Law of Parallelogram of vectors

    Find  
vector a (magnitude)

a > = b

||R1||=vector a+vector b=

√(a2 + b2 +2ab cosθ)

vector b (magnitude) ||R2||=vector a-vector b=

√(a2 + b2 -2ab cosθ)

Angle θ between a & b

(in degree)

R1 + R2
    R1 - R2
    2abcosθ / √(a2+b2  )
    2abcosθ /√(a2+b2sin2θ  )
    perimeter of parallelogram:2(a+b)
    area of parallelogram(absinθ)
    area/perimeter:(sinθ/2)*1/(1/a + 1/b)
    total area swept4* 0π/2 absinθ=4ab
Find   (R1 + R2)(R1 - R2)
l1=b2/a R12 - R22 (4abcosθ)
c1=√(a*a-b*b) tan φ1(angle between R1 & a)

=bsinθ /(a+bcosθ)

e1=c1/a tan φ2(angle between R1 & b)

=asinθ /(b+acosθ)

a/2b φ1 (in degree)
b/2a φ2 (in degree)
(a2+b2)/2ab tan φ3(angle between R2 & a)

=bsinθ /(a-bcosθ)

[(-a2-b2) +(b4/a2)]/2ab tan φ4(angle between R2 & b)

=asinθ /(b-acosθ)

area swept by resultant φ3 (in degree)

area trapped by Locus of P.V of resultant when b rotates around a (πb2)

φ4 (in degree)
area trapped by Locus of P.V of resultant when a rotates around b (πa2) tan φ4 / tan φ2 =(b+acosθ)/(b-acosθ)
√(ab)= (R1/R2)2
angle δ (-π/2 to π/2)    
a2=ksinδ    
b2=k/sinδ    
a2+b2    
  angle in °    
Angle θ=cos-1 (-a/2b) R1/R2
Angle θ=cos-1 (-b/2a) √(R1R2)
Angle θ=cos-1 [(-a2-b2   )/  2ab ] (R1+R2)/2
Angle θ=cos-1 [(-a2-b+b4/a2 )/2ab] 2/(1/R1+1/R2)
Angle θ=cos-1 (-b/a )    
value of R for θ=cos-1 (-a/2b) φ3 / φ1
value of R for θ=cos-1 (-b/2a) φ4 / φ2
 R for θ=cos-1[(-a2-b2 )/  2ab ]

There is no real angle for which R=0

R1aa  (√(bsinθ/cosφ1))
R for θ=cos-1[(-a2-b+b4/a2 )/2ab] R=b2 /a R1ab  (√(a+bcosθ)/sinφ1)
R for θ=cos-1(-b/a)=(a2-b ) R1 = √(bsinθ/cosφ1)* √(a+bcosθ)/sinφ1
a2/2b2 R2aa (√(bsinθ/cosφ3))
R2ab (√(a-bcosθ)/sinφ3)

Distance between Origin of vectors & Focus

R2= √(bsinθ/cosφ3)* √(a-bcosθ)/sinφ3

length of resultant which connects origin of vectors to intersection of semi latus rectum with ellipse (R3)

2bcosθ
value of θ in degree for R3 (R1-R2) / 2bcosθ = x
    a/√(a2+b2sin2θ) =y
    x/y
       

Angle value of x Value of y value of R
Angle θ=0°
Angle θ=30°
Angle θ=45
Angle θ=60°
Angle θ=75
Angle θ=90
Angle θ=120
Angle θ=135
Angle θ=150
Angle θ=165
Angle θ=180
       

 

     
 

