I need some help with this problem. $P:\mathbb{R}^3\to\mathbb{R}^3$ the linear operator such that $u=P(v)$ is the orthogonal projection of v E $\mathbb{R}^3$ on plane $3x+2y+z=0$. Find $P(x,y,z)$
(sorry for bad language, English isn't my main language)
$(3,2,1)$ is normal to the plane
$\frac {(x,y,z)\cdot(3,2,1)}{\|(3,2,1)\|}$ gives the distance of a point in space from the plane.
Then we move from this from this point in space, our distance, in the direction of the normal.
$(x,y,z) - \frac {(x,y,z)\cdot(3,2,1)}{\|(3,2,1)\|}\frac {(3,2,1)}{\|3,2,1\|}\\ (x,y,z) - \frac {(x,y,z)\cdot(3,2,1)}{\|(3,2,1)\|^2}(3,2,1)\\ (x,y,z) - \frac {3x + 2y + z}{14}(3,2,1)\\ (x,y,z) - (\frac {9x + 6y + 3z}{14}, \frac {6x + 4y + 2z}{14}, \frac {3x + 2y + 1z}{14})\\ (\frac {5x - 6y - 3z}{14}, \frac {-6x +10y - 2z}{14}, \frac {-3x - 2y + 13z}{14}) $