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 Multiplication of Vectors ( dot/scalar & cross/vector product ) d1 = |A| ; d2=|B| ; d3=|C| ; d4=|AxB| ;d5=|(AxB)xC| d6=A.(BxC) ; BxC = a61 i +a62 j +a63 k ; y= angle between A and B ; z1= cosine of angle between A and B angle η in degreeangle φ in degreecon 1:θ=η - φcon 2:θ=η + φcon 3:θ=-(η - φ)con 4:θ=-(η + φ)angle θ in degree 1 2 3 4 angleradian sine cos angles  in degree θ1θ2 θ3θ4 i,j,k=x,y,zHere [ (AB)i, (AB)j]=k*(AB)k firstenterk1 afterenteringa11,a12,a13,a13is recomputedsothat a11+a12+a13=0 afterenteringa21,a22,a23 a22,a23isrecomputed based on k1value givenK=2k1 k1 A+B+C= π;   A(°):-B(°): k1 Puta21valuebeforepressing calculate. (number) Squareroot inverseno. square A Enter a11 i Enter a12 j Enter a13 k Find d1 B Enter a21 i Enter a22 j Enter a23 k Find d2 C Enter a31 i Enter a32 j Enter a33 k Find d3 Dimension of Vector space of A B C Det R trR sum1strowsum2ndrow sum3rdrow area1starea2nd area3rd Calculation to make M=Ms a23a   a23b first enter values of A,B M f2 + f + =0 Ms +Ve + -Ve +, (+ /-) discriminant 4 * (ABdis1c2+ABdis2) (ABdis4c2+ABdis3) - - normalizedA  na11 i j k Find nd1 normalizedB  na21 i j k Find nd2 normalizedC  na31 i j k Find nd3 Reciprocal of A(rA) i + j + k Find rd1 Reciprocal of B(rB) i + j + k Find rd2 Reciprocal of C(rC) i + j + k Find rd3 normalizedrA  nra11 i + j + k Find nrd1 normalizedrB  nra21 i + j + k Find nrd2 normalizedrC  nra31 i + j + k Find nrd3 k11=a112/a122+a122/a112 k12=a122/a132+a132/a122 k13=a112/a132+a132/a112 k1=k11-k12-k13 rA.A Find FindAngle(rA,A) in ° yraa Find 1/( d1rd1) Find d1rd1 k21=a212/a222+a222/a212 k22=a222/a232+a232/a222 k23=a212/a232+a232/a212 k2=k21-k22-k23 rB.B Find FindAngle(rB,B) in ° yrbb Find 1/( d2rd2) Find d2rd2 k31=a312/a322+a322/a312 k32=a322/a332+a332/a322 k33=a312/a332+a332/a312 k3=k31-k32-k33 rC.C Find FindAngle(rC,C) in ° yrcc Find 1/( d3rd3) Find d3rd3 nk11=na112/na122+na122/na112 nk12=na122/na132+na132/na122 nk13=na112/na132+na132/na112 nk1=nk11-nk12-nk13 nrA.nA=1/(d1rd1) Find FindAngle(nrA,nA)in ° nyraa Find nd1nrd1 nk21=na212/na222+na222/na212 nk22=na222/na232+na232/na222 nk23=na212/na232+na232/na212 nk2=nk21-nk22-nk23 nrB.nB=1/(d2rd2) Find FindAngle(nrB,nB)in° nyrbb Find nd2nrd2 nk31=na312/na322+na322/na312 nk32=na322/na332+na332/na322 nk33=na312/na332+na332/na312 nk3=nk31-nk32-nk33 nrC.nC=1/(d3rd3) Find FindAngle(nrC,nC)in° nyrcc Find nd3nrd3 Put value of y1y1 should be less than or equal to Put value of f1 Find y2a Find y3a y12 +y2a2 +y3a2= If A=ai+ck, find inverse fill up only f1 Find y2b Find y3b y12 +y2b2 +y3b2= inverse of A :iA1 i j k norm inverse of A :iA2 i j k norm A x iA1 i j k norm A x iA2 i j k norm A.iA1 angle(A,iA1)in ° A.iA2 angle(A,iA2)in ° X1:X2: X3a+ X3b: i* ; X4 X5 X23s: X1s X43s X5s Find A.B angle(A,B) deg (AxB).C=[abc] cos(A,B) Find B.C angle(B,C) deg cos(B,C) Find C.A angle(C,A) deg cos(A,C) E.F Find angle(E,F) deg |E+F| |E-F| |E+F| - |E-F| 2√(|E||F|cos (E,F) AxB Find a41 i Find a42 j Find a43 k Find d4 (AxB)+xy i Find abz42 j Find abz43 k Find d4 (AxB)-xy i Findabz42a j Find abz43a k Find d4 Angle((AxB)+,(AxB)-) in° |AxB|2 |A.B|2 |A|2 +|B|2 |A.B|2+ |AxB|2 (|A||B|)2 (a/2)2(d2-e2-f2)  + (b/2)2(e2-d2-f2)  + (c/2)2(f2-d2-e2)  LHS : abde+bcef+acdf  RHS: [ABxy+ ,  ABz] = (AB)x -  (AB)y =*ABxy- [ABxy- ,  ABz] = (AB)x +  (AB)y [ABxy+,ABxy-] = *(AB)z [ (AB)x, (AB)y]= *(AB)z [ (AB)y, (AB)z]= *(AB)x [ (AB)z, (AB)x]= *(AB)y Put value for Z1:(any arbitrary value) M1=(a11a21+a12a22+a13a23)- Z12 U1=(a12a23-a13a22) / (a13a21-a11a23) T1=(a12a21-a11a22) / (a13a21-a11a23) MUT1a BA vector1 i j k Norm BA vector2 i j k Norm BA vector1.BC vector = BAvec1.(AxB) (see fig below)BA vector2.BC vector CA vector1 i j k Norm CA vector2 i j k Norm A.(AxB) AxBvectorisperpendiculartoplane containingbothA,B B.(AxB) (AxB)xC Find a51 i Find a52 j Find a53 k Find d5 D Find a71 i Find a72 j Find a73 k Find d7 E Find a81 i Find a82 j Find a83 k Find d8 (AxB)xC=D - E, a91 i Find a92 j Find a93 k F Find a101 i Find a102 j Find a103 k Find d10 Ax(BxC)=D - F, a111 i Find a112 j Find a113 k Find d11 (BxC) a121 i Find a122 j Find a123 k Ax(BxC)  a131 i Find a132 j Find a133 k A . ( BxC ) Find a61 i Find a62 j Find a63 k Find d6

