# Find the operator P(x,y,z) [closed]

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)

• Welcome to MSE. People will tend to downvote your question if you don't provide context and/or what you have tried, so try to add that if possible. – N. Bar Nov 16 at 22:23
• @N.Bar hi, thanks. Basically i dont even know how to start this exercise, but it is just what i wrote. No more information, just it. – Pedro Saula Nov 16 at 22:27

$$(3,2,1)$$ 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.
$$(x,y,z) - \frac {(x,y,z)\cdot(3,2,1)}{\|(3,2,1)\|}\frac {(3,2,1)}{\|3,2,1\|}\\ (x,y,z) - \frac {(x,y,z)\cdot(3,2,1)}{\|(3,2,1)\|^2}(3,2,1)\\ (x,y,z) - \frac {3x + 2y + z}{14}(3,2,1)\\ (x,y,z) - (\frac {9x + 6y + 3z}{14}, \frac {6x + 4y + 2z}{14}, \frac {3x + 2y + 1z}{14})\\ (\frac {5x - 6y - 3z}{14}, \frac {-6x +10y - 2z}{14}, \frac {-3x - 2y + 13z}{14})$$