Let \( \vec{a}(t)=\langle-2,4,0\rangle t+\langle 9,-7,7\rangle \) and \( \vec{b}(u)=\langle-4,-2,4\rangle u+\langle 0,-2,-9\rangle \). Find the acute angle (in degrees) between the lines:

To find the acute angle between the lines represented by the parametric equations of \(\vec{a}(t)\) and \(\vec{b}(u)\), we first identify their direction vectors. The direction vector of line \( \vec{a}(t) \) is \(\vec{d_a} = \langle-2, 4, 0\rangle\) and the direction vector of line \( \vec{b}(u) \) is \(\vec{d_b} = \langle-4, -2, 4\rangle\). The acute angle \(\theta\) between two vectors can be found using the formula: \[ \cos(\theta) = \frac{\vec{d_a} \cdot \vec{d_b}}{|\vec{d_a}| |\vec{d_b}|} \] 1. Calculate the dot product \(\vec{d_a} \cdot \vec{d_b}\): \[ \vec{d_a} \cdot \vec{d_b} = (-2)(-4) + (4)(-2) + (0)(4) = 8 - 8 + 0 = 0 \] 2. Calculate the magnitudes of \(\vec{d_a}\) and \(\vec{d_b}\): \[ |\vec{d_a}| = \sqrt{(-2)^2 + 4^2 + 0^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \] \[ |\vec{d_b}| = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \] 3. Substitute back to find \(\cos(\theta)\): \[ \cos(\theta) = \frac{0}{(2\sqrt{5})(6)} = 0 \] Since \(\cos(\theta) = 0\), this means \(\theta = 90^\circ\). Thus, the acute angle between the lines is \(90^\circ\).