How do you show that if you have the assosciated ODEs of a pde: dx/a = dy/b = dz/c = ds, then for constants m,n,p it is also true that (mdx + ndy + pdz)/(ma + mb + pc) = ds?
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If $\displaystyle dx = a ds, dy = b ds, dz = c ds $ then $\displaystyle m dx = m a ds, n dy = n b ds, p dz = p c ds$. Adding gives $\displaystyle m dx + n dy + p dz = (m a + n b + p c)ds$ and dividing gives your answer
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