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I have the following vectors $\vec b=\left(\begin{matrix}-4\\ 3 \end{matrix}\right)$ and $\left(\begin{matrix}x\\ y \end{matrix}\right)$

Now we know that the length of the x,y vector is 1, and that it has the same
direction as $\vec b$

The length teaches us $\sqrt{x^2+y^2}=1$

What does the direction teach us ?

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You can compute $x$ and $y$ from it. Rescaling the vector does not change the direction. So if you know that it has the same direction as $b$ and you know its length, then you get $x$ and $y$. – fabee Jan 17 '13 at 10:14
When $(x,y)$ has the same direction as $\vec b$ then $(x,y)=\lambda \vec b$ for some $\lambda>0$ – Christian Blatter Jan 17 '13 at 10:18
The direction teaches you the ratio, $y/x$. – Gerry Myerson Jan 17 '13 at 11:12
It teaches you all you need to know about $\left(\begin{matrix}x\\ y \end{matrix}\right)$ namely that it is equal to $\vec b/5=\left(\begin{matrix}-4/5\\ 3/5 \end{matrix}\right)$. – Tpofofn Jan 17 '13 at 11:38

asuueme $v=\left(\begin{matrix}x\\ y \end{matrix}\right)$ , $\sqrt{x^2+y^2}=r$ then $x=rsinp$ and $ y=rcosp$ p is angle respect to x axis that p determine direction of this vector let $v=\left(\begin{matrix}-4\\ 3 \end{matrix}\right)$ then p = $arctan\frac{3}{-4}$ and $\left(\begin{matrix}x\\ y \end{matrix}\right)$ has a same drection with $v=\left(\begin{matrix}-4\\ 3 \end{matrix}\right)$ then $arctan\frac{y}{x}=acctan\frac{3}{-4}$ that here$\left(\begin{matrix}x\\ y \end{matrix}\right)$ is unit vector

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