and b .If the vectors 3i j – k and I – 4 j 4 k are parallel , find valueof .If a i 2j k , b i 2j 4k , c 2i 3j k Evaluate:(ii)( a x b ).( a x c )( a x b )x( a x c )Q.14 If a i 3j 7k and b 5i 2j 4k . Find:a.bax b(i)(ii)(iii)Direction cosines of a x bQ.15 Prove that for the vectors a and b(i)ax b(i)(ii)(2 a.b2 a2b2)x ( a x b ) 2 ( a x b )Q.16Prove that for vectors a , b andQ.17Find a vector perpendicular to both the lines AB and CD, where Ais (0, 2, 4), B is (3, –1, 2), C is (2, 0, 1) and D is (4, 2, 0).
Chapter 6Q.18157Vectors and ScalarsFind ( a x b ) x c if a i 2j 3k , b 2i j k ,c i 3j 2k .Q.19Q.20Q.21Q.22Find the sine of the angle and the unit vector perpendicular to each:(i)anda i j kb 2i 3j k(ii)a 2i j kb 3i 4j kGivena 2i j and b j k , if c 12 and c isandperpendicular to both a and b , write the component form of c .Using cross product, find the area of each triangle whose verticeshave the following co-ordinates:(i)(0, 0, 0),(1, 1, 1),(0, 0, 3)(ii)(2, 0, 0), (0, 2, 0), (0, 0, 2)(iii)(1, -1, 1),(2, 2, 2), (4, -2, 1)Find the area of parallelogram determined by the vectorsa and b a i 2j 3k and b 3i 2j k .Answers 6.2Q.1(i) 3; 7i – 2j – 5kQ.3(i)Q.557 22 335 77Q.11 [0, 11]Q.14 (i) –29(ii)–6, –5i – 2i 7k (iii) –3; –i –2j k(ii)0Q.78Q.12 12Q.19 (i)Q.9(i)413 4i 3j k,2126(ii)17Q.4372117 Q.10 52Q.13 (i) 15 (ii) i – 2j k2 39 17,,1814 1814 1814Q.18 5 26155 3i 5j 11k(ii),156155(ii) –2i – 39j – 17kQ.17 7i – j 12k(iii)(iii) Q.20 –4i – 8j 8kQ.21 (i)(iii)3 2sq. unit.2110sq. unit.2(ii)Q.222 3 sq. unit.180 sq. unit
Chapter 6158Vectors and ScalarsSummaryA vector is a quantity which has magnitude as well as direction whilescalar is a quantity which has only magnitude. Vector is denoted as AB orOP .1. If P (x, y, z) be a point in space, then the position vector of Prelative to 0 OP .2. Unit coordinator vectors x, j, k are taken as unit vector s along axisOP xi yj zk.3. Magnitude of a vector. i.e. OP x 2 y2 z24. Unit vector of a (non-zero vector), then a aa5. Direction cosines of OP xi yj zk then,cos xy. Cos OPOP.Cos zOPScalar product:The scalar product of two vector a and b is defined asa . b a b Cos θ1. If a.b 0, vectors are perpendicular.2. i.j j.k i.k zero while i.i j.j k.k 13. a . b (a1i a2j a3k).(b1i b2j b3k) a1b1 a2b2 a3b3Vector product:The vector or cross product of two vectors a and b denoteda x b and is defined as: a x b a b Sin θ n , Sin θ 1. n a x bunit vector.axb2. a x b 0 . a and b are parallel or collinear.3. i x j j x j k x k 0 and i x j k.j x k i, k x i j4. a x b b x a5. a x b (a1i a2j a3k) x (b1i b2j b3k)ij a1 a 2b1 b 2ka3b3axba b
Chapter 6159Vectors and ScalarsShort QuestionsWrite the short answers of the following:Q.1:Q.2:Q.3:Q.4:What is scalar? Give examples.What is a vector? Give example.What is unit vector?Find the formula for magnitude of the vectorQ.5:Find the magnitude of vectorQ.6:Q.7:Q.8:What are parallel vectors?Find α, so that αi (α 1)j 2k 3If cos , cosβ , cosγ are direction cosines of a vector xi yj zk-2i–4j 3k xi yj zk, then show that cos2 α cos2 β co2 1Q.9:Find the unit vector along vector 4i – 3j – 5k.Q.10: Find the unit vector parallel to the sum of the vectors a [2, 4, -5],b [1, 2, 3]Q.11: Given the vectors,a 3i – 2j k , b 2i – 4j – 3kc - i 2j 2k ,Find a b cQ.12: Given the vectors a 3i j – k and b 2i j – k ,find magnitudeof 3 a – bQ.13: Find a vector whose magnitude is 2 and is parallel to5i 3j 2k.Q.14: Define scalar product of two vectors.Q.15 Find a .b ifa i 2j 2k,Q.16: Find (a b) . (a – b) ifa 2i 2j 3k ,bQ.17: Define Vector product.Q.18: Ifa 2i 3j 4k Find a b b 3i – 2j – 4k 2i – j k,b i–j k
Chapter 6160Vectors and ScalarsQ.19: Find the area of parallelogram with adjacent sides,a 7i – j k and b 2j – 3kQ.20: For what value of λ, the vectors 2i – j 2k and 3i 2 λj areperpendicular.Q.21: Under what conditions does the relation a . b a b hold?Q.22: Find scalars x and y such that x (i 2j) y(3i 4j) 7Q.23: Prove that if a i 3j – 2k and b i – j –k, then a and b areperpendicular to each other.Answers7. 1 , – 25.10.3i 6j - 2k711.13.10 i 6j 4k3815.18.7819.21 0o22.4i – 4j 0k–914 sq. unit9.4i - 3j - 5k5 212.5416.820. λ 3-1x 2 , y 5/2