 (A xB) x C =B(A.C)-A(B.C) =D - E        A x (B x C) =B(A.C)-C(A.B)= D - F   ; Vector triple product is not associative i.e.  (A xB) x C ≠  A x (B x C)  Suppose A =ai+bj+ck;               B =di+ej+fk;               C =li+mj+nk; and we represent a matrix as i    j    k a   b   c d  e    f l   m   n D =(dot product of j,k component of A,C)*di +(dot product of i,k component of A,C)*ej +(dot product of i,j component of A,C)*fk E=(dot product of j,k component of B,C)*ai +(dot product of i,k component of B,C)*bj +(dot product of i,j component of B,C)*ck F=(dot product of j,k component of B,A)*li +(dot product of i,k component of B,A)*mj +(dot product of i,j component of B,A)*nk A.(BxC) = (AxB) . C D= (a12a32+a13a33)a21 i  + (a13a33+a11a31)a22 j +(a11a31+a12a32)a23 k E= (a22a32+a23a33)a11 i  + (a23a33+a21a31)a12 j +(a21a31+a22a32)a13 k F= (a12a22+a13a23)a31 i  + (a13a23+a11a21)a32 j +(a11a21+a12a22)a33 k A . (BxC) =(AxB).C (AxB).C =[abc]  a11    a12   a13 a21    a22   a23 a31    a32   a33  A.(BxC) =[cab]a31    a32   a33 a11    a12   a13 a21    a22   a23 *Scalar triple products are cyclic in nature.[abc]=[bca]=[cab]=-[bac]=-[acb]=-[cba] . These represent the volume of the parallelepiped , when we take their      absolute value. The matrix form of scalar triple product is the key as matrices are strongly linked to linear transformations. * if [abc]=[bca]=[cab]=-[bac]=-[acb]=-[cba] = 0, A,B,C vectors are coplanar. * If |E +F| = |E - F| , then E . F =0 or angle between them is 90. |E +F| - |E - F| =2√ [(|E||F|cos (E,F)] .    In Physics, the above relationship is useful in comparing the resultant velocity (E +F) with relative velocity (E-F). In case of parallelograms, |E+F| and |E-F| represent  the 2 diagonals. When E . F =0, the corresponding figure is a rectangle or square or rhombus.In all these cases, diagonals are equal and  perpendicular bisectors. We define reciprocal of a vector A (ai+bj+ck) as rA which is (1/na)i +(1/nb)j+(1/nc)k where n is the dimension of vector space and A.rA =1 i.e. the scalar product of any vector and its reciprocal vector is 1. If A=ai +bj+ck where i,j,k are unit vectors along x,y,z axis respectively, then rA= (1/3a)i +(1/3b)j+(1/3c)k. where n=3 A.rA =1 ;    |A| =  √(a2 + b2 + c2)  |rA| = (1/3) √(1/a2  + 1/b2  + 1/c2) cosθ = 1/( |A| *| rA| ) Angle θ between A,rA is also  given by cosθ = 1/√ [1/n +(1/n2){(a2 / b2  + b2 / a2  ) + (b2 / c2  + c2 / b2 ) +(a2 / c2  + c2 / a2 )} ]          =1/ √[1/3 +(1/9){(a2 / b2  + b2 / a2  ) + (b2 / c2  + c2 / b2 ) +(a2 / c2  + c2 / a2 ) }] Case 1 : when a=b=c, cosθ = 1=1 and  θ=0° i.e. A,rA are parallel and  (a2 / b2  + b2 / a2  )=(b2 / c2  + c2 / b2 )  =(a2 / c2  + c2 / a2 ) =2 Case 2: When out of the 3 components under blue color, any 2 are equal. say (a2 / b2  + b2 / a2  )=(b2 / c2  + c2 / b2 ). Here b is the common component. Then b2 =ca or b=√(ca) or b is the geometric mean of a,c. and  cosθ=1/ √[1/3 +(1/9){2(a2 / b2  + b2 / a2  )  +(a2 / c2  + c2 / a2 )} ] =1/ √[1/3 +(1/9){2(a / c  + c / a  )  +(a / c  + c / a )2 -2} ] = 1/ √[1/3 +(1/9){(a / c  + c / a )2+2(a / c  + c / a  )   +1-3} ] = 1/√[1/3 +(1/9)[{1+(a / c  + c / a )2}-3] ] = 1/{(1/3)[1 + a/c + c/a]} Since maximum value of cosθ is 1, [1 + a/c + c/a]  > = 3 or  (a/c +c/a) > = 2 if (a/c +c/a)  = 2 or x +1/x =2 where x=a/c, then x=1 or a=c and b =√(ca) =a and the case reduces to the first one where θ=0° i.e. A,rA are parallel. Thus case 1 is a special case of case 2. Now let us find out under which circumstances, the angle between A and rA is 45° :- 1/3(1+a/c+c/a)=√2 or (a/c +c/a) =3√2 -1  if we take a/c=x, and solve the quadratic equ. x2 -3.2426x+1=0, we get x=a/c=2.89745, taking c=1, amd b=√(ac) we get b=1.7022. For 2 dimensional vector in say x-y plane, cosθ = 1/√ [1/n +(1/n2)(a2 / b2  + b2 / a2  )  ] =1/(1/2)[a/b+b/a] where n=2. now a/b+b/a ⋝ 2. for a=b, a/b+b/a=2. for 45 degree angle, a/b=√2+1=2.414 Inverse of vector(Right handed) : Inverse of a vector A is defined as a vector iA such that A x iA = unit vector if A =ai+bj+ck iA=  di+ej+fk then I =A x iA = i  j  k a b c d e f = k1i + k2j +k3k where k1=bf-ce k2=cd-af k3=ae-bd and k12 + k22 +k3 2= 1 Suppose we arbitrarily fix k1, k2, then k3 is automatically fixed.abf - ace=ak1 bcd -abf=bk2Adding,ae-bd=(ak1+bk2)/(-c)=k3 k12 + k22 +k3 2= 1 or k12 + k22 + (ak1+bk2)2/(c2)=  1 or ( b2+c2) k22 +(2k1ab)k2+ k12(a2+c2) - c2 =0Ak22  + Bk2   +C =0 where A=( b2+c2) , B=(2k1ab) , C=k12(a2+c2) - c2     ; A is always +ve, B may be +ve or -Ve, C shall be studied below.This is a quadratic equation of k2. Solving, one gets 2 values of k2 and also 2 values of k3. Sum of 2 values of k2 i.e.     k2a+k2b =-B/A = -2k1ab / ( b2+c2)Product of 2 values of k2 i.e. k2a*k2b =C/A =[k12(a2+c2) - c2] / ( b2+c2)                                              k2a -k2b  =√ ( B2-4AC) / A(k2a)2 +(k2b)2 = (B2 -2AC) / A2 ; (k2a)2  -(k2b)2 =(B/A2) *√ ( B2-4AC) √ ( B2-4AC) should be zero or