Example 3. Form the equations of the line passing through the point $A(-2,3,1)$ parallel to the line

$$
\left\{\begin{array}{l}
x-2 y-z-2=0 \\
2 x+3 y-z+1=0
\end{array}\right.
$$
Solution. To write the canonical equations of a line (point $2^{\circ}$), we need a direction vector, which we will determine according to point $6^{\circ}$ (see Example 2):

$$
\vec{l}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & -2 & -1 \\
2 & 3 & -1
\end{array}\right|=5 \vec{i}-\vec{j}+7 \vec{k}
$$

The desired equations are $\frac{x+2}{5}=\frac{y-3}{-1}=\frac{z-1}{7}$;

## Exercises

1. Write the equations of the line passing through the points $M_{1}(2,-1,-3)$ and $M_{2}(3,1,-5)$, and find its direction cosines.
2. Find the angle between the lines

$$
\left\{\begin{array} { l } 
{ x + y - z - 1 = 0 , } \\
{ x - y + 2 z + 1 = 0 }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
x-z+1=0 \\
y+2 z-1=0
\end{array}\right.\right.
$$

3. Write the equations of the line passing through the point $M_{0}(1,-3,2)$ and perpendicular to the lines

$$
\frac{x-2}{3}=\frac{y}{-2}=\frac{z}{1}, \quad \frac{x}{1}=\frac{y+1}{4}=\frac{z+3}{-5}
$$

4. Write the equations of the line passing through the point $M(3,2,-1)$ and intersecting the $O x$ axis at a right angle.
5. Calculate the distance between the lines

$$
\frac{x}{1}=\frac{y-3}{2}=\frac{z-2}{1} \quad \text { and } \quad \frac{x-3}{1}=\frac{y+1}{2}=\frac{z-2}{1}
$$

6. Find the equation of the line passing through the point $M(2,1,-3)$ and forming angles $\alpha=45^{\circ}, \beta=60^{\circ}, \gamma=120^{\circ}$ with the coordinate axes.
7. Given two vertices of a parallelogram $A B C D: C(-2,3,-5)$ and $D(0,4,-7)$, and the point of intersection of the diagonals $M(1,2,-3,5)$. Find the equations of the side $A B$.
8. Check if the three given points lie on the same line: $A(3,5,-5), B(4,6,-4), C(5,7,-3)$.

## Answers

1. $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+3}{-2} ; \quad \cos \alpha=\frac{1}{3}, \quad \cos \beta=\frac{2}{3}, \quad \cos \gamma=-\frac{2}{3}$. 2. $\arccos (-0.5456)$. 3. $\frac{x-1}{3}=\frac{y+3}{8}=\frac{z-2}{7}$. 4. $\frac{x-3}{0}=\frac{y-2}{-2}=\frac{z+1}{1}$. 5. $\frac{5 \sqrt{30}}{6}$. 6. $\frac{x-2}{\sqrt{2}}=\frac{y-1}{1}=\frac{z+3}{-3}$. 7. $\frac{x-4}{2}=\frac{y-1}{1}=\frac{z+2}{-2}$. 8. Yes.

## § 3. Plane and Line in Space

$1^{\circ}$. The angle between a line and a plane is the angle $\varphi\left(0 \leqslant \varphi \leqslant \frac{\pi}{2}\right)$ between the line and its projection on the plane.

Let $\vec{l}\{m, n, p\}$ be the direction vector of the line, and $\vec{N}\{A, B, C\}$ be the normal vector of the plane. Then (Fig. 4.10)

$$
\begin{gathered}
\cos \left(\frac{\pi}{2}-\varphi\right)=\sin \varphi=\frac{\vec{N} \cdot \vec{l}}{|\vec{N}| \cdot|\vec{l}|}= \\
\quad=\frac{A m+B n+C p}{\sqrt{m^{2}+n^{2}+p^{2}} \cdot \sqrt{A^{2}+B^{2}+C^{2}}}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-045.jpg?height=323&width=483&top_left_y=1190&top_left_x=1979)

Fig. 4.10
The condition for the perpendicularity of a line and a plane coincides with the condition for the parallelism of vectors $\vec{l}$ and $\vec{N}: \frac{A}{m}=\frac{B}{n}=\frac{C}{p}$.

The condition for the parallelism of a line and a plane coincides with the condition for the perpendicularity of vectors $\vec{l}$ and $\vec{N}: A m+B n+C p=0$.

$2^{\circ}$. The coordinates of the intersection point of the line $x=x_{0}+m t, y=y_{0}+n t$, $z=z_{0}+p t$ with the plane $A x+B y+C z+D=0$ are determined by substituting the parametric equations of the line into the equation of the plane, finding the value of the parameter $t$, and substituting this value into the parametric equations of the line.
$3^{\circ}$. The coordinates of the intersection point of three planes are determined by solving the system of equations of these planes:

$$
\left\{\begin{array}{l}
A_{1} x+B_{1} x+C_{1} z+D_{1}=0 \\
A_{2} x+B_{2} y+C_{2} z+D_{2}=0 \\
A_{3} x+B_{3} y+C_{3} z+D_{3}=0
\end{array}\right.
$$

## Examples with Solutions