Example. Find the angle between the planes

$$
x+2 y-2 z-7=0, \quad x+y-35=0
$$
Solution. The dihedral angle between the planes is equal to the angle between their normal vectors $\vec{n}_{1}=\{1,2,-2\}$ and $\vec{n}_{2}=\{1,1,0\}$. Therefore, the angle $\varphi$ between the planes is determined by the equality

$$
\cos \varphi=\frac{\left(\vec{n}_{1}, \vec{n}_{2}\right)}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=\frac{1 \cdot 1+2 \cdot 1-2 \cdot 0}{\sqrt{1^{2}+2^{2}+(-2)^{2}} \sqrt{1^{2}+1^{2}}}=\frac{1}{\sqrt{2}}
$$

Thus, $\varphi=\arccos (1 / \sqrt{2})=\pi / 4$.

Answer. The angle between the planes $\varphi=\pi / 4$.

Problem Conditions. Find the angle between the planes.

1. $3 x-y+3=0$,
2. $x-y+3 z-5=0$,
3. $5 x-4 y+3 z-3=0$,
4. $5 x-3 y+2 z+5=0$,
5. $6 x+2 y-4 z+17=0$,
6. $x-y+z \sqrt{2}-5=0$,
7. $y-3 z+5=0$,
8. $6 x+2 y-3 z+1=0$,
9. $2 x+y+2 z-5=0$,
10. $5 x-y+2 z+12=0$, $x-2 y+5 z-10=0$.

$x+z-2=0$.

$4 x-y-z+2=0$.

$3 x+3 y-3 z-8=0$.

$9 x+3 y-6 z-4=0$.

$x+y-z \sqrt{2}+7=0$.

$y+2 z-3=0$.

$x+6 y+2 z-10=0$.

$12 x+16 y-15 z+2=0$.

$3 x+2 y+z+10=0$.

Answers. 1. $\varphi=\arccos \frac{1}{2 \sqrt{3}}$. 2. $\varphi=\arccos \sqrt{\frac{8}{11}}$. 3. $\varphi=\arccos \frac{7}{10}$. 4. $\varphi=\frac{\pi}{2}$. 5. $\varphi=0$. 6. $\varphi=\frac{2 \pi}{3}$. 7. $\varphi=\frac{\pi}{4}$. 8. $\varphi=\frac{\pi}{2}$. 9. $\varphi=\arccos \frac{2}{15}$. 10. $\varphi=\arccos \sqrt{\frac{15}{28}}$.

### 1.10. Canonical Equations of a Line

Problem Statement. Write the canonical equations of the line defined as the intersection of two planes (by general equations)

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

Solution Plan.

1. Check that the vectors $\vec{n}_{1}=\left\{A_{1}, B_{1}, C_{1}\right\}$ and $\vec{n}_{2}=\left\{A_{2}, B_{2}, C_{2}\right\}$ are non-collinear and, therefore, the planes intersect along some line.

The canonical equations of a line with a direction vector $\vec{a}=\{l, m, n\}$ passing through a given point $M_{0}\left(x_{0}, y_{0}, z_{0}\right)$ are of the form

$$
\frac{x-x_{0}}{l}=\frac{y-y_{0}}{m}=\frac{z-z_{0}}{n}
$$

Therefore, to write the equations of the line, it is necessary to find its direction vector and some point on the 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., $\vec{a} \perp \vec{n}_{1}=\left\{A_{1}, B_{1}, C_{1}\right\}$ and $\vec{a} \perp \vec{n}_{2}=\left\{A_{2}, B_{2}, C_{2}\right\}$. Therefore, the direction vector $\vec{a}$ is found using the formula

$$
\vec{a}=\left[\vec{n}_{1}, \vec{n}_{2}\right]=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
A_{1} & B_{1} & C_{1} \\
A_{2} & B_{2} & C_{2}
\end{array}\right|
$$

3. Now choose some point on the line. Since the direction vector of the line is not parallel to at least one of the coordinate planes, the line intersects this coordinate plane. Therefore, a point on the line can be taken as the point of intersection with this coordinate plane.
4. Substitute the found direction vector and point into the line equations (1) and write the answer.