positive . By putting the value of A,B,C, we find that y1=k1 < =√ [(b2+c2)/(a2+b2+c2)] Out of d,e,f , arbitrarily fix any one say f. (In the calculator, it is designated as f1 and k1 is designated as y1) e=(bf-k1)/c  ...... one gets 1 value of e d=(af+k2)/c =af/c +k2/c...... one gets 2 values of d d1=af/c +k2a/cd2=af/c +k2b/cΔd=d1-d2=(k2a -k2b) /c =√ ( B2-4AC) / Ac =(1/c)√[(B/A)2 -(2(√c/√a))2]  where A= ( b2+c2)   , B=2abk1 . C=k12(a2+c2) - c2 ;Study of C:C=k12(a2+c2) - c2 Where both first and second part are +ve if all are real numbers. C=0       if 1/k12 - a2/c2 =  1 where k1 ∈ [-1,1]. Equation in red is that of unit hyperbola. C > 0     if  1/k12 - a2/c2 < 1 C < 0     if  1/k12 - a2/c2 > 1 Let us take X1=B/A , X2=cΔd , X3=2√(C/A) =2√(k2a*k2b) , X4=2c /√(b2+c2) , X5=2k1*√[(a2+c2)/(b2+c2)] Then, X12 =X22 +X32 ;             X52 =X42 +X32 ; Since C can be equal to, greater than, less than zero, X3 can be zero, real or imaginary respectively. Inverse vector iA1 =d1*i +e*j +f*kiA2 =d2*i +e*j +f*kiA1-iA2=(d1-d2)*i =Δd*i; Hence difference between the 2 inverse vectors is manifested only in change of magnitude along 1-axis , in the instant case it being x-axis.So we can write iA1 =(iA1)x + (iA1)y-z plane = (iA1)x + (iA1)⊥   ;iA2= (iA2)x + (iA2)y-z plane = (iA2)x + (iA1)⊥  since  (iA2)⊥= (iA1)⊥   =e*j +f*kNow iA1 - iA2 =(iA1)x -(iA2)x   = d1*i  -d2*i =(d1-d2)*i =Δd12*i Alternatively, if we have 2 vectors differing by only the component of one axis , we can explore possibility of finding a vector of which these 2 vectors are  the inverse ones. In the calculator, instead of d,e,f , we have taken iA11,iA12,iA13 and instead of k1,k2,k3 we have taken y1,y2,y3. We have arbitrarily taken a value of y1, f1. Inverse of 2-D vectors :-A=ai+ck iA=di+fkAxiA= i  j   ka 0   cd 0  f=(cd-af)*j |AxiA| = 1 by definition of inverse vector or  (cd-af) =+ 1 or -1 d1=(af +1) / cd2=(af -1) / c In the calculator, just fill f1 which is f. Leave y1 blank as it is for 3-D vectors.Inverse of 1dimensional vector.Suppose A= ai              iA=diA x  iA =0, hence iA = ejor iA=fkTaking the former A x iA =ae(i x j) =ae* k where ae =± 1 or e = ± 1 / a Now taking one of the components of the vector as residing in imaginary axis,If A=ai +bj+ick where i,j,k are unit vectors along x,y,z axis respectively, then rA= (1/a)i +(1/b)j+(i/c)k. A.rA=1 |A| =√[a2 + b2  - c2] |rA| = √[1/ a2 + 1/ b2   - 1/ c2  ] cosθ = 1/√ [3+K] where K ⋝ - 2 and K= [(a2 / b2  + b2 / a2  ) - {(b2 / c2  + c2 / b2 ) +(a2 / c2  + c2 / a2 )} (a)When K= -2,  θ =0° , cosθ = 1 and a=b=c (b)When K= -1,  θ =45° , cosθ = 1/√2 ( c ) When K= 0,  θ =54.74 ° , cosθ = 1/√3  To achieve case (b), we assume that (b2 / c2  + c2 / b2 ) =(a2 / c2  + c2 / a2 ) which implies c=√(ab) Hence K= (a2 / b2  + b2 / a2  ) - 2(b2 / c2  + c2 / b2 ) = -1 or (a/b + b/a -1)2 =3-1=2 or (a/b +b/a ) =√2 +1 Taking a/b=x, solving the quadratic equation , x = a/b ={(√2 +1) ± √[2√2 - 1] }/2 taking b=1,we get , a =1.8832   , c=√(ab) =1.3723 and    b=1,we get , a =0.5310   , c=√(ab) = 0.7287 To achieve case (c), we assume that (b2 / c2  + c2 / b2 ) =(a2 / c2  + c2 / a2 ) which implies c=√(ab) Hence K= (a2 / b2  + b2 / a2  ) - 2(b2 / c2  + c2 / b2 ) = 0 or (a/b + b/a -1)2 =3  or (a/b +b/a ) =√3 +1 Taking a/b=x, solving the quadratic equation , x = a/b ={(√3 +1) ± √[2√3 ] }/2 taking b=1,we get , a =2.2966   , c=√(ab) =1.5155 or       b=1,we get , a =0.4354   , c=√(ab) =0.6598 ---------------------------------------------------------- nA = (a/|A|) i  + (b/|A|) j + (c/|A|) k where (a/|A| )2 + (b/|A| )2   + (c/|A| )2=1 .....(1) nrA = (1/3a*|rA|) i  + (1/3b*|rA|) j + (1/3c* |rA|)  k Here we have expressed nrA in terms of rA. |nA| =  1|nrA| = 1nA . nrA= 1/( |A| *| rA| )cosθ =nA * nrA /( |nA| *|n rA| ) = 1/( |A| *| rA| ) Alternatively, nA = (a/|A|) i  + (b/|A|) j + (c/|A|) k rnA= (1/3a)|A| i  + (1/3b) |A| j + (1/3c) |A| k Here we have expressed nrA in terms of A. nA.rnA=1 |nA| =  1|rnA| =( |A| *| rA| ) cosA =nA * rnA /( |nA| *|rnA| ) = 1/( |A| *| rA| )where A is the angle between nA and rnA.