Example 1.33. Given a line

\[
\begin{aligned}
3 x-2 y-z+4 & =0 \\
x-4 y-3 z-2 & =0
\end{aligned}
\]

Find its projection on the plane

\[
5 x+2 y+2 z-7=0 .
\]
Solution. We need to find a plane that passes through the given line and is perpendicular to the given plane; then the desired projection will be determined as the intersection of this plane with the given one.

Let's form the equation of the pencil of planes passing through the given line:

$$
\begin{aligned}
& (3 x-2 y-z+4)+\lambda(x-4 y-3 z-2)=0 \text { or } \\
& (3+\lambda) x-(2+4 \lambda) y-(1+3 \lambda) z+(4-2 \lambda)=0 .
\end{aligned}
$$

This plane must be perpendicular to the given plane, which can be written as:

$$
5(3+\lambda)-2(2+4 \lambda)-2(1+3 \lambda)=0
$$

from which $\lambda=1$.

Then the equation of the plane passing through the given line and perpendicular to the given plane will be:

$$
\begin{gathered}
(3+1) x-(2+4 \cdot 1) y-(1+3 \cdot 1) z+(4-2 \cdot 1)=0 \text { or } \\
2 x-3 y-2 z+1=0 .
\end{gathered}
$$

The projection of the given line onto the given plane is determined as the line of intersection of the planes:

$$
\begin{gathered}
2 x-3 y-2 z+1=0 \\
5 x+2 y+2 z-7=0
\end{gathered}
$$

Let's write this line in canonical form. Find any point on the line. For this, let's set, for example, $x_{0}=1$, and the system will be written as:

$$
\begin{aligned}
3 y_{0}+2 z_{0} & =3, \\
y_{0}+z_{0} & =1 .
\end{aligned}
$$

From this, $y_{0}=1, z_{0}=0$, i.e., the point $M(1 ; 1 ; 0)$ belongs to the desired line.

The direction vector of the line $\bar{S}=(l ; m ; n)$ will be found from the condition that it is perpendicular to the normal vectors $\bar{N}_{1}=(2 ;-3 ;-2)$ and $\bar{N}_{2}=(5 ; 2 ; 2)$ of the planes defining the desired line.

As $\bar{S}$, we take the vector product of vectors $\bar{N}_{1}$ and $\bar{N}_{2}$, i.e.,

$$
\bar{S}=\bar{N}_{1} \times \bar{N}_{2}=\left|\begin{array}{rrr}
\bar{i} & \bar{j} & \bar{k} \\
2 & -3 & -2 \\
5 & 2 & 2
\end{array}\right|=-2 \bar{i}-14 \bar{j}+19 \bar{k}=(-2 ;-14 ; 19)
$$

Then the desired equation in canonical form will be:

$$
\frac{x-1}{-2}=\frac{y-1}{-14}=\frac{z}{19} \text {. }
$$

## Section 2

## DIFFERENTIAL CALCULUS

## 2.1. Functions, Limit, Continuity

One of the fundamental concepts of mathematical analysis is the concept of the limit of a function.

Definition. A number $A$ is called the limit of the function $y=f(x)$ as $x \rightarrow a$, if for any positive arbitrarily small number $\varepsilon$ there exists $\delta(\varepsilon)>0$ such that for $0<|x-a|<\delta(\varepsilon)$ the inequality $|f(x)-A|<\varepsilon$ holds. In this case, we write $\lim f(x)=A$.

${ }^{x \rightarrow a}$ When calculating limits of functions, the following properties of limits are used:

1. $\lim _{x \rightarrow a} c=c$, where $c$ - const,
2. $\lim _{x \rightarrow a} x=a$.
3. $\lim _{x \rightarrow a} c f(x)=c \lim _{x \rightarrow a} f(x)$.
4. If $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} \phi(x)$ exist, then
a) $\lim _{x \rightarrow a}[f(x) \pm \phi(x)]=\lim _{x \rightarrow a} f(x) \pm \lim _{x \rightarrow a} \phi(x)$,

b) $\lim _{x \rightarrow a} f(x) \cdot \phi(x)=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} \phi(x)$,

c) $\lim _{x \rightarrow a} \frac{f(x)}{\phi(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} \phi(x)}$, if $\lim _{x \rightarrow a} \phi(x) \neq 0$,

d) $\lim _{x \rightarrow a}[f(x)]^{\phi(x)}=\lim _{x \rightarrow a} f(x)^{\lim _{x \rightarrow a} \phi(x)}$.

For all basic elementary functions, at any point in their domain of definition, the equality holds:

$$
\lim _{x \rightarrow a} f(x)=f \lim _{x \rightarrow a} x=f(a)
$$

that is, the limit of the function is found by direct substitution of the limiting value of the argument.