* Suppose an equation is R2 =a2 +b2+2absin2φ . How to convert it to an equation of law of parallelogram of addition of vectors ? we write R2 =a2 +b2+2absin2φ =R2 =a2 +b2+2abcos (π/2 - 2φ). Taking π/2 - 2φ =θ,  R2 =a2 +b2+2abcosθ....(1) Assuming one end of Resultant vector being at origin, the other end is at a point x=a+bcosθ & y=bsinθ...(2) . What equation does the locus of the resultant vector satisfy ? From (2), eliminating θ, we get (x-a)2 +(y-0)2 =b2 which is the equation of a circle with origin at (0,0) and having radius b and center at(a,0).This equation resulted when a coincides with x-axis, a>b and the smaller vector b rotates around a (a being the axis). We can also coincide b with the x-axis and let a rotate around which creates a symmetrical equation by replacing a with b and vice versa. The equation becomes (x-b)2 +(y-0)2 =a2.The center lies at (b,0) with origin at (0,0)

* What curve does the resultant vector draw when a, b are constants and angle θ varies from 0° to 360° i.e completes one cycle. Since the equation carves out  symmetrical shape from (0 to 180) & (180 to 360), we explore the behavior of the equation from 0 to 180 .

When the angle is 0 degree, R= a+b

When the angle is 90 degree, R= a2 +b2 .

When the angle is 180 degree, R=a-b

Total spectrum covered in x-axis (a+b)+(a-b)=2a

Corollary of above is x=a+bcosθ . Hence xmax =a+b at cosθ=1 or θ=0 and xmin =a-b at cosθ=-1 or θ=180. Similarly y =bsinθ . Hence  ymax =b at sinθ=1 or θ=90 and ymin =-b at sinθ=-1 or θ=270

When θ=90 , y=b, x=a. Hence, the trajectory swept by the resultant traps an enclosed space which is  a circle whose  radius is b , and origin is  (0,0) which is outside the circle and  with near vertex at (a-b,0) and farthest vertex at (a+b,0) and center is at (a,0). Area trapped is πb2 and circumference is 2πb.

The above conclusion was because of equation (x-a)2 +(y-0)2 =b2. For the equation (x-b)2 +(y-0)2 =a2,i.e. when b is in x-axis and a is rotated around b, one can observe from the accompanying figure that the origin lies within the circle, every resultant intersects the circle at 2 points which are 180 degree apart. But , unlike the above case, there are no tangents from origin to the circle. Area trapped is πa2 and circumference is 2πa.

When (a/b) >π /2 .......area swept by by the resultant > area trapped by the position vector of resultant on rotation of b

When (a/b) =π /2 .......area swept by by the resultant = area trapped by the position vector of resultant on rotation of b

when π /2> (a/b)>=1 ....area swept by by the resultant is less than area trapped by the position vector of resultant on rotation of b

In all the above cases, area swept by by the resultant is less than area trapped by the position vector of resultant on rotation of a

    2ab=πa2 => a/b= π / 2 which is not true

   2ab=πb2 => (a/b)min= π / 2 for the equality

Difference of area trapped on rotation of a and b=Δab=πa2-πb2 =

(a)Minimum Value of  Δab  =0 when b=a.

(b) Δab  =a2 (π   -  4/π)=1.87a  when b=2a/π=0.6365a

(c ) Δab  is greater than  πb2   but less than πa2  when b < 2a/π

The angle R makes with a is φ1 .

(d) Δab  is directly proportional to ( a2-b2) which is discrete and not continuous since a, b are discrete.

tanφ1= bsinθ / (a+bcosθ)

The angle R makes with b is φ2 .

tanφ2= asinθ / (b+acosθ)

Now we can put a as a=(π/2)b*ncosθ1 where n is any integer n=1,2,3,..... and θ1 can take any value within specified range. we can refine this formula to a=(π/2)b[1+ncosθ1] where n=0,1,2,3,......We have thus quantized the value of a .