---------------------------------from (1), takung c/|A| = cosθ sin2θ  + cos2θ  =1 where  (a/|A| )2 + (b/|A| )2=sin2θ = r2 =r2 ( cos2φ + sin2φ )  ;  (c/|A| )2  = cos2θ  = 1 - r2    ; So  (a/|A| )=rcosφ      (b/|A| )=rsinφ     (c/|A| )  = cos2θ  = √(1 - r2 ) and φ can be any angle with respect to a reference line from common origin. Now nA =|r| (cosφ i + sinφ j) +√(1 -  r2 )k =|r|r + √(1 -  r2 )k and  r is the basis unit vector along r. r=cosφ i + sinφ j nA = nA⊥+ nAz    ;where nA⊥=|r|r=|r| (cosφ i + sinφ j) nAz  =√(1 -  r2 )k So any normalized vector can be represented as vector sum of a radius vector in any plane and a perpendicular vector to the plane from the same origin and |r| =1       ∊ [0,1]. When |r| < 1, the magnitude of normalized vector -------??? In 3 dimensional space, nA=|r| cosφ i +|r| sinφ j +√(1 -  r2 )k when k=0,  nA=|r| cosφ i +|r| sinφ j and |nA| = r  and |r| = 1 So nA=cosφ i + sinφ j In general, A=|r| cosφ i +|r| sinφ j where |r| can be greater than 1 or less than 1. When it is 1, the vector is normalized. nA=cosφ i + sinφ j =a11*i +a12*j rnA=(1/2cosφ)i +(1/2sinφ)j where nA.rnA =1 |nA| =1 |rnA| =√ [1/(4cos2φsin2φ)] =1/sin2φ Angle between nA,rnA is cos A =sin2φ and can have any value depending on the value of φ .  cos A =sin2φ =2sinφcosφ =2*sinφ*cos(-φ) =2*sinφ*sin(π/2 -(-φ))=2*sinφ*sin( φ + π/2) When φ = 45 ° , nA=(1/√2)i +(1/√2)j rnA=(1/√2)i +(1/√2)j i.e both coincide. Relation between dot product and cross product of 2 vectors A and B :(A.B)2 +|AxB|2 =|A|2 * |B| 2 ;or CA2 + CB2 =BA2 ; (A.B) > |AxB| if (a/2)2(d2-e2-f2)  + (b/2)2(e2-d2-f2)  + (c/2)2(f2-d2-e2)  >   abde+bcef+acdf From the figure above, we get the length of the 3 sides of the triangle. How to get the vectors ? For CB , the vector is AxB CB=(bf-ce)i+(cd-af)j+(ae-bd)k What will be the vector form of CA ? The vector is say  xi+yj+zk  has to satisfy 2 conditions :--   condition 1: x(bf-ce) + y(cd-af) +z(ae-bd) =0 since both vectors are to be orthogonal. or y = T - Ux where T =z(bd-ae)/(cd-af)   and U=(bf-ce)/(cd-af) condition 2 :  x2 +y2 + z2 =(A.B)2=(ad+be+cf)2 ; or  x2 +y2 = (ad+be+cf)2 - z2 ; or  x2 + (T-Ux)2 =M2 where M2=(ad+be+cf)2 -z2 ; Taking an arbitrary value of z, this reduces to a quadratic equation of x. (U2 +1)x2 -2UTx +(T2 -M2) =0 Solving x=(UT ± √(M2[U2 +1]-T2) )/(U2 +1) y= T-Ux z=value you have assigned Now, vector  BA =CA -CB =[x-(bf-ce) ] i +[y-(cd-af)]j+[z-(ae-bd)]k tan CAB =(A.B) /|AxB| In the right angle triangle, the magnitude of resultant vector equals the magnitude of relative vector. Only the direction of both varies and are at right angles to each other. Lagrange Identity:(AxB).(CxD) =(A.C)(B.D)-(A. D)(B.C) Jacobi Identity: Ax(BxC) +Bx(CxA) +Cx(AxB) = 0 Co-planarity of 4 vectors : (AxB)x(CxD)=0 Commutation and Anti- Commutation Rules of Vectors : Let vector A =ai+bj+ck       vector B =di+ej+fk vector AB=AxB=(AB)x i + (AB)y j  + (AB)z k = (bf-ce)i +(cd-af)j + (ae-bd)k[ (AB)x, (AB)y] = (AB)xx (AB)y - (AB)y   x (AB)x  =2*( (AB)x*(AB)y /(AB)z   )*(AB)z=2*[(bf-ce)(cd-af)/(ae-bd)] (AB)z; [ (AB)x, (AB)y] =2*[(bf-ce)(cd-af)/(ae-bd)] (AB)z; [ (AB)x, (AB)y] =0 if b/c =e/f   or a/c = d/f[ (AB)y, (AB)z] =2*[(cd-af)(ae-bd)/(bf-ce)] (AB)x; [ (AB)y, (AB)z] =0 if a/b=d/e   or a/c = d/f [ (AB)z, (AB)x] =2*[(ae-bd)(bf-ce)/(cd-af)] (AB)y;  [ (AB)z, (AB)x] =0 if a/b=d/e  or b/c =e/fwhere (AB)x represents with bold letters a vector  and (AB)x   represents with non-bold letters the scalar component of the vector. If one takes any of the conditions for commutations, 2 out of the 3 commute, but the 3rd one commutation becomes infinity.Exa- let us assume that a/c = d/f condition is satisfied. Then  [ (AB)x, (AB)y] =0  and [ (AB)y, (AB)z] =0  But [ (AB)z, (AB)x] =infinity.Hence none of the sets commute when they represent physical quantities. This is the characterstic of axial vectors.But [AB2 , (AB)x]  =0 as  AB2= null vector     [AB2 , (AB)y]  =0      [AB2 , (AB)z]  =0 [ (AB)x, (AB)y] =2*[ (AB)x*(AB)y/  (AB)z]  (AB)z;  where  (AB)x =(bf-ce) , (AB)y =(cd-af) , (AB)z =(ae-bd) [ (AB)y, (AB)z] =2*[ (AB)y*(AB)z/  (AB)x]  (AB)x;  [ (AB)z, (AB)x] =2*[ (AB)z*(AB)x/  (AB)z](AB)y;  suppose we have a case where  k (constant) = 2*[ (AB)x*(AB)y/  (AB)z]    = 2*[ (AB)y*(AB)z/  (AB)x]       =  2*[ (AB)z*(AB)x/  (AB)z] which implies that  (AB)x= ±  (AB)y  =  ± (AB)z or (bf-ce) = ± (cd-af) =  ± (ae-bd) [ (AB)x, (AB)y] =k*(AB)z;  where  (AB)x =(bf-ce) , (AB)y =(cd-af) , (AB)z =(ae-bd) [ (AB)y, (AB)z] =k* (AB)x;  [ (AB)z, (AB)x] =k*(AB)y;  and k is the x/y/z component of Generalizing, [ (AB)i, (AB)j] =k* εijk (AB)k;  i , j , k=1,2,3 and εijk  is structure constant k=2*[(bf-ce)(cd-af)/(ae-bd)] =2*[(cd-af)(ae-bd)/(bf-ce)]=2*[(ae-bd)(bf-ce)/(cd-af)] or k/2= [(bf-ce)(cd-af)/(ae-bd)] =[(cd-af)(ae-bd)/(bf-ce)]=[(ae-bd)(bf-ce)/(cd-af)]          =(bf-ce) = ±(cd-af)= ±(ae-bd) There are 2 sub conditions in order for k to become constant (a) (bf-ce) = (cd-af)= (ae-bd) (b)(bf-ce) = (cd-af)= -(ae-bd) condition (a)  implies that (a+b) / c = (d+e) / f (b+c) / a = (e+f) / d (a+c) / b = (d+f) / e The trivial condition that d=a, e=b,f=c makes (AB)x,(AB)y, (AB)z , each zero and hence not acceptable. Now in general, let cd-af=k/2=k1  or f=(cd-k1)/a but bf-ce=k1 or f=(ce+k1)/b Equating, (cd-k1)/a=(ce+k1)/b or e=bd/a -(k1/ac)(a+b) Now putting these values in ae-bd=k1, we get k1(a+b+c) =0 Hence either k1=0 which is a trivial solution as earlier pointed out, or a+b+c=0 That is you fix up the value of k1 and also fix up the value of d, and then if a+b+c=0 is satisfied, the commutation coefficient remain same at k=2k1. Alternatively, you put a value of d, and choose a,b,c such that a+b+c=0, then arrive at a value of k1 and k. For condition (b), value of f and e remain as above and putting these values in -(ae-bd) =k1, we get k1(a+b-c) =0 That is you fix up the value of k1 and also fix up the value of d, and then if a+b-c=0 is satisfied, the commutation coefficient remain same at k=2k1. Alternatively, you put a value of d, and choose a,b,c such that a+b-c=0, then arrive at a value of k1 and k. It is easy to prove that when a+b+c=0, d=-(e+f) * In A+B+C=π, while putting a+b+c=0, we have taken a=sin2A -4sinAsinBsinC/3 b=sin2B -4sinAsinBsinC/3 c=sin2C -4sinAsinBsinC/3 this is from the identity sin2A+sin2B+sin2C=4sinAsinBsinC Proof hints : sin2A+sin2B+sin2C =2sin(A+B)cos(A-B) +2sinCcosC = 2sinC[cos(A-B) -cos(A+B)]=4sinA.sinB*sinC Defining  ABxy+=[ABx + ABy]  =[ABxi + AByj] ; ABxy+  is called raising vector ABxy- =[ABx - ABy]   = [ABxi - AByj]  ;ABxy-   is called lowering vector Together ABxy+      and ABxy-     are called ladder vectors. In a matrix equation,   ABxy+     = ABx       ABy  *     i =    M1 *   i   ABxy-         ABx     -ABy        j                  j   or M1     = ABx       ABy               ABx     -ABy                       Det M1=(-2ABxABy) [ ABxy+, ABxy-] =- 4(ABx * ABy/ ABz )ABz { ABxy+, ABxy-}= 0 [ ABxy+, ABz] = (2ABz * ABy/ ABx ) ABx  -    (2ABz * ABx/ ABy )ABy =k*ABxy- if both blue and red components become same. [ ABxy-, ABz] = -(2ABz * ABy/ ABx ) ABx  -    (2ABz * ABx/ ABy )ABy =k1*ABxy+ if both blue and red components become same and  k1=-k   In a matrix equation,  ABxy+                     =                      ABx                       ABy                 *      i  =M2  *   i  [ ABxy+, ABz]                       2ABz(ABy)        2ABz(-ABx)                       j                 j or M2  =              ABx                     ABy                                        2ABz(ABy)        2ABz(-ABx)                      Det M2 = -2ABz[ AB2x + AB2y  ] where bold are the vectors and non bold are scalar components. When components of A and B vectors are less than or equal to |1| AxB =(AB)x i + (AB)y j  + (AB)z k = (bf-ce)i +(cd-af)j + (ae-bd)k Let (bf-ce) = (AB)x=sinθ       (cd-af) =(AB)y=-cosθ  (AB)2x+(AB)2y=(bf-ce)2  +(cd-af)2   =1 or f2(a2+b2) +c2(d2+e2) -2cf(be+ad) =1 We take a=cosφ  ; b=-sinφ  ;                d=sinφ  ; e=cosφ   ;  Putting these values,  f2+c2 =1, we put f2=cos2η             c2=sin2η Then following combination of f and c arise :-- (a) cos η , sin η(+,+) (b) -cos η , -sin η(-,-)  (c) cos η , -sin η(+,-)  (d) -cos η , sin η(-,+)  Which combinations are feasible ? Let us find out:- (a)    (AB)x= (bf-ce) = -sin(η +φ) = sinθ or sin(η +φ) =sin(-θ)   since -sinθ =sin(-θ)     or  (η +φ) =  (-θ)           (AB)y= (cd-af) = -cos(η +φ) = -cosθ        or  (η +φ) =  (θ)          (AB)z= (ae-bd) =cos2φ + sin2φ =1        Red and blue equations represent the same numerical value of θ but opposite chirality. They shall have same chirality only when  θ = 0 degree. Hence in that case, η =- φ   (b)    (AB)x= (bf-ce) = sin(η +φ) = sinθ     or  (η +φ) =  θ           (AB)y= (cd-af) = cos(η +φ) = -cosθ =cos(π -θ)        or  (η +φ) =  (π-θ)          (AB)z= (ae-bd) =cos2φ + sin2φ =1        Red and blue equations represent the complentary value of θ . They shall have same value only when  θ = π/2 degree. Hence in that case, η + φ =π/2 (c)    (AB)x= (bf-ce) = sin(η - φ) = sinθ     or  (η -φ) =  θ           (AB)y= (cd-af) = -cos(η -φ) = -cosθ        or  (η -φ) =  θ          (AB)z= (ae-bd) =cos2φ + sin2φ =1        Red and blue equations represent the same numerical value of θ and same chirality. (d)    (AB)x= (bf-ce) = sin( φ -η) = sinθ     or  (φ - η) =  θ           (AB)y= (cd-af) = cos(φ - η) = -cosθ =cos(π -θ)            or  (φ - η) =π -  θ          (AB)z= (ae-bd) =cos2φ + sin2φ =1        Red and blue equations represent the complementary value of θ .They shall have same value only when  θ = π/2 degree. Hence in that case,  φ -η=π/2 let us put various values of the angle and try to ascertain the commutation