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.) 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 use the dot product formula: \[ \mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos(\theta) \] We know that \( \mathbf{v} \cdot \mathbf{w} = 8 \). First, we need to calculate the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \). 1. Calculate \( \|\mathbf{v}\| \): \[ \|\mathbf{v}\| = \sqrt{(-4)^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \] 2. Calculate \( \|\mathbf{w}\| \): \[ \|\mathbf{w}\| = \sqrt{1^2 + 4^2 + 4^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \] Now, substituting the values into the dot product formula: \[ 8 = \sqrt{21} \cdot \sqrt{33} \cdot \cos(\theta) \] Now, calculate \( \sqrt{21} \cdot \sqrt{33} = \sqrt{693} \). Substituting into the equation gives: \[ 8 = \sqrt{693} \cos(\theta) \] Now, solve for \( \cos(\theta) \): \[ \cos(\theta) = \frac{8}{\sqrt{693}} \] Now we'll find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{8}{\sqrt{693}}\right) \] Using a calculator, we compute \( \theta \) in degrees. First, evaluate \( \frac{8}{\sqrt{693}} \): Using a calculator, we approximate \( \sqrt{693} \approx 26.347 \): \[ \frac{8}{\sqrt{693}} \approx 0.3035 \] Now find \( \theta \): \[ \theta \approx \cos^{-1}(0.3035) \approx 72.34^\circ \approx 72.3^\circ \text{ (rounded to the nearest tenth)} \] Thus, the angle between \( \mathbf{v} \) and \( \mathbf{w} \) is \[ \theta \approx 72.3^\circ \]