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Let $U$ be the line intersection of the planes $x + y + z = 0$ and $4x + y + z = 0$. Let $T:\mathbb{R}^3\rightarrow U$ be the projection of $\mathbb{R}^3$ onto U. Find the formula for $T(x, y, z)$

This is a problem in my textbook. Finding the equation of the line is easy: The equation of line is: $[0 -1 1] * t$

But second part, finding the formula, I don't understand so much. And I cannot find connection in the phrase "onto U" to solve this problem.

Thanks :)

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The line of intersection is $(0,1,-1)t$, which in this case is also a $1$-dim subspace of $\mathbb{R^3}$, span$\{(0,1,-1)\}$. You need to find the projection of $\mathbb{R^3}$ onto this subspace. The projection of a vector $u$ onto a vector $v$ is $$ proj_v u=\frac{\langle u,v\rangle}{\langle v,v\rangle}v $$

When $u=(x,y,z)$ and $v=(0,1,-1)$, you get: $$ T(x,y,z)=\frac{y-z}{\sqrt{2}}(0,1,-1)=(0,\frac{y-z}{\sqrt{2}},-\frac{y-z}{\sqrt{2}}) $$

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Can you tell me clearer please :) –  hqt Nov 9 '12 at 16:59
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