**Law of Parallelogram of vectors**

* Suppose an equation is R 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).The
center lies at (b,0) with origin at (0,0)^{2} +(y-0)^{2} =a^{2}* 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= a 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 x
When θ=90° , y=b, x=a. Hence, the trajectory swept by the resultant
2πb.
The above conclusion was because of equation (x-b),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. ^{2} +(y-0)^{2} =a^{2}But , unlike the above case, there
are no tangents from origin to the circle. Area trapped is
πa and circumference is ^{2}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
2ab=πb
Difference of area trapped on rotation of a and b=Δ
(a)Minimum Value of Δ
(b)
Δ
(c )
Δ The angle R makes with a is φ1 . (d)
Δ 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
* 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 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
x
ab=k 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=k a,b shall have the same sign,and extremum values are a=k , b=k & a=-k, b=-k.
as a+b= x+k
Which means 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|| =√ [ a
area =∫||A||da =∫ √[x
If x=1 & y=1, area =∫ √[2+4a(1-a
So area =1/√2 [-cos2θ/2 +θ +sin2θ/2]
If x=1,y=-1
area=√2∫ [(sinθ -cosθ)cosθ ]
dθ=√2 [-cos2θ/2 -θ -sin2θ/2] =(1/√2)(3+π ) * Average area swept by
resultant vector is area 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) = [ |
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