Example. Write the canonical equations of the line given as the line of intersection of two planes (by general equations)

$$
\left\{\begin{array}{l}
2 x+3 y+z-8=0 \\
x-2 y-2 z+1=0
\end{array}\right.
$$
Solution.

1. Let's check that the vectors $\vec{n}_{1}=\{2,3,1\}$ and $\vec{n}_{2}=\{1,-2,-2\}$ are non-collinear (see problem 1.2). We have

$$
\frac{2}{1} \neq \frac{3}{-2}
$$

The vectors $\vec{n}_{1}=\{2,3,1\}$ and $\vec{n}_{2}=\{1,-2,-2\}$ are non-collinear, as their coordinates are not proportional. Therefore, the two planes intersect along a line.

2. Since the line belongs to both planes simultaneously, its direction vector $\vec{a}$ is orthogonal to the normal vectors of both planes, i.e., $\vec{a} \perp \vec{n}_{1}=\{2,3,1\}$ and $\vec{a} \perp \vec{n}_{2}=\{1,-2,-2\}$. Therefore, the direction vector $\vec{a}$ is found using the formula

$$
\vec{a}=\left[\vec{n}_{1}, \vec{n}_{2}\right]=\left|\begin{array}{rrr}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & 1 \\
1 & -2 & -2
\end{array}\right|=-4 \vec{i}+5 \vec{j}-7 \vec{k}
$$

3. Now let's choose any point on the line. Since the direction vector of the line is not parallel to any of the coordinate planes, the line intersects all three coordinate planes.

Therefore, as a point on the line, we can take the point of its intersection, for example, with the plane $y=0$. The coordinates of this point are found by solving the system of three equations

$$
\left\{\begin{array}{l}
2 x+z-8=0 \\
x-2 z+1=0 \\
y=0
\end{array}\right.
$$

We get $x_{0}=3, y_{0}=0$ and $z_{0}=2$, i.e., $M_{0}(3,0,2)$.

4. Substituting the found direction vector and point into the equations of the line (1), we get

$$
\frac{x-3}{-4}=\frac{y}{5}=\frac{z-2}{-7} .
$$

Answer. The canonical equations of the line are

$$
\frac{x-3}{-4}=\frac{y}{5}=\frac{z-2}{-7} .
$$

Problem conditions. Write the canonical equations of the line defined as the intersection of two planes.

1. $\left\{\begin{array}{l}x+y+z-2=0, \\ x-y-3 z+6=0 .\end{array}\right.$
2. $\left\{\begin{array}{l}2 x-3 y+2 z+2=0, \\ 2 x+3 y+z+14=0 .\end{array}\right.$
3. $\left\{\begin{array}{l}x-2 y+2 z-4=0, \\ 2 x+2 y-2 z-8=0 .\end{array}\right.$
4. $\left\{\begin{array}{l}x+y+z-2=0, \\ x-y-3 z+2=0 .\end{array}\right.$
5. $\left\{\begin{array}{l}2 x+3 y+z+3=0, \\ x-3 y-2 z+3=0 .\end{array}\right.$
6. $\left\{\begin{array}{l}x+y-z-4=0, \\ x-y+2 z=0\end{array}\right.$
7. $\left\{\begin{array}{l}x+5 y+2 z+5=0, \\ x-y-z-1=0 .\end{array}\right.$
8. $\left\{\begin{array}{l}2 x+2 y-2 z+1=0, \\ 3 x-2 y+3 z+4=0 .\end{array}\right.$
9. $\left\{\begin{array}{l}4 x+y-3 z+4=0, \\ 2 x-y+2 z+2=0 .\end{array}\right.$
10. $\left\{\begin{array}{l}x-y-z-2=0, \\ x-3 y+z+4=0 .\end{array}\right.$

Answers.

1. $\frac{x+2}{1}=\frac{y-4}{-2}=\frac{z}{1}$.
2. $\frac{x+4}{9}=\frac{y+2}{-2}=\frac{z}{-12}$.
3. $\frac{x-4}{0}=\frac{y}{1}=\frac{z}{1}$.
4. $\frac{x}{1}=\frac{y-2}{-2}=\frac{z}{1}$.
5. $\frac{x+2}{3}=\frac{y}{-5}=\frac{z}{9}$.
6. $\frac{x-2}{1}=\frac{y-3}{-3}=\frac{z}{-2}$.
7. $\frac{x}{7}=\frac{y+1}{-1}=\frac{z}{2}$.
8. $\frac{x+1}{1}=\frac{y-1 / 2}{-6}=\frac{z}{-5}$.
9. $\frac{x+1}{1}=\frac{y}{14}=\frac{z}{6}$.
10. $\frac{x-5}{2}=\frac{y-3}{1}=\frac{z}{1}$.

### 1.11. Point of Intersection

## of a Line and a Plane

PROBLEM STATEMENT. Find the point of intersection of the line

$$
\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}
$$

and the plane

$$
A x+B y+C z+D=0
$$

SOLUTION PLAN.

1. Check that the line is not parallel to the plane. This means that the direction vector of the line $\vec{a}=\{l, m, n\}$ and the normal vector of the plane $\vec{n}=\{A, B, C\}$ are not orthogonal, i.e., their dot product is not zero:

$$
A l+B m+C n \neq 0
$$

In this case, there is a unique point of intersection of the line and the plane.

2. To find the point of intersection of the line and the plane, in general, we need to solve a system of three equations with three unknowns (two equations of the line and one equation of the plane). However, it is more convenient to use the parametric equations of the line.

Let

$$
\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}=t
$$

Then the parametric equations of the line are

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

3. Substituting these expressions for $x, y$ and $z$ into the equation of the plane and solving it for $t$, we find the value of the parameter $t=t_{0}$ at which the line and the plane intersect.
4. Substituting the found value $t_{0}$ into the parametric equations of the line, we get the desired coordinates of the point of intersection:

$$
\left\{\begin{array}{l}
x_{0}=l t_{0}+x_{1} \\
y_{0}=m t_{0}+y_{1} \\
z_{0}=n t_{0}+z_{1}
\end{array}\right.
$$

We write the answer in the form: the line and the plane intersect at the point $\left(x_{0}, y_{0}, z_{0}\right)$.