I am currently referring a book on Partial Differential Equations and I came across the following statement:
The curves whose equations are solutions of $$\frac{dx}{P(x,y,z)}=\frac{dy}{Q(x,y,z)}=\frac{dz}{R(x,y,z)}$$ are orthogonal to the system of the surfaces whose equation satisfies $$P(x,y,z)\,dx+Q(x,y,z)\,dy+R(x,y,z)\,dz=0$$
I am unable to understand this as no explanation was given. Any explanation on why this is true would be helpful.
On
Geometrically the equation $$P(x,y,z)\,dx+Q(x,y,z)\,dy+R(x,y,z)\,dz=0\tag1$$
means a straight line whose direction cosines are proportional to $~dx,~dy,~dz~$, is perpendicular to a line whose direction cosines are proportional to $~P(x,y,z),~Q(x,y,z),~R(x,y,z)~$.
Therefore, a point that is moving subject to the condition expressed by $~(1)~$ must go in a direction at right angles to a line whose direction cosines are proportional to $~P(x,y,z),~Q(x,y,z),~R(x,y,z)~$.
On the other hand, geometrically, the equation $$\frac{dx}{P(x,y,z)}=\frac{dy}{Q(x,y,z)}=\frac{dz}{R(x,y,z)}\tag2$$
means a straight line whose direction cosines are proportional to $~dx,~dy,~dz~$, is parallel to a line whose direction cosines are porpotional to $~P(x,y,z),~Q(x,y,z),~R(x,y,z)$.
Therefore, a point that is moving subject to the conditions expressed by $~(2)~$ must go in a direction parallel to a line whose direction cosines are proportional to $~P(x,y,z),~Q(x,y,z),~R(x,y,z)$.
Hence, the curves traced out by points that are moving subject to the condition $(1)$ are orthogonal to the curves traced out by points that are moving subject to the conditions $(2)$.
The former curves are any of the curves upon the surfaces represents by $(1)$; therefore the curves represents by $(2)$ are normal to the surfaces represented by $(1)$. If $(1)$ is not integrable, there is no family of surfaces which is orthogonal to all the lines that form the locus of equation $(2)$.
Heuristically, the equation $$\frac{dx}{P(a,b,c)}=\frac{dy}{Q(a,b,c)}=\frac{dz}{R(a,b,c)}$$ means a vector $(s_1,s_2,s_3)$ tangent to this curve at point $(a,b,c)$ satisfies $$\frac{s_1}{P(a,b,c)}=\frac{s_2}{Q(a,b,c)}=\frac{s_3}{R(a,b,c)}.$$ After all, tangent vector represents small change, so you just put the "components of small change" into $dx,dy,dz$.
Similarly, the equation $$P(a,b,c)\,dx+Q(a,b,c)\,dy+R(a,b,c)\,dz=0$$ means a vector $(t_1,t_2,t_3)$ tangent to this surface at point $(a,b,c)$ satisfies $$P(a,b,c)\,t_1+Q(a,b,c)\,t_2+R(a,b,c)\,t_3=0.$$
Multiply both sides of this equation by $\frac{s_1}{P(a,b,c)}(=\frac{s_2}{Q(a,b,c)}=\frac{s_3}{R(a,b,c)})$, and you get $$s_1t_1+s_2t_2+s_3t_3=0,$$ meaning the curve and the surface intersect orthogonally.