# 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

and

$y\quad if\quad y\leq -a~ and~ y\geq a\\ \\ 0\quad if\quad -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
4 years, 4 months ago

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## Comments

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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, $\cos^2x + \cos^2y + \cos^2z + 2\cos x\cos y + 2\cos y\cos z + 2 \cos x\cos y +$

$\sin^2x + \sin^2y + \sin^2z + 2\sin x\sin y + 2\sin y\sin z + 2 \sin x\sin y = a^2 + b^2$

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

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

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

$0 = a^2 + b^2 \implies a = 0\quad \& \quad b = 0$

- 4 years, 4 months ago

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Thanks for this man.

- 4 years, 4 months ago

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- 4 years, 4 months ago

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The answer to q 1 is very easy: $a=b=0$.

- 4 years, 4 months ago

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Ans 2 q2 seems to be zero too.

- 4 years, 4 months ago

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can u explain in brief the first part

- 4 years, 4 months ago

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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

- 4 years, 4 months ago

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