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

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

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

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

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

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

Thanks for this man.

- 3 years ago

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

- 3 years ago

Ans 2 q2 seems to be zero too.

- 3 years ago

can u explain in brief the first part

- 3 years 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

- 3 years ago