Answer:
The approximate angle of [tex]\vec u = \frac{1}{2}\cdot \vec v[/tex], where [tex]\vec v = \langle 10, 15\rangle[/tex], is 56º.
Step-by-step explanation:
Let [tex]\vec v = \langle 10, 15\rangle[/tex] and [tex]\vec u = \frac{1}{2}\cdot \vec v[/tex], then the resulting vector is described below:
[tex]\vec u = \frac{1}{2}\cdot \langle 10,15\rangle[/tex]
[tex]\vec u = \left\langle 5,\frac{15}{2} \right\rangle[/tex]
From Trigonometry, we find that direction angle of this vector is defined by the following inverse trigonometric expression:
[tex]\theta = \tan^{-1}\left(\frac{u_{y}}{u_{x}} \right)[/tex] (1)
Where:
[tex]u_{x}[/tex] - x-Component of [tex]\vec u[/tex], dimensionless.
[tex]u_{y}[/tex] - y-Component of [tex]\vec u[/tex], dimensionless.
If we know that [tex]u_{x} = 5[/tex] and [tex]u_{y} = \frac{15}{2}[/tex], then the direction angle of the vector is:
[tex]\theta = \tan^{-1}\left(\frac{\frac{15}{2}}{5} \right)[/tex]
[tex]\theta =\tan^{-1} \frac{15}{10}[/tex]
[tex]\theta = \tan^{-1} \frac{3}{2}[/tex]
[tex]\theta \approx 56.310^{\circ}[/tex]
The approximate angle of [tex]\vec u = \frac{1}{2}\cdot \vec v[/tex], where [tex]\vec v = \langle 10, 15\rangle[/tex], is 56º.
