Find the family of surfaces normal to the vector lines of…

Find the family of surfaces normal to the vector lines of \[ \vec A = (2xy-3yz)\,\vec i+(x^2-3xz)\,\vec j-3xy\,\vec k. \]

Step 1 · Set up and combine the characteristic system

With $P=y(2x-3z)$, $Q=x(x-3z)$, $R=-3xy$, pairing ratios 1,3 and 2,3 and dividing by $-y$, $-x$ respectively gives \[ 3x\,dx+(2x-3z)\,dz=0 \quad(\mathrm A), \qquad 3y\,dy+(x-3z)\,dz=0 \quad(\mathrm B). \]

Step 2 · First integral from $2(\mathrm B)-(\mathrm A)$

\[ 6y\,dy-3x\,dx-3z\,dz=0 \implies d(2y^2-x^2-z^2)=0 \implies u_1 = 2y^2-x^2-z^2 = C_1. \]

Step 3 · Second integral from equation (A)

Equation (A) alone, $3x\,dx+(2x-3z)\,dz=0$, is homogeneous in $x,z$: with $z=vx$, \[ x\frac{dv}{dx} = -\frac{3v^2-2v-3}{3v-2}, \qquad \frac{(3v-2)\,dv}{3v^2-2v-3} = -\frac{dx}{x}. \] The quadratic $3v^2-2v-3=0$ has roots $\mu_{1,2}=\dfrac{1\mp\sqrt{10}}{3}$. Partial fractions give exponents $a=\dfrac{10+\sqrt{10}}{20}$, $b=\dfrac{10-\sqrt{10}}{20}$ (with $a+b=1$), and integrating and translating back from $v=z/x$ to $x,z$ (the homogeneity in $a+b=1$ cancels the leftover power of $x$): \[ u_2 = (z-\mu_1x)^{a}(z-\mu_2x)^{b} = C_2. \]

Verification of $u_1$

\[ du_1 = 4y\,dy-2x\,dx-2z\,dz = \tfrac23\bigl[2\cdot3y\,dy-3x\,dx-3z\,dz\bigr]. \] Substituting $3y\,dy=(3z-x)\,dz$ from (B) and $3x\,dx=(3z-2x)\,dz$ from (A): \[ du_1 = \tfrac23\bigl[2(3z-x)-(3z-2x)-3z\bigr]dz = \tfrac23\cdot0\,dz = 0. \checkmark \]

Verification of $u_2$ (numerical check)

At $(x,z)=(2,5)$, direct computation confirms $\partial_xu_2/(3x) = \partial_zu_2/(2x-3z)$ to high precision, confirming $u_2$ solves equation (A). \checkmark