* Area swept by  the resultant vector  is the area of the triangle formed by vector a,b and R. This is absinθ/2. So total area swept is 4 times the area swept when θ moves from zero degree to 90 degree. Hence the equn.- A=dA=4/2 90°0°absinθ =-2ab[cosθ2 -cosθ1]90°   =-2ab*(-1) =2ab (integral of sinθ is -cosθ). We can visualize the area as that of a rectangle with length 2a and breadth b or a circle with radius r= √[(2/π)(ab)] = √[(0.63694)(ab)] =√[(ab*sin2θ)]=sinθ*√(ab) where θ=0.924135 radian or 52.93° .

sinθ*√(ab) can be put as √[(asinθ)(bsinθ)]=√ [(projection of a on y-axis)*(projection of b on y-axis)] where we have reduced 2-D vector into 1-D vector (of course with loss of information).

* If we think of a rectangle with sides 2a,b, then we can inscribe an ellipse  with major axis 2a and minor axis b with center coinciding with the center of the rectangle (considered as origin ). The equation of such an ellipse is       x2/a2 +4y2/b2=1 and area is πab /2 =(π/2)ab=ab*cosec2θ where θ=0.924135 radian or 52.93 . Can we find a transformation matrix that transforms to the locus to the ellipse from the locus of resultant vector ?? and find the behaviour of resultant vector vis-a-vis the components.If we inscribe a circle , maximum area will be  πb2 /4. The ratio of area of the recangle and that of the ellipse with major axis 2a & minor axis b is π irrespective of the value of a and b. The ratio of the area of ellipse to the circle inscribed in it shall be major axis/minor axis.

* Suppose we have different value of a,b such that ab=constant=c2 , then the average of a,b shall be the geometric mean c. Then the total area swept by resultant vector =2ab=2c2 =(√2c)*(√2c). If we construct a right angled isosceles  triangle with side c, the hypotenuse shall be √2c and the square constructed on the hypotenuse will have the same area as total area swept by resultant vector.We can take 

ab=k2

a=ksinδ

b=k/sinδ

which are defined for all values of δ except δ=0;

At δ=0, we can also define the value with following trick:-

ab=  ksinδ * k/sinδ =k (sinδ/δ) * k (δ/sinδ) . as δ->0, sinδ/δ =1 and hence ab=k2 

a,b shall have the same sign,and extremum values are  a=k , b=k  & a=-k, b=-k.

as a+b= x+k2 /x = extremum value => d(x+k2 /x)/dx=0 or x=+k or -k.

Which means a can oscillate within the range δ=π/2 to -π/2  ( csinδ lies in[-k,+k]) and accordingly k/sinδ also varies .

a+b shall vary between 2k & -2k with condition that both shall have the same sign..

* Let

a    b  * x =     ax+by

b    a    y         bx+ay i.e. a vector say A with components as column matrix.

||A|| =√ [ a2x2 + b2y2 + 2abxy +b2x2 + a2y2 + 2abxy    ] =√[(a2+b2)(x2+y2)+4abxy]=√[(x2+y2)+2ab*2xy ] as  a2+b2=1

area =∫||A||da =∫ √[x2+y2+2xy*2a(1-a2)1/2]  da.

If x=1 & y=1, area =∫ √[2+4a(1-a2)1/2]  da=√2∫ √[1+2a(1-a2)1/2]  da=√2∫ [(sinθ +cosθ)cosθ ]  dθ by taking a=sinθ

So area =1/√2 [-cos2θ/2 +θ +sin2θ/2]π/20

=(1/√2)(1+π/2 )

If x=1,y=-1

area=√2∫ [(sinθ -cosθ)cosθ ]  dθ=√2 [-cos2θ/2 -θ -sin2θ/2]π/20

=(1/√2)(3+π )

* Average area swept by resultant vector is areaav =(1/2)ab0π/2sinθdθ / 0π/2dθ =ab/π. Had the area been swept in a linear manner, areaav =(0+2ab)/2 = ab.

Here the deviation =ab -ab/π =ab (1 - 1/π)

Again perimeter of area swept is taken as perimeter of parallelogram divided by 2 = a+b (this is not actually correct).

hence area swept / perimeter = absinθ / (a+b) = [ab/(a+b)] sinθ . Figure in red looks like 2 resistances connected in parallel.