414. The line $L$ passes through the point $(2 ; 1 ;-1)$ and the point of intersection of the line

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

with the plane

$$
x-y+z-1=0
$$

Find the angle formed by the line $L$ with the plane

$$
x+2 y-z+3=0
$$
Solution. The line (A) and the plane (B) intersect, since the condition of parallelism (14) is not satisfied, i.e., $2 \cdot 1 + (-1) \cdot (-1) + 1 \cdot 1 \neq 0$. Let's find the point of intersection of the line (A) with the plane (B). Write the equations of the line (A) in parametric form:

$$
\frac{x-3}{2}=t, \frac{y}{-1}=t, z+1=t
$$

from which

$$
x=2 t+3, y=-t, z=t-1
$$

Substitute the obtained values of $x, y, z$ into the equation (B) and find $t$:

$$
2 t+3-(-t)+t-1-1=0 \text{ or } 4 t+1=0
$$

from which $t=-\frac{1}{4}$.

Therefore, $x=\frac{5}{2}, y=\frac{1}{4}, z=-\frac{5}{4}$.

Now write the equations of the line $L$ passing through the points $(2 ; 1 ;-1)$ and $\left(\frac{5}{2} ; \frac{1}{4} ;-\frac{5}{4}\right)$:

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

or

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

or

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

The angle between the line $L$ and the plane (C) will be found using formula (12):

$$
\begin{gathered}
\sin \varphi=\frac{|1 \cdot 2+2(-3)+(-1)(-1)|}{\sqrt{1^{2}+2^{2}+(-1)^{2}} \sqrt{\overline{2}^{2}+(-3)^{2}+(-1)^{2}}}= \\
=\frac{3}{\sqrt{6} \sqrt{14}}=\frac{\sqrt{21}}{14} \approx 0.3274
\end{gathered}
$$

From trigonometric tables, we find $\varphi \approx 19^{\circ} 7^{\prime}$. 415. Find the distance from the point $M(2 ; 3 ; 1)$ to the line

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

Solution. Through the given point $M$, draw a plane $P$ perpendicular to the given line. Then find the point of intersection $N$ of this plane with the line (A). The segment $M N$ will lie in the plane $P$, perpendicular to the line (A), hence it will serve as the perpendicular dropped from the point $M$ to the given line, with the point $N$ being the foot of the perpendicular. The shortest distance from the point $M$ to the given line will be the length of the perpendicular $M N$, dropped from the point $M$ to the line (A).

Write the equation of the plane passing through the point $M$:

$$
A(x-2)+B(y-3)+C(z-1)=0
$$

Determine the coefficients $A, B, C$ so that the plane (B) is perpendicular to the line (A). Use the condition of perpendicularity of a line to a plane (15):

$$
\frac{A}{1}=\frac{B}{-3}=\frac{C}{-2}
$$

We can set $A=1, B=-3, C=-2$. Substituting these values into the equation (B), we get the equation of the plane $P$:

$$
x-2-3(y-3)-2(z-1)=0
$$

or

$$
x-3 y-2 z+9=0
$$

Now find the point of intersection of this plane with the given line. Write the equations (A) in parametric form:

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

from which

$$
x=t-5 ; y=-3 t+4 ; z=-2 t+3
$$

Substituting these values into the equation of the plane ( $\mathrm{B}^{\prime}$ ), we get

$$
t-5-3(4-3 t)-2(3-2 t)+9=0
$$

or

$$
14 t-14=0
$$

from which $t=1$.

Therefore, the point of intersection of the plane ( $\mathrm{B}^{\prime}$ ) and the line ( $\mathrm{A}$ ) will have coordinates

$$
x=1-5=-4 ; y=-3+4=1 ; z=-2+3=1
$$

Now find the distance between the two points $M(2 ; 3 ; 1)$ and $N(-4 ; 1 ; 1)$ using formula (1) § 1 ch. III:

$M N=\sqrt{(2+4)^{2}+(3-1)^{2}+(1-1)^{2}}=\sqrt{40}=2 \sqrt{10}$.