Question 6 Find the (exact) direction cosines and (rounded to 1 decimal place) direction angles of $\vec{v}=\langle 3,3,2\rangle$ \[ \begin{array}{l} \cos \alpha= \\ \cos \beta= \\ \cos \gamma= \\ \alpha= \\ \beta= \\ \gamma= \end{array} \] Hint: Need some help? Review direction cosines and direction angles 5 in Section 2.3 (The Dot Product). Submit Question Jump to Answer
Solution
Step 1 ：Given the vector \(\vec{v}=\langle 3,3,2\rangle\), we can find the direction cosines and direction angles.
Step 2 ：The direction cosines are given by the formulas: \[\cos \alpha = \frac{x}{\sqrt{x^2 + y^2 + z^2}}\], \[\cos \beta = \frac{y}{\sqrt{x^2 + y^2 + z^2}}\], \[\cos \gamma = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\], where (x, y, z) are the components of the vector.
Step 3 ：Substituting the given values, we get \(\cos \alpha = \frac{3}{\sqrt{3^2 + 3^2 + 2^2}} = 0.64\), \(\cos \beta = \frac{3}{\sqrt{3^2 + 3^2 + 2^2}} = 0.64\), and \(\cos \gamma = \frac{2}{\sqrt{3^2 + 3^2 + 2^2}} = 0.43\).
Step 4 ：The direction angles are the angles themselves, which can be found by taking the inverse cosine (arccos) of the direction cosines.
Step 5 ：Calculating the direction angles, we get \(\alpha = \arccos(0.64) = 50.2^\circ\), \(\beta = \arccos(0.64) = 50.2^\circ\), and \(\gamma = \arccos(0.43) = 64.8^\circ\).
Step 6 ：Final Answer: The direction cosines of the vector \(\vec{v}=\langle 3,3,2\rangle\) are \(\cos \alpha = \boxed{0.64}\), \(\cos \beta = \boxed{0.64}\), \(\cos \gamma = \boxed{0.43}\). The direction angles of the vector \(\vec{v}=\langle 3,3,2\rangle\) are \(\alpha = \boxed{50.2^\circ}\), \(\beta = \boxed{50.2^\circ}\), \(\gamma = \boxed{64.8^\circ}\).
