Given $\vec{a}=[2,5,5]$ and a direction vector $\vec{u}=[-3,5,-3]$, find two vector components $\overrightarrow{a_{1}}$ and $\overrightarrow{a_{2}}$ of the vector $\vec{a}$ such that $\vec{a}=\overrightarrow{a_{1}}+\overrightarrow{a_{2}}$ and $\overrightarrow{a_{1}} \| \vec{u}$ and $\overrightarrow{a_{2}} \perp \vec{u}$.
\[
\begin{array}{l}
\overrightarrow{a_{1}}= \\
\overrightarrow{a_{2}}=
\end{array}
\]
Enter the answers a thit of scalar components. Examples: $[1,-2,3]$ or $[1 / 2,-2 / 3,0]$

Step 1 ：Given the vector $\vec{a} = [2, 5, 5]$ and the direction vector $\vec{u} = [-3, 5, -3]$, we want to find two vector components $\overrightarrow{a_{1}}$ and $\overrightarrow{a_{2}}$ such that $\vec{a} = \overrightarrow{a_{1}} + \overrightarrow{a_{2}}$, with $\overrightarrow{a_{1}} \parallel \vec{u}$ and $\overrightarrow{a_{2}} \perp \vec{u}$.

Step 2 ：First, we find the parallel component $\overrightarrow{a_{1}}$ using the projection formula: $\overrightarrow{a_{1}} = \frac{\vec{a} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u}$.

Step 3 ：Calculating the dot products, we get $\vec{a} \cdot \vec{u} = 2(-3) + 5(5) + 5(-3) = -6 + 25 - 15 = 4$ and $\vec{u} \cdot \vec{u} = (-3)^2 + 5^2 + (-3)^2 = 9 + 25 + 9 = 43$.

Step 4 ：Thus, $\overrightarrow{a_{1}} = \frac{4}{43} \vec{u} = \frac{4}{43} [-3, 5, -3] = [-0.27906977, 0.46511628, -0.27906977]$.

Step 5 ：Next, we find the perpendicular component $\overrightarrow{a_{2}}$ by subtracting $\overrightarrow{a_{1}}$ from $\vec{a}$: $\overrightarrow{a_{2}} = \vec{a} - \overrightarrow{a_{1}} = [2, 5, 5] - [-0.27906977, 0.46511628, -0.27906977] = [2.27906977, 4.53488372, 5.27906977]$.

Step 6 ：\(\boxed{\overrightarrow{a_{1}} = [-0.27906977, 0.46511628, -0.27906977]}\)

Step 7 ：\(\boxed{\overrightarrow{a_{2}} = [2.27906977, 4.53488372, 5.27906977]}\)