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To solve the problem of finding the acute angle between the two given lines, we need to use the directional vectors of the lines and apply the formula for the angle between two vectors.

### Step 1: Identify the directional vectors
From the given parametric equations of the lines:
- For \(\vec{a}(t)\), the direction vector is:
  \[
  \vec{d}_a = \langle -4, 2, 4 \rangle
  \]
- For \(\vec{b}(u)\), the direction vector is:
  \[
  \vec{d}_b = \langle -4, -1, 3 \rangle
  \]

### Step 2: Use the formula for the angle between two vectors
The angle \(\theta\) between two vectors \(\vec{d}_a\) and \(\vec{d}_b\) can be found using the dot product formula:
\[
\cos \theta = \frac{\vec{d}_a \cdot \vec{d}_b}{|\vec{d}_a| |\vec{d}_b|}
\]
Where:
- \(\vec{d}_a \cdot \vec{d}_b\) is the dot product of the two vectors.
- \(|\vec{d}_a|\) and \(|\vec{d}_b|\) are the magnitudes (lengths) of the vectors.

### Step 3: Calculate the dot product
The dot product of \(\vec{d}_a = \langle -4, 2, 4 \rangle\) and \(\vec{d}_b = \langle -4, -1, 3 \rangle\) is:
\[
\vec{d}_a \cdot \vec{d}_b = (-4)(-4) + (2)(-1) + (4)(3)
= 16 - 2 + 12 = 26
\]

### Step 4: Calculate the magnitudes of the vectors
The magnitude of \(\vec{d}_a = \langle -4, 2, 4 \rangle\) is:
\[
|\vec{d}_a| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6
\]
The magnitude of \(\vec{d}_b = \langle -4, -1, 3 \rangle\) is:
\[
|\vec{d}_b| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}
\]

### Step 5: Calculate \(\cos \theta\)
Substituting the dot product and magnitudes into the formula:
\[
\cos \theta = \frac{26}{6 \times \sqrt{26}} = \frac{26}{6\sqrt{26}}
\]
Simplify:
\[
\cos \theta = \frac{13}{3\sqrt{26}}
\]

### Step 6: Find the angle \(\theta\)
To find the angle, we take the inverse cosine (arccos) of the result. Let me compute the value of \(\theta\).The acute angle between the two lines is approximately \( 31.81^\circ \).

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How do you calculate the cross product of two vectors?
2. What is the geometric interpretation of the dot product?
3. Can you explain how to parametrize a line given two points?
4. What is the significance of the magnitude of a vector in physics?
5. How do you determine if two lines are perpendicular or parallel using vectors?

**Tip:** The dot product is positive when the angle between two vectors is acute and negative when it's obtuse.

Find the acute angle (in degrees) between the lines: Let d̂(t) = (-4, 2, 4)t + (-6, 7, 3) and b̂(u) = (-4, -1, 3)u + (8, 4, 9).

Explore how to calculate the acute angle between two lines using their directional vectors. The problem uses concepts of dot product and vector magnitudes to find the angle, applicable for college-level mathematics.

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