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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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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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