Example 1. Find the vector lines of the vector field

$$
\mathbf{a}=[\mathbf{c}, \mathbf{r}]
$$

where $\mathbf{c}$ - a constant vector.
Solution. We have

$$
\mathbf{c}=c_{1} \mathbf{1}+c_{2} \mathbf{j}+c_{3} \mathbf{k}, \quad \mathbf{r}=x \mathbf{1}+y \mathbf{j}+z \mathbf{k}
$$

so that

$$
\mathrm{x}=[\mathbf{c}, \mathrm{r}]=\left|\begin{array}{ccc}
1 & i & \mathbf{k} \\
c_{1} & c_{2} & c_{3} \\
x & y & z
\end{array}\right|=\left(c_{2} z-c_{3} y\right) \mathbf{1}+\left(c_{3} x-c_{1} z\right) j+\left(c_{1} y-c_{2} x\right) \mathbf{k}
$$

The differential equations of the vector lines are

$$
\frac{d x}{c_{2} z-c_{3} y}=\frac{d y}{c_{3} x-c_{1} z}=\frac{d z}{c_{1} y-c_{2} x}
$$

Multiply the numerator and denominator of the first fraction by $x$, the second by $y$, and the third by $z$, and add term by term. Using the property of proportions, we get

$$
\frac{d x}{c_{2} z-c_{3} y}=\frac{d y}{c_{3} x-c_{1} z}=\frac{d z}{c_{1} y-c_{3} x}=\frac{x d x+y d y+z d z}{0} \text{. }
$$

From this,

$$
x d x+y d y+z d z=0
$$

which means,

$$
x^{2}+y^{2}+z^{2}=A_{1}, \quad A_{1}=\text{ const }>0
$$

Now, multiply the numerator and denominator of the first fraction (3) by $c_{1}$, the second by $c_{2}$, and the third by $c_{3}$, and add term by term, we get

$$
\frac{d x}{c_{2} z-c_{3} y}=\frac{d y}{c_{3} x-c_{1} z}=\frac{d z}{c_{1} y-c_{2} x}=
$$

$$
=\frac{c_{1} d x+c_{2} d y+c_{3} d z}{0}
$$

from which

$$
c_{1} d x+c_{2} d y+c_{3} d x=0
$$

and, consequently,

$$
c_{1} x+c_{2} y+c_{3} z=A_{2}, \quad A_{2}=\text{ const. }
$$

The sought equations of the vector lines are

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-18.jpg?height=280&width=274&top_left_y=952&top_left_x=1388)

Fig. 14

$$
\left\{\begin{aligned}
x^{2}+y^{2}+z^{2} & =A_{1} \\
c_{1} x+c_{2} y+c_{1} z & =A_{2}
\end{aligned}\right.
$$

These equations show that the vector lines are obtained by the intersection of spheres with a common center at the origin with planes perpendicular to the vector $\mathbf{c}=c_{1} \mathbf{1}+c_{2} \mathbf{j}+c_{3} \mathbf{k}$. This implies that the vector lines are circles, the centers of which lie on a line passing through the origin in the direction of the vector $\mathbf{c}$. The planes of the circles are perpendicular to the specified line (Fig. 14).