coefficients of various components of vector AB. Unit Vectors in spherical co-ordinates:- Let a unit vector iin 3 dimension make an angle  η with the Z-axis & is called inclination and its projection on x-y plane make an angle φ with the x axis known as azimuth. then the radial unit vector     r = sinηcosφ*i +sinηsinφ*j +cosη*k           unit vector     η =cosηcosφ*i +cosηsinφ*j -sinη*k           unit vector      φ =-sinφ*i +cosφ*j             r2+η2+φ2  =3 and r  ∈  [0,∞], φ ∈ [0,2π] , η ∈ [0, π] Here A=r    B=η      C=φ; The above 3 are mutually orthogonal, each having unit norm and hence they can be treated as basis vectors in spherical coordinates. Position of a point i.e a vector with initial point at origin  in 3-D spherical coordinate is A=rr any Vector X = r*r   + η *η + φ*φ where r,η,φ are components  and r,η,φ are the unit vectors. In the calculator , if you put the value of η,φ and press key to calculate unit vectors in spherical coordinates, A,B,C vectors become r,η,φ respectively. Ellipsoids are represented through modified spherical co-ordinates. any Vector X = (r/√a)*r   + (r/√b) *η +(r/√c) *φ where () are components of unit vectors and r,η,φ are the unit vectors and a,b,c are constants. or  X= (r/√a)sinηcosφ*i +(r/√b)sinηsinφ*j+(r/√c)cosη*k in modified spherical coordinates . They satisfy the equation x2 + y2 + z2 =r2 where x=(r/√a)sinηcosφ   , y =(r/√b)sinηsinφ  and z=(r/√c)cosη In cartesian coordinate system, the unit vector in 3-D space is a combination of 3 sides.(x,y,z) In spherical coordinate system, the unit vector in 3-D space is a combination of 1 sides.(r) and 2 angles ( η,φ) Hence, there must be another co-ordinate system whose unit vector in 3-D space shpuld have  2 sides and 1 angle. This reminds us of the fact that a unique triangle can be constructed if 3sides are given or 2 angle and 1 side is given or 2 sides and 1 angle is given. and any triangle with n times the above where n is any number shall be a magnification of the above triangle. Hence we must find co-ordinate system whose unit vector in 3-D space shpuld have  2 sides and 1 angle. Indeed, such a system exists known as cylindrical coordinate system. In such a system, the unit vector     r = cosφ*i +sinφ*j unit vector    φ =-sinφ*i +cosφ*j unit vector    z = -k or +k Here A=r    B=φ      C=z; any Vector X  = r*r   + φ*φ  +z*z Matrix R= cartesian coordinate a11   a12    a13 a21   a22    a23 a31    a32    a33 spherical coordinate sinηcosφ     sinηsinφ       cosη    * r   where r is the length cosηcosφ     cosηsinφ    -sinη  -sinφ            cosφ               0 For basis vectors r=1 (a11+a12+a13)2 =1+sin2ηsin2φ+(sinφ +cosφ)sin2η (a21+a22+a23)2 =1+cos2ηsin2φ-(sinφ +cosφ)sin2η (a31+a32+a33)2 =1-sin2φ  (a11+a12+a13)2+(a21+a22+a23)2 +(a31+a32+a33)2 =3 Jacobian in spherical co-ordinates J = a11         r*a21       rsinη*a31          = sinηcosφ      r*cosηcosφ            -sinφ(rsinη) a12        r*a22        rsinη*a32             sinηsinφ       r*cosηsinφ              cosφ(rsinη) a13        r*a23           0                        cosη               r* -sinη                   0 The elements of the J matrix are given by Jij =∂ƒi /  ∂xj g=JTJ =symmetric covariant metric tensor Determinant of J=|J|= r2sinφ cylinderical coordinate rcosφ     rsinφ       0       where r,m   are the length -rsinφ    rcosφ      0         0          0            m1 For basis vectors r=1,m=1 all the coordinates are reflected in the above calculator as aij where i,j=1,2,3 Sum of the row vectors such as a11+a12+a13,  a21+a22+a23 , a31+a32+a33 each represent the perimeter of a triangle if area enclosed is non-zero real positive. Jacobian for cylindrical to spherical = (r sinq)-1 (r2 sinq) = r. This result is correct, but it is not intuitively obvious. Determinant of J for cylindrical coordinates=|J|= r Determinant of J for polar coordinates=|J|= r Scalar Matrix DefinitionIf A , B are 2 vectors as above, A.B=Ms= 1      1     1 a      b     c d     e     f determinant  of Ms=1*ad+1*be+1*cfIf A=ai+bj+ck    B=di+ej+fkdet AxB = M |AxB|2 =M2 =(bf-ce)2+(cd-af)2+(ae-bd)2 A.B =Ms =ad+bc+ef Now M2 -M2s   > or = or <  0 . Putting the values, we get f2(a2+b2-c2)-4c(ad+be)f+(e2-d2) (a2-b2)+c2(d2+e2)-4abde = 0 This is a quadratic equation in f and solving for f, we get the value of f. Discriminant is 16c2(ad+be)2 -4(a2+b2-c2)[(e2-d2) (a2-b2)+c2(d2+e2)-4abde] =4*[c2{4(ad+be)2+1} - (a2+b2)][c2(d2+e2)+{(e2-d2) (a2-b2)-4abde}] =4*[ABdis1*c2 +ABdis2][ABdis4*c2 +ABdis3] where ABdis1 ={4(ad+be)2+1} =always positive ABdis4 =(d2+e2) = always positive