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  

(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.

 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 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=bk2

Adding,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 =0

Ak2²  + 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/c

d2=af/c +k2b/c

Δd=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*k

iA2 =d2*i +e*j +f*k

iA1-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*k

Now 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+fk

AxiA=

i  j   k

a 0   c

d 0  f

=(cd-af)*j

|AxiA| = 1 by definition of inverse vector

or  (cd-af) =+ 1 or -1

d1=(af +1) / c

d2=(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=di

A x  iA =0, hence

iA = ej

or

iA=fk

Taking the former

A x iA =ae(i x j) =ae* k where ae =± 1 or

e = ± 1 / a

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| = 1

nA . 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_⊥+ nA_z    ;where

nA_⊥=|r|r=|r| (cosφ i + sinφ j)

nA_z  =√(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.

(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.

(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

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/f

where (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* ε_ijk (AB)_k;  i , j , k=1,2,3 and ε_ijk  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

 AB_xy⁺=[AB_x + AB_y]  =[AB_xi + AB_yj] ; AB_xy⁺  is called raising vector

AB_xy^- =[AB_x - AB_y]   = [AB_xi - AB_yj]  ;AB_xy^-   is called lowering vector

Together AB_xy⁺      and AB_xy^-     are called ladder vectors.

In a matrix equation,

  AB_xy⁺     = AB_x       AB_y  *     i =    M1 *   i

  AB_xy^-         AB_x     -AB_y        j                  j  

or

M1     = AB_x       AB_y 

             AB_x     -AB_y                      

Det M1=(-2AB_xAB_y)

[ AB_xy⁺, AB_xy^-] =- 4(AB_x * AB_y/ AB_z )AB_z

{ AB_xy⁺, AB_xy^-}= 0

[ AB_xy⁺, AB_z] = (2AB_z * AB_y/ AB_x ) AB_x  -    (2AB_z * AB_x/ AB_y )AB_y

=k*AB_xy^- if both blue and red components become same.

[ AB_xy^-, AB_z] = -(2AB_z * AB_y/ AB_x ) AB_x  -    (2AB_z * AB_x/ AB_y )AB_y

=k1*AB_xy⁺ if both blue and red components become same and  k1=-k

In a matrix equation,

 AB_xy⁺                     =                      AB_x                       AB_y                 *      i  =M2  *   i

 [ AB_xy⁺, AB_z]                       2AB_z(AB_y)        2AB_z(-AB_x)                       j                 j

or

M2  =              AB_x                     AB_y                

                       2AB_z(AB_y)        2AB_z(-AB_x)                     

Det M2 = -2AB_z[ AB²_x + AB²_y  ]

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)²_x+(AB)²_y=(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 J_ij =∂ƒ_i /  ∂x_j

g=J^TJ =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

If A , B are 2 vectors as above,

A.B=M_s=

1      1     1

a      b     c

d     e     f

determinant  of M_s=1*ad+1*be+1*cf

If A=ai+bj+ck

   B=di+ej+fk

det AxB = M

|AxB|² =M² =(bf-ce)²+(cd-af)²+(ae-bd)²

A.B =M_s =ad+bc+ef

Now M² -M²_s   > 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=M_s under above circumstances. Put value of a11,a12,a13,a21,a22,a23. Press calculate . a23 will be recalculated.

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 = L_z where z axis is taken as perpendicular to the plane of uniform circular motion.  Both  L_z and ω are along the z-axis

Work done : W =F.S =F.(θ x r ) = -F.(r x θ ) =-(F x r).θ =(r x F).θ = τ .θ

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' * r̂

Now r̄' =r'r̂ +rr̂' =r'r̂ +rφ'*φ̂  =v_rr̂ +v_φφ̂

         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  =a_rr̂ +a_φφ̂

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.

U_eff =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.

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(B_real + B_img ) =(AxB_real ) +(AxB_img ) =kB =kB_real  +kB_img  ;

B_real= di  +                  abd(a²+b²)/(k²b²-a²c²) * j +acd(a²+b²)/(k²b²-a²c²) *k ;

B_img= 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,

(AxB_real ) =kB_img........(6)

(AxB_img ) =kB_real........(7)

or

i          j          k     =k(0 *i  +ie2*j + if2*k)...........(6a)

a         b         c

d        e1        f1

and

i          j          k     =k(d*i  +e1*j + f1*k)...........(7a)

a         b         c

0        ie2        if2

Important 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.

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]

or

cos( 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]

If A is a real vector in 3-D,

A =xi+yj+zk where x,y,z are real numbers.

If A is a complex vector , then in general, x,y,z are complex numbers.

A =(a+ib)i +(c+id)j +(e+if)k  where a,b,c.d.e.f are real numbers.

Suppose, we say that vector

v=(2,i,2-i). Then

a=2,b=0, c=0,d=1,e=2,f=-1

If u,v are real vectors, u.v=v.u

But if u,v are complex vectors, u.v=u.v* =(v.u)*  where * indicates complex conjugate. Hence u.v is not necessarily equal to v.u.

 if k is a complex number, u,v both are complex vectors,

⊛(ku).v=k(u.v)

⊛u.(kv) =k*(u.v)

⊛u.u=0 iff u=0

Eucledean Norm:

let u=(2+i,0,4-5i)

     v=(1+i,2+i,0)

||u|| = √[(2*2+1*1)+0+4*4+-5*-5 ] =√46

||v||=√[(1*1+1*1)+(2*2+1*1)+0] =√7

d(u,v) =||u-v|| =||1,-2-i.4-5i|| =√47

Two component complex vectors  are spinors of rank 1.

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 vectors

A=i + j

B=i+ j

||A|| = √2

||B|| = √2

<A,B> = ||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

If A=a1i+a2j+a3k

   B=b1i+b2j+b3k

A x B =(a2b3-a3b2)i +(a3b1-a1b3)j+(a1b2-a2b1)k=

| 0  -a3  a2 | *  b1 =M * i

 a3  0  -a1       b2          j

|-a2 a1  0 |      b3          k

M is an skew-symmetric matrix.