Question

If $(y-z)\propto \frac{1}{x},(z-x)\propto \frac{1}{y}$ and $(x-y)\propto \frac{1}{z}$, then find the sum of three variation constants.

Answer 1

$(y-z)\propto \frac{1}{x}\Rightarrow y-z=k_1\cdot \frac{1}{x}$ (where $k_1\ne 0=$ variation constant.)
$\Rightarrow k_1=x(y-z)......\left(1\right)$
$(z-x)\propto \frac{1}{y},\therefore (z-x)=k_2\cdot \frac{1}{y}$ (where $k_2\ne 0=$ variation constant.)
$\therefore k_2=y(z-x)......\left(2\right)$
Also, $(x-y)\propto \frac{1}{z},\therefore x-y=k_3\cdot \frac{1}{z}$ (where $k_3\ne 0=$ variation constant.)...(3)
$\therefore k_3=z(x-y)$
Now, adding $\left(1\right)+\left(2\right)+\left(3\right)$ we get,
$$\begin{array}{cc}{k}_1+k_2+k_3& =x(y-z)+y(z-x)+z(x-y)\\ & =xy-xz+yz-xy+zx-yz=0\\ \therefore k_1+k_2& +k_3=0.\end{array}$$
Hence the sum of three variation constants $=0$.