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

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

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

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

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

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

- 6 years, 3 months ago

Thanks for this man.

- 6 years, 3 months ago

- 6 years, 3 months ago

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

- 6 years, 3 months ago

Ans 2 q2 seems to be zero too.

- 6 years, 3 months ago

can u explain in brief the first part

- 6 years, 3 months ago

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

- 6 years, 3 months ago