Find the direction angles of the vector \(\Vect{x} = (3,4)\) in \(\RNrSpc{2}\).
We have \(| \Vect{x} | = 5\), and so,
\[\omega_1 = \arccos \frac{3}{5} \approx 53.1\quad \text{degrees}\]To find \(\omega_2 \) we now have two possible approaches. The first approach just uses the formula for direction angles again:
\[\omega_2 = \arccos \frac{4}{5} \approx 36.9\quad \text{degrees}\]In the second approach we observe that \(\Vect{x}\) is a first quadrant vector. Therefore
\(\omega_1 + \omega_2 = 90\), and so, \(\omega_2 \approx 90 - 53.1 = 36.9\quad \text{degrees}\).
Find the cosines of the direction angles of \(\Vect{u} = (-1,1,1)\) in \(\RNrSpc{3}\).
We have \(\Vect{u} = \sqrt{3}\), and so
| \(\cos \omega_1\) | \(=\) | \(- \dfrac{1}{ \sqrt{3} }\) |
| \(\cos \omega_2\) | \(=\) | \(\dfrac{1}{ \sqrt{3} }\) |
| \(\cos \omega_3\) | \(=\) | \(\dfrac{1}{ \sqrt{3} }\) |