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Let $u=[\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0]^T$ a versor in $R^3$. Without making any calculation explain what is the angle between $\sqrt{2}u$ and $e_2=[0,1,0]^T$.

How should I do it?

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Even a very rough sketch should give you the answer. Note that you don’t need to calculate the length of $u$; you just need its direction, and you can ignore the third coordinate, since it’s $0$ for both vectors. – Brian M. Scott Jan 18 '12 at 17:27
up vote 3 down vote accepted

$u$ and $e_2$ are in the unitary circle of $R^2$. If you draw those vectors, you will show that $e_2$ is in the y-axis and the $u$ is the first bisector (between x-axis & y-axis). So the angle is $45°$

Other solution, $\sqrt(2)u$ is the diagonal of the square ((0,0),(1,0),(1,1),(0,1)) and $e_2$ is a part of this square. The angle between the diag & a part of square is $45°$

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thank you very much – Andrew Jan 18 '12 at 18:01

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