Example 6. Find the equation of the plane passing through the points $M_{1}(1,2,0), M_{2}(1,-1,2)$ and $M_{3}(0,1,-1)$.
Solution. According to point $4^{\circ}$, the equation of the desired plane is determined by the equality

$$
\left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
0 & -3 & 2 \\
-1 & -1 & -1
\end{array}\right|=0
$$

We expand the determinant (Chapter I) by the elements of the first row:

$$
(x-1) 5-(y-2) 2+z(-3)=0, \quad \text { or } \quad 5 x-2 y-3 z-1=0
$$

## Exercises

1. Form the equation of the plane that intercepts the segment $O B=5$ on the $O y$ axis and is perpendicular to the vector $\vec{n}=\{3,-2,4\}$.
2. Form the equation of the plane passing through the bases of the perpendiculars dropped from the point $A(2,3,-5)$ to the coordinate axes.

Hint. Use point $4^{\circ}$.

3. Write the equation of the plane if it is known that the point $A(2,1,-3)$ is the base of the perpendicular dropped from the origin to this plane.
4. Write the equation of the plane passing through the points $A(2,-1,4)$ and $B(3,2,-1)$ and perpendicular to the plane $x+y+2 z-3=0$.

Hint. Use the normal vector $\vec{N}=\overrightarrow{A B} \times \vec{n}$, where $\vec{n}=\{1,1,2\}$.

5. Find the length of the perpendicular dropped from the point $M(2,3,-5)$ to the plane $4 x-2 y+5 z-12=0$.
6. Find the angle between the plane $2 x+y-2 z-3=0$ and the coordinate plane $O y z$.
7. Find the distance between the parallel planes $x-y+z-1=0$ and $2 x-2 y+2 z-5=0$.

Hint. Take any point on one of the planes and apply the formula from point $7^{\circ}$.

8. Write the equation of the plane passing through the point $A(1,-1,1)$ and perpendicular to the planes $2 x-y+z-1=0$ and $x+2 y-z+1=0$.

Hint. Use point $3^{\circ}$, where $\vec{N}=\vec{N}_{1} \times \vec{N}_{2}$.

9. Form the equation of the plane passing through the points $A(2,-3,2), B(-1,-2,-1), C(-3,1,4)$.

## Answers

1. $3 x-2 y+4 z-12=0$. 2. $15 x+10 y-6 z-60=0$. 3. $2 x+y-3 z-14=0$. 4. $11 x-7 y-2 z-21=0$. 5. $\frac{7 \sqrt{5}}{3}$. 6. $\cos \varphi=\frac{2}{3}, \varphi \approx 48^{\circ} 11^{\prime}$. 7. $\frac{\sqrt{3}}{2}$. 8. $x-3 y-5 z+1=0$. 9. $2 x+3 y-z+7=0$.

## § 2. Line in Space

$1^{\circ}$. A line in space can be defined as the intersection of two planes. The system of equations

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

defines the general equations of the line.

$2^{\circ}$. The canonical equations of a line

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

define a line passing through the point $M_{0}\left(x_{0}, y_{0}, z_{0}\right)$ parallel to the vector $\vec{l}=\{m, n, p\}$, which is called the direction vector of the line (Fig. 4.7).

$3^{\circ}$. The parametric equations of a line are given by $x=x_{0}+m t$, $y=y_{0}+n t, z=z_{0}+p t$, where the parameter $t$ varies in the interval $-\infty<t<\infty$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-044.jpg?height=134&width=317&top_left_y=913&top_left_x=103)

Fig. 4.7

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-044.jpg?height=398&width=366&top_left_y=1117&top_left_x=78)

Fig. 4.8

$4^{\circ}$. The equations of a line passing through two given points $M_{1}\left(x_{1}, y_{1}, z_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}, z_{2}\right)$ are

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

(if the denominator of any fraction is zero, then its numerator is also zero).

$5^{\circ}$. To convert the general equations of a line to canonical form, one should:

- take two points on the line, for which one variable should be assigned two numerical values and solve the system of equations for the other variables (or take two values of the parameter $t$);
- write the equations of the line passing through two points (point $4^{\circ}$).

$6^{\circ}$. The direction vector $\vec{l}$ of a line given by general equations (Fig. 4.8)

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

is given by: $\vec{l}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2}\end{array}\right|$, which is the vector product of the normal vectors $\vec{N}_{1}=\left\{A_{1}, B_{1}, C_{1}\right\}$ and $\vec{N}_{2}=\left\{A_{2}, B_{2}, C_{2}\right\}$.

$7^{\circ}$. The angle between two skew lines, given by canonical equations

$$
\frac{x-x_{1}}{m_{1}}=\frac{y-y_{1}}{n_{1}}=\frac{z-z_{1}}{p_{1}} \quad \text { and } \quad \frac{x-x_{2}}{m_{2}}=\frac{y-y_{2}}{n_{2}}=\frac{z-z_{2}}{p_{2}}
$$

is the angle $\varphi$ (Fig. 4.9) between the direction vectors of these lines (see Chapter III, § 2). This angle can be determined using the cosine:

$$
\cos \varphi=\frac{\vec{l}_{1} \vec{l}_{2}}{\left|\vec{l}_{1}\right|\left|\vec{l}_{2}\right|}=\frac{m_{1} m_{2}+n_{1} n_{2}+p_{1} p_{2}}{\sqrt{m_{1}^{2}+n_{1}^{2}+p_{1}^{2}} \sqrt{m_{2}^{2}+n_{2}^{2}+p_{2}^{2}}}
$$

The condition for the lines to be perpendicular is the same as the condition for their direction vectors to be perpendicular:

$$
\vec{l}_{1} \vec{l}_{2}=0, \quad \text { or } \quad m_{1} m_{2}+n_{1} n_{2}+p_{1} p_{2}=0
$$

The condition for the lines to be parallel is the same as the condition for their direction vectors to be collinear:

$$
\overrightarrow{l_{1}}=\lambda \overrightarrow{l_{2}}, \quad \text { or } \quad \frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}=\frac{p_{1}}{p_{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-044.jpg?height=343&width=291&top_left_y=622&top_left_x=2189)

Fig. 4.9

## Examples with Solutions