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Over $i^2$ there is $U=${$(x,y)|y=1/5x$}.

Let $v=(52,52)$ vector.

Let $p\in U^\perp$ such that $v-p\in U$ so what is the value of $||p||$?

I know the answer is $8 \sqrt{26}$ but I can't see how to get it...

(hope everything is clear since im not english speaker)

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I don't understand the notation in the first line – Belgi Jan 10 '13 at 22:47
which part exactly?? – Anna Jan 10 '13 at 22:50
$i^2U$ [spacefiller] – Belgi Jan 10 '13 at 22:52
Do you mean "let $p\in U^\perp$"? – Git Gud Jan 10 '13 at 22:58
Do you mean $v-p\in U$? – Git Gud Jan 10 '13 at 23:09
up vote 3 down vote accepted

Unsure of what's being asked I suggest the following:

Notice that $U=\{(x,\frac{1}{5}x) | x\in \mathbb{R}\}$.

Now what's $U^\perp$? Check that it is $U^\perp=\{(-x,5x) | x\in \mathbb{R}\}$.

Now take $p\in U^\perp$ such that $v-p\in U$. Because $p\in U^\perp$ you know that $p=(-5\alpha, \alpha)$, for some $\alpha \in \mathbb{R}$. So we get $v-p=(52+5\alpha, 52 -\alpha) $. But now we want $v-p\in U$.

Because $\{(5x,x) | x\in \mathbb{R}\}=\{(x,\frac{1}{5}x) | x\in \mathbb{R}\} = U$, we know that $v-p\in U$ if, and only if, $5\cdot(52+5\alpha)=52-\alpha$.

Solving for $\alpha$ in order to get $v-p\in U$, follows that $\alpha=-8$.

Therefore $p=(40,-8)$ and $||p||=\sqrt {40^2+8^2}=\sqrt{8^2\cdot5^2+8^2}=\sqrt{8^2\cdot(25+1)}=\sqrt{8^2}\sqrt{26}=8\sqrt{26}$.

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thank you for your answer actually this is exactly where I was stuck (52,52)-(-5x,x)=? – Anna Jan 10 '13 at 23:44
I wrote the solution farther, but I didn't finish it. I must go to bed now. Good luck. – Git Gud Jan 10 '13 at 23:49
how can I make it ∈U? this is where I was stuck from beginning – Anna Jan 11 '13 at 7:25
I solved it on the answer. – Git Gud Jan 11 '13 at 17:32

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