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?
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?
$u$ 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°