1. **State the problem:** Find the angle between vectors $\mathbf{u} = [-1, 3, 4]$ and $\mathbf{w} = [-2, -1, 3]$. 2. **Formula used:** The angle $\theta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$$ where $\mathbf{a} \cdot \mathbf{b}$ is the dot product and $\|\mathbf{a}\|$ is the magnitude (length) of vector $\mathbf{a}$. 3. **Calculate the dot product $\mathbf{u} \cdot \mathbf{w}$:** $$\mathbf{u} \cdot \mathbf{w} = (-1)(-2) + (3)(-1) + (4)(3) = 2 - 3 + 12 = 11$$ 4. **Calculate the magnitudes:** $$\|\mathbf{u}\| = \sqrt{(-1)^2 + 3^2 + 4^2} = \sqrt{1 + 9 + 16} = \sqrt{26}$$ $$\|\mathbf{w}\| = \sqrt{(-2)^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$$ 5. **Calculate $\cos \theta$:** $$\cos \theta = \frac{11}{\sqrt{26} \times \sqrt{14}} = \frac{11}{\sqrt{364}} = \frac{11}{2\sqrt{91}}$$ 6. **Find the angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{11}{2\sqrt{91}} \right)$$ 7. **Interpretation:** This angle $\theta$ is the measure in radians (or degrees if converted) between vectors $\mathbf{u}$ and $\mathbf{w}$. **Final answer:** $$\boxed{\theta = \cos^{-1} \left( \frac{11}{2\sqrt{91}} \right)}$$