= Solution
Use a parameter $t$. The characteristic equations are
$$
\boxed{\dot x=x+cy-bz,\quad\dot y=y+az-cx,\quad\dot z=z+bx-ay.}
$$
For a <graph of a function> $z=f(x,y)$ satisfying the equation, solve the first two equations after substituting $z=f(x,y)$. The <chain rule> and the <partial differential equation> give $d f(x(t),y(t))/dt=f+bx-ay$, exactly the third equation. Uniqueness for the resulting <linear ordinary differential equation> system shows that its characteristic through the initial point agrees with this curve on the <graph of a function>. This proves invariance as long as the <graph of a function> remains defined along the curve. A claim about the entire curve additionally requires a <surface> continued along that flow; a restricted <graph of a function> patch alone cannot imply global continuation.
Let $\mathbf k=(a,b,c)$, $\boldsymbol\ell=(l,m,n)$ and $\mathbf r=(x,y,z)$. Then $\dot{\mathbf r}=\mathbf r-\mathbf k\times\mathbf r$. Since $|\mathbf k|=1$,
$$
\frac d{dt}(\mathbf k\cdot\mathbf r)=\mathbf k\cdot\mathbf r,\qquad\frac d{dt}|\mathbf r|^2=2|\mathbf r|^2.
$$
Along the characteristic starting from $s\boldsymbol\ell$ these give $\mathbf k\cdot\mathbf r=e^ts(\mathbf k\cdot\boldsymbol\ell)$ and $|\mathbf r|=e^t|s|$. Consequently the characteristic <surface> generated by the given line satisfies
$$
\boxed{|ax+by+cz|=|al+bm+cn|\sqrt{x^2+y^2+z^2}.}
$$
To see that the whole displayed <cone> is generated, not just contained in it, decompose $\boldsymbol\ell=\kappa\mathbf k+\mathbf v$ with $\mathbf v\perp\mathbf k$, $\kappa=\mathbf k\cdot\boldsymbol\ell$. The full solution is
$$
\mathbf r(s,t)=se^t\bigl(\kappa\mathbf k+\cos t\,\mathbf v-\sin t\,(\mathbf k\times\mathbf v)\bigr).
$$
When $0<|\kappa|<1$, variation of $t$ gives all azimuths about the axis, while the freely chosen $s$ supplies every positive radius on both halves. This is precisely the two-sided circular <cone>, with
$$
\boxed{\text{vertex }(0,0,0),\quad\text{axis }\mathbb R(a,b,c),\quad\text{semi-angle }\arccos|al+bm+cn|.}
$$
The edge cases matter: $\kappa=0$ gives the <plane> $ax+by+cz=0$ (semi-angle $\pi/2$), and $|\kappa|=1$ generates only the axis (semi-angle zero), not a two-dimensional <surface>. In the latter case the initial line is itself characteristic, so it does not determine a unique integral <surface>. The nondegenerate <cone> is smooth away from its vertex and can be used as a <graph of a function> only on patches whose <tangent plane> is not vertical. Thus the printed geometric conclusion has its ordinary characteristic-<surface> meaning in the nondegenerate case, with these domain and degeneracy qualifications; it cannot be asserted for every global smooth <graph of a function> without them.
Back to article page