ABdis2=-(a2+b2)= always negative ABdis3={(e2-d2) (a2-b2)-4abde} may be positive or negative If the discriminant becomes negative, increase c by integers. Whether c is positive or negative does not affect the discriminant since it is in a square form. Only its magnitude matters. steps to do in calculator for M=Ms under above circumstances. Put value of a11,a12,a13,a21,a22,a23. Press calculate . a23 will be recalculated. In uniform circular motion,  ω= v/r= (1/r)* v  where (1/r) is the reciprocal vector and not inverse vector.Application of vector and scalar triple products :- We find out angular momentum of a point mass in uniform circular motion . Angular momentum L =r x p (where r is radius vector and p is linear momentum)= r x (mv)=r x m(ω x r) where ω  is angular velocity =m[r x (w x r)]=m [(r.r)ω -(r.ω)r] = mr2ω -0 =Iω where I=mr2 .since r and ω are at right angles to each other . Here L = Lz where z axis is taken as perpendicular to the plane of uniform circular motion.  Both  Lz and ω are along the z-axisWork done : W =F.S =F.(θ x r ) = -F.(r x θ ) =-(F x r).θ =(r x F).θ = τ .θ Acceleration in spherical Co-ordinates : r̄ = r r̂           where r̂  =sinηcosφ i   +  sinηsinφ j +  cosη  k Since we confine our selves into x-y plane r̂  =cosφ i   +  sinφ j φ̂ =-sinφ i  +  cosφ j r̂'  =-sinφ*φ' i   +  cosφ*φ' j =φ'*φ̂ φ̂ '  =-cosφ*φ' i  - sinφ*φ' j =-φ' * r̂Now r̄' =r'r̂ +rr̂' =r'r̂ +rφ'*φ̂  =vrr̂ +vφφ̂          r̄'' =(r''r̂ +r'r̂ '  ) +(rφ''+r'φ')φ̂  +rφ'φ̂ ' =(r''r̂ +r'φ'φ̂ ) +(rφ''φ̂ )+r'φ'φ̂  +rφ'φ̂ ' =r''r̂ +2r'φ'φ̂ +rφ''φ̂  +rφ'φ̂ '               =r''r̂ +2r'φ'φ̂ +rφ''φ̂  +rφ'(-r̂ φ') =(r'' -rφ'2) r̂  + (2r'φ'+rφ'')φ̂  =arr̂ +aφφ̂ When we  talk of a Central Force which is radial in nature, (2r'φ'+rφ'') =0  or d/dt(r2φ') =0 or r2φ' =constant or  r2ω =constant = Angular momentum L/m  =h =|rxv| =v*r⊥ω = h / r2 ; since h is constant, and denominator is always positive, ω monotonically increases or decreases with h being positive or negative.L/m is called the specific angular momentum.r̄'' =(r'' -rφ'2) r̂  + (2r'φ'+rφ'')φ̂  or r̄''+rφ'2 r̂ -2r'φ'φ̂   = r''r̂ +rφ''φ̂  =ā Here second order derivatives are shifted to RHS. We call ā as co-ordinate version of acceleration. On the left side , both pink and green color entries are treated as fictitious forces with of course m multiplied. Pink one is called centrifugal force and green one is called coriolis force. These forces, though fictitious , are not zero in inertial frame of reference unlike normal fictitious forces. these newly defined forces are called the "coordinate" centrifugal force and the "coordinate" Coriolis force to separate them from the "state-of-motion" forces. so observers in the same frame of reference standing on different street corners see different "forces" even though the actual events they witness are identical. How can a physical force (be it fictitious or real) be zero in one frame S, but non-zero in another frame S' identical, but a few feet away? Even for exactly the same particle behavior the expression rφ'2  is  is different in every frame of reference, even for very trivial distinctions between frames. In short, if we take  rφ'2     as "centrifugal force", it does not have a universal significance: it is unphysical. The absurdity of the behavior of rφ'2  indicates that one must say that   rφ'2 is not centrifugal force, but simply one of two terms in the acceleration. This view, that the acceleration is composed of two terms, is frame-independent: there is zero centrifugal force in any and every inertial frame. It also is coordinate-system independent: we can use Cartesian, polar, or any other curvilinear system: they all produce zero. Apart from the above physical arguments, of course, the derivation above, based upon application of the mathematical rules of differentiation, shows the radial acceleration does indeed consist of the two terms (r'' -rφ'2). That said, the next subsection shows there is a connection between these centrifugal and Coriolis terms and the fictitious forces that pertain to a particular rotating frame of reference (as distinct from an inertial frame). Angular momentum contributes an effective potential energy.Ueff =U(r) + L2 /2mr2 ;For a free particle , kinetic momentum is p=mv ; But in general, the canonical momentum (total momentum) is a sum of kinetic momentum and potential momentum (if potential energy is velocity dependent ). For example, for a particle moving in a uniform magnetic field, p=mv+qA where A is vector action potential. B=∇xA E=∇φ -∂A/∂t Hamiltonian for e.m field is H=(1/2m) [p-qA]2 +qφ In general, neither canonical momentum nor kinetic momentum is a conserved quantity. If A is a vector and B is also a vector, under what conditions [A,B] =kB where k is a constant .Under normal situation, [A,B] is perpendicular to the plane of A,B where as k2B is in the plane of A,B. hence the equation is not satisfied. But we have to find special conditions under which it is satisfied. A= ai + bj+ck B=di + ej +fk [A,B] = 2AxB or (1/2)[A,B] =AxB= i     j    k a   b   c d   e   f =      kB which implies that bf - ce=kd ------(1) cd - af=ke ------(2) ae -bd=kf ------(3) multiplying (1) by a and (2) by b and adding, e=d(bc-ka) /(kb+ac)..........