Chapter 6161Vectors and ScalarsObjective Type QuestionQ.1Each questions has four possible answers. Choose the correctanswer and encircle it.1.2.Magnitude of the vector 2i – 2j – k is:(a) 4(b) 3(c) 2Unit vector of i j k is:(a) i j k(c)3.1( i j k)31(d)( i j k)2(b)1( i j k)3Unit vector of i – 2j – 2k is:1( i – 2j – 2k)31(d)(i – 2j – 2k)2(a) i – 2j – 2k(b)1( i – 2j – 2k)3If i, j and k are orthogonal unit vectors, then j x i is:(c)4.(d) 1(b) k(a) k(d) 1(c) 15.The magnitude of a vector i 3j 5k is:6.(a) 3(b) 25(c) 35(d) 35In l, m and n are direction cosine of a vector, then:7.(b) l 2 m 2 n 2 1(a) l 2 m 2 n 2 1(c) l 2 m 2 n 2 1(d) l 2 m 2 n 2 1If θ is the angle between the vector a and b , then cos θ is:(a) a . b(c)a . ba8.9.(b)(d)a . ba ba . bbIf a a1j a2j a3k, b b1i b2j b3k, then a . b is:(a) a1b1j a2b2j a3b3k(b) a1b1 a2b2 a3b3(c) a1b2j a2b3j a3b1k(d) None of thesea . b 0 implies that a and b are:(a) Perpendicular(b) Parallel
Chapter 6162Vectors and Scalars(c) Non-parallel(d) Oblique10. If a i j k and b i j mk are perndicular then m willbe equal to:(a) 1(b) 2(c) 1(d) 311. a . b is a:(a) Vector quantity(b) Scalar quantity(c) Unity(d) None of these12.a . a is equal to:(b) a2(a) 1(c) a(d) None of these13. If a 2i 3j k and b i 2j 7k then a . b is equal to:(a) 1(b) 2(c) 3(d) 414. If a x b 0 then a and b is:(a) Non-parallel(b) Parallel(c) Perpendicular(d) None of these15. The cross product of two vectors a and b is:(a) ab cos θ(b) ab sin θ(c) ab sin θ n̂(d) ab cos θ n̂16. If n̂ is the unit vector in the direction of a x b , then n̂ is:(a)a x b(b)a b17.a . ba ba x b(c)a b sin θa x ba x ba x b is area of the figure called:(a) Triangle (b) Rectangle (c) Parallelogram18. a x b is equal to:(a) b x a (b) b x a(c) a x ba x b is a:(a) Vector quantity(c) Unity(d) Sector(d) b x a19. If a and b are collinear vectors, then:(a) a x b 0(b) a . b 0(c) a b 0(d) a b 020.(d)(b) Scalar quantity(d) None of these
Chapter 6163Vectors and 8.bbaa4.9.14.19.baba5.10.15.20.dccb
2. Subtraction of Vectors: If a vector . is to be subtracted from a vector , the difference vector . can be obtained by adding vectors and . The vector . is a vector which is equal and parallel to that of vector but its arrow-head points in opposite direction. Now the vectors . and . can be added by the head-to-tail rule. Thus the line . AC
Chapter 6 139 Vectors and Scalars (ii) Vectors Addition is Associative: i.e. a b c a b c where . a , b . and . c . are any three vectors. (iii) O is the identity in vectors addition: Fig.9. For every vector . a O a Where . O. is the zero vector. Remarks: Non-parallel vectors are not added or subtracted by the .
Scalars & Vectors 5 23. Three coplanar vectors in arbitrary units are given by A 4i 2j 3k , B i j 3k and C 4i 5j 3k , the resultan
Units of Chapter 3 Vectors and Scalars Addition of Vectors—Graphical Methods Subtraction of Vectors, and Multiplication of a Vector by a Scalar Adding Vectors by Components Unit Vectors Vector Kinematics Projectile Motion Solving
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
9. Parallel and Collinear Vectors: The vectors. and. are parallel if for any real number n, n . If (i) n 0 then the vectors . and . have the same direction. (ii) n 0 then . and . have opposite directions. Now, we can also define collinear vectors which lie along the same straight line or having their directions parallel to one another. 10.
I think of atomic vectors as “just the data” Atomic vectors are the building blocks for augmented vectors Augmented vectors Augmented vectors are atomic vectors with additional attributes attach
2.2. Linear property of 2-vectors. The Clifford algebra is not a graded multi vector algebra, however it is a graded vector space [4]: it can be decomposed as sum of linear subspaces of homogeneous grade. The twists, or 2-vectors, form a vector subspace within the Clifford algebra over the scalars (0-vectors) and also over the dual scalars,
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
