Question 6
Find the (exact) direction cosines and (rounded to 1 decimal place) direction angles of $\overrightarrow{v}=⟨3,3,2⟩$
$$\begin{array}{l}\cos \alpha =\\ \cos \beta =\\ \cos \gamma =\\ \alpha =\\ \beta =\\ \gamma =\end{array}$$
Hint: Need some help? Review direction cosines and direction angles 5 in Section 2.3 (The Dot Product).
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Step 1 ：Given the vector $\overrightarrow{v}=⟨3,3,2⟩$, 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 $\overrightarrow{v}=⟨3,3,2⟩$ are $\cos \alpha =\boxed{{\displaystyle 0.64}}$, $\cos \beta =\boxed{{\displaystyle 0.64}}$, $\cos \gamma =\boxed{{\displaystyle 0.43}}$. The direction angles of the vector $\overrightarrow{v}=⟨3,3,2⟩$ are $\alpha =\boxed{{\displaystyle {50.2}^{\circ }}}$, $\beta =\boxed{{\displaystyle {50.2}^{\circ }}}$, $\gamma =\boxed{{\displaystyle {64.8}^{\circ }}}$.