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 two lines defined by the vector functions: \[ \vec{a}(t) = \langle -2, 4, 0 \rangle t + \langle 9, -7, 7 \rangle \] \[ \vec{b}(u) = \langle -4, -2, 4 \rangle u + \langle 0, -2, -9 \rangle \] we follow these steps: ### 1. Identify the Direction Vectors The direction vectors of the lines are the coefficients of \( t \) and \( u \): - For \( \vec{a}(t) \), the direction vector is \( \vec{d}_a = \langle -2, 4, 0 \rangle \). - For \( \vec{b}(u) \), the direction vector is \( \vec{d}_b = \langle -4, -2, 4 \rangle \). ### 2. Compute the Dot Product The dot product of the direction vectors is: \[ \vec{d}_a \cdot \vec{d}_b = (-2)(-4) + (4)(-2) + (0)(4) = 8 - 8 + 0 = 0 \] ### 3. Compute the Magnitudes of the Direction Vectors Calculate the magnitudes of \( \vec{d}_a \) and \( \vec{d}_b \): \[ \|\vec{d}_a\| = \sqrt{(-2)^2 + 4^2 + 0^2} = \sqrt{4 + 16 + 0} = \sqrt{20} = 2\sqrt{5} \] \[ \|\vec{d}_b\| = \sqrt{(-4)^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \] ### 4. Find the Angle Between the Lines The cosine of the angle \( \theta \) between the two lines is given by: \[ \cos \theta = \frac{\vec{d}_a \cdot \vec{d}_b}{\|\vec{d}_a\| \|\vec{d}_b\|} = \frac{0}{2\sqrt{5} \times 6} = 0 \] When \( \cos \theta = 0 \), the angle \( \theta \) is: \[ \theta = 90^\circ \] ### Conclusion The acute angle between the two lines is **90 degrees**. **Answer:** 90^\circ