Example. Find the coordinates of the projection $P^{\prime}$ of the point $P(1,2,-1)$ onto the plane $3 x-y+2 z-4=0$.
SOLUTION.

1. We compose the equations of the line passing through 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: $\vec{a}=\vec{n}=$
$=\{3,-1,2\}$. Then the canonical equations of the line are

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

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

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

Then the parametric equations of the line are

$$
\left\{\begin{array}{l}
x=3 t+1 \\
y=-t+2 \\
z=2 t-1
\end{array}\right.
$$

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 intersects the plane:

$$
3(3 t+1)-1(-t+2)+2(2 t-1)-27=0 \Longrightarrow t_{0}=2
$$

4. Substituting the found value $t_{0}=2$ into the parametric equations of the line, we get $x_{0}=7, y_{0}=0, z_{0}=1$.

Thus, the intersection point of the line and the plane, and therefore the projection of point $P$ onto the plane, has coordinates $(7,0,1)$.

Answer. The projection $P^{\prime}$ has coordinates $(7,0,1)$.

Conditions of the problems. Find the coordinates of the projection of point $P$ onto the plane.

1. $P(1,0,1)$,
2. $P(-1,0,-1)$,
3. $P(2,1,0)$,
4. $P(0,2,1)$,
5. $P(-1,2,0)$,
6. $P(2,-1,1)$,
7. $P(1,1,1)$,
8. $P(1,2,3)$,
9. $P(0,-3,-2)$,
10. $P(1,0,-1)$, $4 x+6 y+4 z-25=0$.

$2 x+6 y-2 z+11=0$.

$y+z+2=0$.

$2 x+4 y-3=0$.

$4 x-5 y-z-7=0$.

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

$x+4 y+3 z+5=0$.

$2 x+10 y+10 z-1=0$.

$2 x+10 y+10 z-1=0$.

$2 y+4 z-1=0$.

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

### 1.13. Symmetry with Respect to a Line or Plane

Problem Statement. Find the coordinates of point $Q$, symmetric to point $P\left(x_{P}, y_{P}, z_{P}\right)$ with respect to the line

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

Plan of Solution. The desired point $Q$ lies on the line perpendicular to the given one and intersecting it at point $P^{\prime}$. Since point $P^{\prime}$ divides the segment $P Q$ in half, the coordinates $x_{Q}, y_{Q}$ and $z_{Q}$ of point $Q$ are determined from the conditions

$$
x_{P^{\prime}}=\frac{x_{P}+x_{Q}}{2}, \quad y_{P^{\prime}}=\frac{y_{P}+y_{Q}}{2}, \quad z_{P^{\prime}}=\frac{z_{P}+z_{Q}}{2}
$$

where $x_{P}, y_{P}, z_{P}$ are the coordinates of point $P$ and $x_{P^{\prime}}, y_{P^{\prime}}, z_{P^{\prime}}$ are the coordinates of its projection $P^{\prime}$ onto the given line.

1. We find the projection of point $P$ onto the given line, i.e., point $P^{\prime}$ (see problem 1.12). For this:

a) we compose the equation of the plane passing through point $P$ and perpendicular to the given line. As the normal vector $\vec{n}$ of this plane, we can take the direction vector of the given line, i.e., $\vec{n}=\vec{a}=\{l, m, n\}$. We get

$$
l\left(x-x_{P}\right)+m\left(y-y_{P}\right)+n\left(z-z_{P}\right)=0
$$

b) we find the coordinates of the intersection point $P^{\prime}$ of this plane with the given line. For this, we write the equations of the line in parametric form

$$
\left\{\begin{array}{l}
x=l t+x_{0} \\
y=m t+y_{0} \\
z=n t+z_{0}
\end{array}\right.
$$

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 intersects the plane;

c) we substitute the found value $t_{0}$ into the parametric equations of the line and obtain the desired coordinates of point $P^{\prime}$.

2. The coordinates of point $Q$, symmetric to point $P$ with respect to the given line, are determined from the conditions (1). We get

$$
x_{Q}=2 x_{P^{\prime}}-x_{P}, \quad y_{Q}=2 y_{P^{\prime}}-y_{P}, \quad z_{Q}=2 z_{P^{\prime}}-z_{P}
$$

Remark. The problem of finding the coordinates of a point symmetric to a given one with respect to a plane is solved similarly.