Step 1: Understanding the Concept:
The equations of the lines are given in parametric form.
The coefficients of the parameter '$t$' represent the components of the direction vectors of the lines.
The angle between two lines is the angle between their direction vectors.
Step 2: Key Formula or Approach:
If a line is given by $\vec{r} = \vec{a} + t\vec{v}$, then $\vec{v}$ is its direction vector.
The angle $\theta$ between two vectors $\vec{v}_1$ and $\vec{v}_2$ is given by:
\[ \cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|} \]
Step 3: Detailed Explanation:
From the first line equations: $x = -2 + 2t, y = 3 - 4t, z = -4 + t$, the direction vector is:
$\vec{v}_1 = 2\hat{i} - 4\hat{j} + 1\hat{k}$
From the second line equations: $x = -2 - t, y = 3 + 2t, z = -4 + 3t$, the direction vector is:
$\vec{v}_2 = -1\hat{i} + 2\hat{j} + 3\hat{k}$
Now calculate the dot product of the two direction vectors:
\[ \vec{v}_1 \cdot \vec{v}_2 = (2)(-1) + (-4)(2) + (1)(3) = -2 - 8 + 3 = -7 \]
Calculate the magnitudes of the direction vectors:
\[ |\vec{v}_1| = \sqrt{2^2 + (-4)^2 + 1^2} = \sqrt{4 + 16 + 1} = \sqrt{21} \]
\[ |\vec{v}_2| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
Use the cosine formula to find the acute angle:
\[ \cos \theta = \frac{|-7|}{\sqrt{21} \cdot \sqrt{14}} = \frac{7}{\sqrt{3 \cdot 7} \cdot \sqrt{2 \cdot 7}} \]
\[ \cos \theta = \frac{7}{\sqrt{3 \cdot 2 \cdot 7 \cdot 7}} = \frac{7}{7\sqrt{6}} = \frac{1}{\sqrt{6}} \]
Therefore, the acute angle $\theta$ is:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \]
Step 4: Final Answer:
The acute angle is $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$.