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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)

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  • $\begingroup$ 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. $\endgroup$ – N. Bar Nov 16 at 22:23
  • $\begingroup$ @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. $\endgroup$ – Pedro Saula Nov 16 at 22:27
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$(3,2,1)$ 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}) $

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  • $\begingroup$ Just that solve the problem? It looks simple to do, if it is just it $\endgroup$ – Pedro Saula Nov 16 at 22:36

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