Let \( \vec{a}(t)=-4,4,-3\rangle t+\langle 1,2,-1\rangle \) and \( \vec{b}(u)=\langle 2,-3,-1\rangle u+\langle-3,8,-4\rangle \). Find the acute angle (in degrees) between the lines:

To find the acute angle between the lines represented by the vectors \( \vec{a}(t) \) and \( \vec{b}(u) \), we first need to determine the direction vectors of each line. The direction vector of \( \vec{a}(t) \) is \( \vec{A} = \langle -4, 4, -3 \rangle \), and the direction vector of \( \vec{b}(u) \) is \( \vec{B} = \langle 2, -3, -1 \rangle \). The cosine of the angle \( \theta \) between the two direction vectors can be found using the dot product formula: \[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \] 1. Calculate \( \vec{A} \cdot \vec{B} \): \[ \vec{A} \cdot \vec{B} = (-4)(2) + (4)(-3) + (-3)(-1) = -8 - 12 + 3 = -17 \] 2. Calculate \( |\vec{A}| \) and \( |\vec{B}| \): \[ |\vec{A}| = \sqrt{(-4)^2 + 4^2 + (-3)^2} = \sqrt{16 + 16 + 9} = \sqrt{41} \] \[ |\vec{B}| = \sqrt{(2)^2 + (-3)^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \] 3. Now, substitute these values back into the cosine formula: \[ \cos(\theta) = \frac{-17}{\sqrt{41} \sqrt{14}} \] 4. Calculate \( \theta \): Since we are interested in the acute angle, we take the absolute value of the cosine: \[ \cos(\theta) = \left| \frac{-17}{\sqrt{41} \sqrt{14}} \right| = \frac{17}{\sqrt{41} \sqrt{14}} \approx \frac{17}{\sqrt{574}} \approx \frac{17}{23.99} \approx 0.708 \] Now use the arccos function to find \( \theta \): \[ \theta \approx \arccos(0.708) \approx 45.56^\circ \] So, the acute angle \( \theta \) between the lines is approximately \( 45.6^\circ \).