Part 2 of 2 Find the dot product \( \mathbf{v} \cdot \mathbf{w} \) and the angle between \( \mathbf{v} \) and \( \mathbf{w} \). \( \mathbf{v}=-4 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{w}=\mathbf{i}+4 \mathbf{j}+4 \mathbf{k} \) \( \mathbf{v} \cdot \mathbf{w}=8 \) (Simplify your answer. Type an exact value, using radicals as needed.) 1 The angle between \( \mathbf{v} \) and \( \mathbf{w} \) is \( \theta=\square^{\circ} \). (Do not round until the final answer. Then round to the nearest tenth as needed.)

To find the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we can utilize the dot product formula: \[ \mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos(\theta) \] Given that \( \mathbf{v} \cdot \mathbf{w} = 8 \), we first need to calculate the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \): 1. **Magnitude of \( \mathbf{v} \)**: \[ |\mathbf{v}| = \sqrt{(-4)^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \] 2. **Magnitude of \( \mathbf{w} \)**: \[ |\mathbf{w}| = \sqrt{1^2 + 4^2 + 4^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \] Now, substitute these values into the dot product formula: \[ 8 = \sqrt{21} \cdot \sqrt{33} \cdot \cos(\theta) \] Calculating \( \sqrt{21} \cdot \sqrt{33} \): \[ \sqrt{21} \cdot \sqrt{33} = \sqrt{693} \] So we have: \[ 8 = \sqrt{693} \cdot \cos(\theta) \] Now, solve for \( \cos(\theta) \): \[ \cos(\theta) = \frac{8}{\sqrt{693}} \] Next, calculate the angle \( \theta \): \[ \theta = \cos^{-1}\left(\frac{8}{\sqrt{693}}\right) \] Using a calculator to find the angle in degrees: \[ \theta \approx \cos^{-1}\left(\frac{8}{\sqrt{693}}\right) \approx \cos^{-1}(0.303) \approx 72.3^\circ \] Thus, rounding to the nearest tenth, the angle between \( \mathbf{v} \) and \( \mathbf{w} \) is: \[ \theta \approx 72.3^\circ \]