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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
2 years, 3 months 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, \( \cos^2x + \cos^2y + \cos^2z + 2\cos x\cos y + 2\cos y\cos z + 2 \cos x\cos y + \)

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

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

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

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

\( 0 = a^2 + b^2 \implies a = 0\quad \& \quad b = 0 \) Siddhartha Srivastava · 2 years, 3 months ago

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@Siddhartha Srivastava Thanks for this man. Raghav Vaidyanathan · 2 years, 3 months ago

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The answer to q 1 is very easy: \(a=b=0\). Raghav Vaidyanathan · 2 years, 3 months ago

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@Raghav Vaidyanathan Ans 2 q2 seems to be zero too. Raghav Vaidyanathan · 2 years, 3 months ago

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@Raghav Vaidyanathan can u explain in brief the first part Tanishq Varshney · 2 years, 3 months 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 · 2 years, 3 months ago

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