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

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

Step 1 · Set up and diagonalise

\[ \frac{dx}{x}=\frac{dy}{z}=\frac{dz}{y+2z}=dt \ \Longleftrightarrow\ \dot{\mathbf r}=A\mathbf r,\quad A=\begin{pmatrix}1&0&0\\0&0&1\\0&1&2\end{pmatrix}. \] Since the $(1,1)$ entry decouples, $\det(A-\lambda I)=(1-\lambda)(\lambda^2-2\lambda-1)=0$, giving $\lambda_1=1$, $\lambda_2=1-\sqrt2$, $\lambda_3=1+\sqrt2$, with eigenvectors $\mathbf v_1=(1,0,0)$, $\mathbf v_2=(0,1,1-\sqrt2)$, $\mathbf v_3=(0,1,1+\sqrt2)$.

Step 2 · Write $\xi_1=x$, $\xi_2,\xi_3$ in terms of $y,z$

With $\xi_i=c_ie^{\lambda_it}$: $x=\xi_1$, $y=\xi_2+\xi_3$, $z=(1-\sqrt2)\xi_2+(1+\sqrt2)\xi_3$. Solving the last two for $\xi_2,\xi_3$: \[ \xi_2 = \frac{(1+\sqrt2)y-z}{2\sqrt2}, \qquad \xi_3 = \frac{z-(1-\sqrt2)y}{2\sqrt2}. \]

Step 3 · First integral: $\lambda_2+\lambda_3=2\lambda_1$

Since $\lambda_2+\lambda_3=2=2\lambda_1$, the ratio $\xi_2\xi_3/\xi_1^2$ is independent of $t$. Substituting and simplifying the numerator ($[(1+\sqrt2)y-z][z-(1-\sqrt2)y]=y^2+2yz-z^2$): \[ u_1 = \frac{y^2+2yz-z^2}{x^2} = C_1. \]

Step 4 · Second integral: $\xi_1^{\lambda_2}/\xi_2$

Since $\xi_1^{\lambda_2}/\xi_2^{\lambda_1}=x^{1-\sqrt2}/\xi_2$ is independent of $t$: \[ u_2 = \frac{x^{1-\sqrt2}}{(1+\sqrt2)y-z} = C_2. \]

Verification

With $N=y^2+2yz-z^2$, $\nabla N=(0,2y+2z,2y-2z)$: \[ \vec A\cdot\nabla N = z(2y+2z)+(y+2z)(2y-2z) = 2N, \] so $\vec A\cdot\nabla u_1 = 2N/x^2-2N/x^2=0$. \checkmark For $u_2$, with $\xi=(1+\sqrt2)y-z$, direct computation of $\vec A\cdot\nabla u_2$ reduces (after factoring out $x^{1-\sqrt2}/\xi^2$) to \[ (1-\sqrt2)\xi-(1+\sqrt2)z+(y+2z) = 0 \checkmark \] after substituting $\xi=(1+\sqrt2)y-z$ and simplifying.