Let $\vec{r}(t)=\langle-3,1,3\rangle+\langle 4,0,5\rangle t+$ and $\vec{s}(t)=\langle 8,4,-4\rangle+\langle 3,0,-3\rangle t$.
Find the acute angle between the lines (in radians)

Step 1 ：Given the lines with vector equations \(\vec{r}(t)=\langle-3,1,3\rangle+\langle 4,0,5\rangle t\) and \(\vec{s}(t)=\langle 8,4,-4\rangle+\langle 3,0,-3\rangle t\), we are asked to find the acute angle between these lines.

Step 2 ：The direction vectors of the lines are given by the coefficients of t in the vector equations of the lines. In this case, the direction vectors are \(\langle 4,0,5\rangle\) and \(\langle 3,0,-3\rangle\).

Step 3 ：The dot product of two vectors \(\vec{a}=\langle a_1,a_2,a_3\rangle\) and \(\vec{b}=\langle b_1,b_2,b_3\rangle\) is given by \(a_1b_1+a_2b_2+a_3b_3\). In this case, the dot product of the direction vectors is -3.

Step 4 ：The magnitude of a vector \(\vec{a}=\langle a_1,a_2,a_3\rangle\) is given by \(\sqrt{a_1^2+a_2^2+a_3^2}\). The magnitudes of the direction vectors are approximately 6.4031242374328485 and 4.242640687119285.

Step 5 ：The cosine of the angle between two vectors is given by the dot product of the vectors divided by the product of their magnitudes. Therefore, the cosine of the angle between the direction vectors is approximately -0.11043152607484655.

Step 6 ：The angle itself can be found by taking the arccosine of the cosine of the angle. Therefore, the acute angle between the lines is approximately 1.6814535479687922 radians.

Step 7 ：Final Answer: The acute angle between the lines is \(\boxed{1.6814535479687922}\) radians.