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[webfactor]

Question 11..... Part (B) May/june2007... Paper 1
Awaiting for a response...

 

[usman]

Plx make clear wich question u r referrin' 2, is it in Paper 1(i.e. the 1 abt 3D vectors) or Paper 2(i.e. the 1 abt prob density function)?

 

[webfactor]

Sorry, its Ques 11, part b. paper 1

 

[anzaar]

Your sol for Q.10 part B Paper 1
find a vector from S to l3.
say v=(1,-2,8)-(-1,3,2)
v=(2,-5,6)
find d1;director vector of line l1
d1=(-2,5,13)-(1,-1,-2)=(-3,6,15)
find d2; direction vector of line l2
which is d2=(1,-1,-3)
take vector product of d1& d2
d=d1xd2=(-3,6,15)x(1,-1,-3)
d=(-3,6,-3)
Take its unit vector d^=(-3,6,-3)/sqr.rt(54)
take vector product of v and d^
vxd^=(2,-5,6)x(-3,6,-3)
=(21,-12,-3)/sq.rt(54)
take its magnitude.
ans is Sq.rt(11)

 

[webfactor]

its ques 11 part B paper 1

 

[usman]

This may help u:
http://www.physicsforums.com/showthread.php?t=235914

 

[webfactor]

It din't solve... Can some1 post the whole solution? How to get the 2^n(5-Cos@)


How do we get cos???? If we are taking the imaginry part???? Please solve it and post it... thanx

 

[usman]

We want to show that
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)].
-------------------------
Note that e^(iθ) = cos θ + i sin θ.
So, De Moivre's Theorem yields for any positive integer n
e^(inθ) = (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ).

Next, note that
Σ(n=1 to N) sin(nθ) / 2^n
= Im [Σ(n=1 to N) e^(inθ) / 2^n], by the first observation
= Im {Σ(n=1 to N) [e^(iθ)/2]^n}.

Thus, we need to compute the imaginary part of the finite geometric series
Σ(n=1 to N) [e^(iθ)/2]^n
= (e^(iθ)/2) Σ(k=0 to N-1) [e^(iθ)/2]^k, by re-indexing the sum
= (e^(iθ)/2) (1 - (e^(iθ)/2)^N) / (1 - e^(iθ)/2)
= e^(iθ) (1 - (e^(iθ)/2)^N) / (2 - e^(iθ))
= e^(iθ) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))]

Multiply numerator and denominator by the conjugate (2 - e^(-iθ)):
==> e^(iθ)(2 - e^(-iθ)) (2^N - e^(iNθ)) / [2^N (2 - e^(iθ))(2 - e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (e^(iθ) + e^(-iθ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 2 (2 cos θ))]
= (2e^(iθ) - 1) (2^N - e^(iNθ)) / [2^N (5 - 4 cos θ)].

Now, it's a matter of simplifying the numerator.
(2e^(iθ) - 1) (2^N - e^(iNθ))
= 2^(N+1) e^(iθ) - 2^N - 2e^(i(N+1)θ) + e^(iNθ)
= 2^(N+1) (cos θ + i sin θ) - 2^N - 2 (cos(N+1)θ + i sin(N+1)θ) + (cos(Nθ) + i sin(Nθ))
= [2^(N+1) cos θ - 2^N - 2 cos(N+1)θ + cos(Nθ)] + i [2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ)]

Its imaginary part is 2^(N+1) sin θ - 2 sin(N+1)θ) + sin(Nθ).

Therefore, we have obtained
Σ(n=1 to N) sin(nθ) / 2^n = [2^(N+1) sin θ + sin (Nθ) - 2 sin(N+1)θ] / [2^N (5 - 4 cos θ)]
as required.

I hope this helps!

 

[webfactor]

Can you tell me how come this changes into 2cosθ?
(e^(iθ) + e^(-iθ))???
you have taken the imaginary part?

 

[webfactor]

Thanks so much............ IT helped.... thanx alot... may I knw ur city???

 

[usman]

Can you tell me how come this changes into 2cosθ?
(e^(iθ) + e^(-iθ))???
you have taken the imaginary part?

This is because:
cos(θ) = Re {e^iθ} = [{e^(iθ) + e^(-iθ)}/2] is an identity
Now, rearranging this identity
cos(θ) = [e^(iθ) + e^(-iθ)]/2
We get this,
2cos(θ) = e^(iθ) + e^(-iθ)

I hop this helps. bTw m 4m islamabad