(4) Putting the value of e in (3), we get f=-d(a2+b2)/(ac+kb),,,,,,,,,,(5) putting the value of e,f in (1) , we get k = ±i√(a2+b2+c2) and k2 = -(a2+b2+c2) Substituting the value of k in (4), we get e=abd(a2+b2)/(k2b2-a2c2) + i* cd(a2+b2)√(a2+b2+c2)/(k2b2-a2c2) =e1+ie2 where e1,e2 are real Similarly f=acd(a2+b2)/(k2b2-a2c2) - i* bd(a2+b2)√(a2+b2+c2)/(k2b2-a2c2) =f1-if2 where f1,f2 are real (1/2)[A,B] =AxB=Ax(Breal + Bimg ) =(AxBreal ) +(AxBimg ) =kB =kBreal  +kBimg  ; Breal= di  +                  abd(a2+b2)/(k2b2-a2c2) * j +acd(a2+b2)/(k2b2-a2c2) *k ; Bimg= 0i  +i*cd(a2+b2)√(a2+b2+c2)/(k2b2-a2c2) * j - i* bd(a2+b2)√(a2+b2+c2)/(k2b2-a2c2) *k ; Since k is imaginary, (AxBreal ) =kBimg........(6) (AxBimg ) =kBreal........(7) ori          j          k     =k(0 *i  +ie2*j + if2*k)...........(6a) a         b         cd        e1        f1 andi          j          k     =k(d*i  +e1*j + f1*k)...........(7a)a         b         c 0        ie2        if2Important Observation (1) A is real vector with a,b,c being real numbers. (2) B is a spinor with d being real and we are at liberty to choose the value of d.(3) e and f are complex numbers with real and imaginary parts. (4) k is also an imaginary number and its value is determined by the value of a,b,c. (calculator in blue font. )Suppose A, B are vectors given by A= ai + bj+ck B= di + ej +fk Angle between A, B given by θ and cosθ = A.B / (|A| * |B|) =(ad + be + cf) /  (|A| * |B|)  = (ad + be ) /  (|A| * |B|)  +  cf /  (|A| * |B|) = cos φ  + cos η or cos φ =(ad + be ) /  (|A| * |B|) ;  If |A| =|B| = 1, then  cos φ =(ad + be ) cos η = cf  /  (|A| * |B|) ;            If |A| =|B| = 1, then   cos η = cf If we take a=cosθ1 sinθ2 b=cosθ1 cosθ2 c=sinθ1 d=cosθ3 sinθ4 e=cosθ3 cosθ4 f=sinθ3 then A,B become normalized vectors cos φ =(ad + be ) =cos(θ2 - θ4)*cosθ1cosθ3 cos η = cf =sinθ1*sinθ3 cosθ  = cos φ  + cos η . Since nature is symmetric to sin and cos, We can write sinθ  = sin φ  + sin η which leads to upon squaring and adding cos(φ - η) =cos( η - φ) =-1/2   or  φ - η =   120° or η - φ = 120° which exhibits symmetry with respect to interchange of η and  φ Moreover  η and  φ take only those values which satisfy the above equation. If on the RHS, we take the unit as h , it becomes a spin equation. h   = h/2π Now sin φ = √[(|A| * |B|)2  - (ad + be )2] / (|A| * |B|) ;  If |A| =|B| = 1, then  sin φ =√[(1  - (ad + be )2]          sin η =√[(|A| * |B|)2  - c2f2] / (|A| * |B|) ;             If |A| =|B| = 1, then   sin η =√[1  - c2f2] We know cf=cos η  and                 (ad + be )=cosφ =cos(  η-120 )cos( η-φ) =cos φcos η  + sin φ sin η =    (ad + be )cf   + √[(1  - (ad + be )2]√[1  - c2f2] = (1/4)*cos (θ2 - θ4)*sin2θ1 *sin2θ3  +√[(1  - sin2θ1*sin2θ3]*√[(1  - cos2(θ2 - θ4)*cos2θ1cos2θ3] orcos( η-φ) =1/4   +   (√3/2)* (-√3/2)=-1/2     if  θ2 =θ4 and  θ1=θ3 =45° and the blue and red signatures are of opposite sign with same numerical value.If θ2 =θ4, cos( η-φ) = (1/4)*sin2θ1 *sin2θ3  +√[(1  - sin2θ1*sin2θ3]*√[(1  - cos2θ1cos2θ3] Spinors: These are somewhat like tensors. They allow a more general treatment of notion of invariance under rotation and Lorentz Boost. To every tensor of rank k, there corresponds a spinor of rank 2k. Some kinds of tensors can be associated with spinors of same rank. A general 4 vector would correspond to a Hermitian spinor of rank 2, which is represented by a 2x2 Hermitian matrix. Spinors arise naturally in discussions of Lorentz group. In fact, spinor is the most basic sort of mathematical object that can be lorentz transformed.Suppose vectorsA=i + jB=i+ j||A|| = √2 ||B|| = √2 = ||A|| ||B|| cosθ =2cosθ, maximum value being 2 and minimum absolute value 0. If vector A and vector B are added to say vector C ||C|| =√ (2+2+4 cosθ) =2√(1+ cosθ)=2√2 when cosθ has maximum value In Bell Inequality,  Δ  ‌≤ 2  (in case of hidden variable theory) where as otherwise  Δ  ‌≤ 2√2