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Just a small thing 762 here,the question is about projection to the plane $T: x+2z-3y+3=0$ in the direction of the vector $\bar{i}+\bar{j}-\bar{k}$. Now I understand this in a way

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where $k\in\mathbb R$ but I got into the dead-end with 3. Initially, I thought $\mathbb R^{4}$ but it mentions $f: \mathbb R^{3}\mapsto \mathbb R^{3}$. Now I need some mapping $(1\times3) (3\times1)$ to get scalar, thinking. I must be misunderstanding something with "projection to the plane in the direction of the vector". Perhaps, this way?

$$\bar{n}* (\bar{r}-\bar{r}_{0})=-3$$

where $\bar{n}=(1,-3,2)$ is the normal vector to the plane. Now

$$\bar{n}*\left(\bar{x}_{1}+k (1,1,-1)\right)=-3$$

where is the $\bar{r}_{0}=k(1,1,-1)$ so $1(a+k)-3(b+k)+2(c-k)=-3$. Not sure yet whether this was the correct way of doing, thinking...trying to visualize...

Perhaps Related

  1. http://en.wikipedia.org/wiki/Plane_(geometry)
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1 Answer 1

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I'll try to do it without getting away too much from your point of view.

You have a plane $T : x + 2y - 3z + 3 = 0$, which I can rewrite as $$ \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \begin{bmatrix} x & y & z \end{bmatrix} = -3. $$ Now you are projecting in the direction $\begin{bmatrix} 1 & 1 & -1 \end{bmatrix}$, so you want to find $\alpha$ such that for arbitrary $x,y,z$ real, we have that $$ \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \left( \begin{bmatrix} x & y & z \end{bmatrix} + \alpha \begin{bmatrix} 1 & 1 & -1 \end{bmatrix} \right) = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \begin{bmatrix} x + \alpha & y + \alpha & z - \alpha \end{bmatrix} = -3. $$ But the last equation can be re-written as $$ (x+\alpha) + 2(y + \alpha) -3(z - \alpha) = x + 2y - 3z + 6\alpha = -3. $$ Therefore, $$ \alpha = \frac{x+2y-3z+3}6 $$ gives you the unique $\alpha$ for which this is possible. If you actually need to compute the algorithm (matrix) which finds the projection as a function of $x$,$y$ and $z$, just compute the vector $[x+\alpha, y+\alpha,z-\alpha]$ and see the matrix there. This would give $$ \begin{bmatrix} x + \alpha & y + \alpha & z - \alpha \end{bmatrix} = \begin{bmatrix} \frac{7x + 2y - 3z + 3}6 & \frac{x+8y-3z+3}6 & \frac{-x-2y+3z-3}6 \end{bmatrix} $$ or, written in column form (as standard) $$ \begin{bmatrix} x + \alpha \\ y + \alpha \\ z - \alpha \end{bmatrix} = \begin{bmatrix} \frac{7x + 2y - 3z + 3}6 \\ \frac{x+8y-3z+3}6 \\ \frac{-x-2y+3z-3}6 \end{bmatrix} = \begin{bmatrix}\frac 76 & \frac 26 & \frac {-3}6 \\ \frac 16 & \frac 86 & \frac {-3}6 \\ \frac {-1}6 & \frac{-2}6 & \frac 36 \end{bmatrix} \begin{bmatrix} x \\ \\ \\ y \\ \\ \\ z \end{bmatrix} + \begin{bmatrix} \frac 36 \\ \frac 36 \\ \frac{-3}6 \end{bmatrix} = P \left( \begin{bmatrix} x \\ \\ \\ y \\ \\ \\ z \end{bmatrix} \right). $$ where $P$ would be the "projection map".

Hope that helps,

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There is a small change in the plane $x+2z-3y-3=0$, not $x+2y-3z-3=0$, but I think it is good idea not to provide the identical calculation so learners can also calculate things :) ...anyway back to calculating and drawing...anyway it seems that I was in the right direction, thanks for that +1 –  hhh Mar 7 '12 at 7:21
    
You made the typo or I did? It's not clear ; in your comment there is twice the same equation. But I provide computations when I feel the OP is lost and has worked ; if he wants to he can do them again and go through by himself. ^_^ –  Patrick Da Silva Mar 7 '12 at 7:29
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