## problem statement

Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.

$M(3 ;-3 ;-1)$

$2 x-4 y-4 z-13=0$
## Solution

Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:

$$
\vec{s}=\vec{n}=\{2 ;-4 ;-4\}
$$

Then the equation of the desired line is:

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

Let's find the point $M_{0 \text { of intersection of the line and the plane. }}$.

We write the parametric equations of the line.

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

$\left\{\begin{array}{l}x=3+2 t \\ y=-3-4 t \\ z=-1-4 t\end{array}\right.$

Substitute into the equation of the plane:

$2(3+2 t)-4(-3-4 t)-4(-1-4 t)-13=0$

$6+4 t+12+16 t+4+16 t-13=0$

$36 t+9=0$

$t=-0.25$

Find the coordinates of the point of intersection of the line and the plane:

$$
\left\{\begin{array}{l}
x=3+2 \cdot(-0.25)=2.5 \\
y=-3-4 \cdot(-0.25)=-2 \\
z=-1-4 \cdot(-0.25)=0
\end{array}\right.
$$

We get:

$M_{0}(2.5 ;-2 ; 0)$

Since $M_{0}$ is the midpoint of the segment $M M^{\prime}$, then

$$
\begin{aligned}
& x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2.5-3=2 \\
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-2)-(-3)=-1 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0-(-1)=1
\end{aligned}
$$

We get:

$M^{\prime}(2 ;-1 ; 1)$