What about these

Q1) If \(cos(x-y)+cos(y-z)+cos(z-x)=\frac{-3}{2}\) and \(cosx+cosy+cosz=a\) and \(sinx+siny+sinz=b\). Find \(a\) and \(b\)

Q2) Two non-conducting infinite wires have shapes as shown above. The linear charge densities of wire (1) and wire (2) are

\(x\quad if\quad x\leq -a~ and~ x\geq a\\ \\ 0\quad if\quad -a<x<a)\)

and

\(y\quad if\quad y\leq -a~ and~ y\geq a\\ \\ 0\quad if\quad -a<y<a)\)

respectively where \((x,y)\) are coordinates of points on the wire. An electron is released from origin, the unit vector along the direction of velocity of the electron just after its release will be in terms of \(\hat { i } \) and \(\hat { j } \)

Note by Tanishq Varshney
3 years ago

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\( cosx + cosy + cosz = a \)

\( \cos^2x + \cos^2y + \cos^2z + 2\cos x\cos y + 2\cos y\cos z + 2 \cos x\cos y = a^2\)

Likewise, \( \sin^2x + \sin^2y + \sin^2z + 2\sin x\sin y + 2\sin y\sin z + 2 \sin x\sin y = b^2\)

Adding, \( \cos^2x + \cos^2y + \cos^2z + 2\cos x\cos y + 2\cos y\cos z + 2 \cos x\cos y + \)

\( \sin^2x + \sin^2y + \sin^2z + 2\sin x\sin y + 2\sin y\sin z + 2 \sin x\sin y = a^2 + b^2 \)

\( \implies 1 + 1 + 1 + 2(\cos x\cos y + \cos y\cos z + \cos x\cos y + \sin x\sin y + \sin y\sin z + \sin x\sin y) = a^2 + b^2 \)

\( \implies 3 + 2(\cos(x-y) + \cos(y-z) + \cos(z-x)) = a^2 + b^2 \)

\( \implies 3 + 2* \frac{-3}{2} = a^2 + b^2 \)

\( 0 = a^2 + b^2 \implies a = 0\quad \& \quad b = 0 \)

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Thanks for this man.

Raghav Vaidyanathan - 3 years ago

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The answer to q 1 is very easy: \(a=b=0\).

Raghav Vaidyanathan - 3 years ago

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Ans 2 q2 seems to be zero too.

Raghav Vaidyanathan - 3 years ago

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can u explain in brief the first part

Tanishq Varshney - 3 years ago

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@Tanishq Varshney for first one The simplest way is to assume all the three angle differences equal, which ofcourse then must be 120 degree, so just consider 1 , w ,w^2 , add em up

Mvs Saketh - 3 years ago

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