Example. Find the point of intersection of the line

$$
\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{-1}
$$

and the plane

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

1. We have

$$
(\vec{a}, \vec{n})=2 \cdot 2+0 \cdot(-3)+(-1) \cdot 1=3 \neq 0
$$

Therefore, the direction vector of the line and the normal vector of the plane are not orthogonal, i.e., the line and the plane intersect at a unique point.

2. Let

$$
\frac{x-1}{2}=\frac{y+1}{0}=\frac{z}{-1}=t .
$$

Then the parametric equations of the line are

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

3. Substituting these expressions for $x, y$ and $z$ into the equation of the plane, we find the value of the parameter $t$ at which the line and the plane intersect:

$$
2(2 t+1)-3(-1)+1(-t)-8=0 \Longrightarrow t_{0}=1
$$

4. Substituting the found value $t_{0}=1$ into the parametric equations of the line, we get

$$
x_{0}=3, \quad y_{0}=-1, \quad z_{0}=-1
$$

Answer. The line and the plane intersect at the point $(3,-1,-1)$.

Conditions of the Problem. Find the point of intersection of the line and the plane.

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

Answers. 1. $(1,2,3)$. 2. $(1,-1,4)$. 3. $(2,-1,3)$. 4. $(4,0,-1)$. 5. $(1,1,2)$. 6. $(-1,0,-4)$. 7. $(-1,1,-2)$. 8. $(3,2,1)$. 9. $(-1,-1,-3)$. 10. $(-1,2,-1)$.

### 1.12. Projection of a Point

## onto a Plane or a Line

Problem Statement. Find the coordinates of the projection $P^{\prime}$ of the point $P\left(x_{P}, y_{P}, z_{P}\right)$ onto the plane $A x+B y+C z+D=0$.

Solution Plan. The projection $P^{\prime}$ of the point $P$ onto the plane is the base of the perpendicular dropped from the point $P$ to this plane.

1. We write the equations of the line passing through the point $P$ and perpendicular to the given plane. For this, we take the normal vector of the plane as the direction vector of the line: $\vec{a}=\vec{n}=\{A, B, C\}$. Then the canonical equations of the line are

$$
\frac{x-x_{P}}{A}=\frac{y-y_{P}}{B}=\frac{z-z_{P}}{C} .
$$

2. We find the coordinates of the point of intersection $P^{\prime}$ of this line with the given plane (see problem 1.11). Let

$$
\frac{x-x_{P}}{A}=\frac{y-y_{P}}{B}=\frac{z-z_{P}}{C}=t
$$

Then the parametric equations of the line are

$$
\left\{\begin{array}{l}
x=A t+x_{P} \\
y=B t+y_{P} \\
z=C t+z_{P}
\end{array}\right.
$$

3. Substituting $x, y, z$ 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 obtain the desired coordinates of the point $P^{\prime}$.

NOTE. The problem of finding the coordinates of the projection of a point onto a line is solved similarly.