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).
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Answer
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 \(\alpha = \boxed{50.2^\circ}\), \(\beta = \boxed{50.2^\circ}\), \(\gamma = \boxed{64.8^\circ}\).
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 \(\alpha = \boxed{50.2^\circ}\), \(\beta = \boxed{50.2^\circ}\), \(\gamma = \boxed{64.8^\circ}\).
