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

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

Step 1 · Set up the linear system

\[ \frac{dx}{2x-y}=\frac{dy}{3y-z}=\frac{dz}{x-2y}=dt \quad\Longleftrightarrow\quad \dot{\mathbf r}=A\mathbf r,\qquad A=\begin{pmatrix}2&-1&0\\0&3&-1\\1&-2&0\end{pmatrix}. \]

Step 2 · Find the eigenvalues

\[ \det(A-\lambda I)=0 \implies \lambda^3-5\lambda^2+4\lambda+3=0. \] This cubic has no rational roots; sign changes locate three distinct real roots $\lambda_1\approx-0.461$, $\lambda_2\approx1.761$, $\lambda_3\approx3.700$.

Step 3 · Find the eigenvectors

For each $\lambda_i$, the eigenvector (up to scale) is \[ \mathbf v_i = \bigl(1,\ 2-\lambda_i,\ (3-\lambda_i)(2-\lambda_i)\bigr)^{\!\top}, \] using rows 1 and 2 of $(A-\lambda_iI)\mathbf v=0$; row 3 is automatically satisfied since $\lambda_i$ solves the characteristic cubic.

Step 4 · Diagonalise and form the first integrals

Writing $\xi_i=c_ie^{\lambda_it}$, the general solution is $x=\sum\xi_i$, $y=\sum(2-\lambda_i)\xi_i$, $z=\sum(3-\lambda_i)(2-\lambda_i)\xi_i$; inverting this (non-singular, Vandermonde-type) linear system expresses each $\xi_i=p_ix+q_iy+r_iz$ as an explicit linear form in $x,y,z$. Since $\xi_i=c_ie^{\lambda_it}$, the combinations $\xi_1^{\lambda_2}/\xi_2^{\lambda_1}$ and $\xi_1^{\lambda_3}/\xi_3^{\lambda_1}$ are independent of $t$, giving two first integrals.

Verification

Along a field line, $\dot\xi_i=\lambda_i\xi_i$, so \[ \frac{d}{dt}\!\left(\xi_1^{\lambda_2}\xi_2^{-\lambda_1}\right) = \lambda_1\lambda_2\xi_1^{\lambda_2}\xi_2^{-\lambda_1} -\lambda_1\lambda_2\xi_1^{\lambda_2}\xi_2^{-\lambda_1} = 0, \checkmark \] and identically for the second combination with $\lambda_3$ in place of $\lambda_2$.