Example 1.25. Form the equation of the plane passing through the point $M(2 ; 3 ; 5)$ and perpendicular to the vector $\bar{N}=4 \bar{i}+3 \bar{j}+2 \bar{k}$
Solution. It is sufficient to use the equation of the plane passing through the given point and perpendicular to the given vector:

$$
4(x-2)+3(y-3)+2(z-5)=0, \quad \text { i.e. } \quad 4 x+3 y+2 z-27=0 .
$$

## Line.

1. A line can be defined by the equations of two - intersecting planes

$$
\begin{aligned}
& A_{1} x+B_{1} y+C_{1} z+D_{1}=0 \\
& A_{2} x+B_{2} y+C_{2} z+D_{2}=0,
\end{aligned}
$$

intersecting along this line.

2. By sequentially eliminating $x$ and $y$ from the previous equations, we obtain the equations $x=a z+c, y=b z+d$. Here, the line is defined by two planes projecting it onto the $x o z$ and $yoz$ planes.
3. If two points $M\left(x_{1}, y_{1}, z_{1}\right)$ and $N\left(x_{2}, y_{2}, z_{2}\right)$ are given, then the equations of the line passing through them will be of the form:

$$
\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}
$$

4. The so-called canonical equations $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=$ $=\frac{z-z_{1}}{n}$ define a line passing through the point $M\left(x_{1}, y_{1}, z_{1}\right)$ and parallel to the vector $\bar{S}=l \bar{i}+m \bar{j}+n \bar{k}$. In particular, these equations can be written in the form:

$$
\frac{x-x_{1}}{\cos \alpha}=\frac{y-y_{1}}{\cos \beta}=\frac{z-z_{1}}{\cos \gamma}
$$

where $\alpha, \beta$ and $\gamma$ are the angles formed by the line with the coordinate axes.

The direction cosines of the line are found using the formulas

$$
\cos \alpha=\frac{l}{\sqrt{l^{2}+m^{2}+n^{2}}}, \cos \beta=\frac{m}{\sqrt{l^{2}+m^{2}+n^{2}}}, \cos \gamma=\frac{n}{\sqrt{l^{2}+m^{2}+n^{2}}} .
$$

5. From the canonical equations of the line, it is not difficult to transition to the parametric equations of the line by introducing a parameter $t$:

$$
\begin{aligned}
& x=l t+x_{1} \\
& y=m t+y_{1} . \\
& z=n t+z_{1}
\end{aligned}
$$

6. The angle between two lines given by their canonical equations $\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}$ and $\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}}$ is determined by the formula

$$
\cos \phi=\frac{l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \cdot \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}
$$

the condition for the parallelism of two lines:

$$
\frac{l_{1}}{l_{2}}=\frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}
$$

the condition for the perpendicularity of two lines:

$$
l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0
$$

7. The necessary and sufficient condition for the coplanarity of two lines given by their canonical equations:

$$
\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
l_{1} & m_{1} & n_{1} \\
l_{2} & m_{2} & n_{2}
\end{array}\right|=0
$$

If the values $l_{1}, m_{1}, n_{1}$ are not proportional to the values $l_{2}, m_{2}$, $n_{2}$, then the specified relation is the necessary and sufficient condition for the intersection of two lines in space.

8. The angle of intersection of a line $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ with a plane $A x+B y+C z+D=0$ is determined by the formula

$$
\sin \phi=\frac{A l+B m+C n}{\sqrt{A^{2}+B^{2}+C^{2}} \sqrt{l^{2}+m^{2}+n^{2}}}
$$

the condition for the parallelism of a line and a plane:

$$
A l+B m+C n=0,
$$

the condition for the perpendicularity of a line and a plane:

$$
\frac{A}{l}=\frac{B}{m}=\frac{C}{n}
$$

9. To determine the point of intersection of a line $\frac{x-x_{0}}{l}=\frac{y-y_{0}}{m}=$ $=\frac{z-z_{0}}{n}$ with a plane $A x+B y+C z+D=0$, it is necessary to solve their equations simultaneously, for which one should use the parametric equations of the line $x=l t+x_{0}, y=m t+y_{0}, z=n t+z_{0}$ :

a) if $A l+B m+C n \neq 0$, then the line intersects the plane at one point

b) if $A l+B m+C n=0$ and $A x_{0}+B y_{0}+C z_{0}+D \neq 0$, then the line is parallel to the plane

c) if $A l+B m+C n=0$ and $A x_{0}+B y_{0}+C z_{0}+D=0$, then the line lies in the plane.