Find the dot product \( \mathbf{v} \cdot \mathbf{w} \) and the angle between \( \mathbf{v} \) and \( \mathbf{w} \). \[ \mathbf{v}=-4 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{w}=\mathbf{i}+4 \mathbf{j}+4 \mathbf{k} \] \( \mathbf{v} \cdot \mathbf{w}=\square \) (Simplify your answer. Type an exact value, using radicals as needed.)
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To find the dot product \( \mathbf{v} \cdot \mathbf{w} \) and the angle between \( \mathbf{v} \) and \( \mathbf{w} \), we start with the dot product formula: \[ \mathbf{v} \cdot \mathbf{w} = v_x w_x + v_y w_y + v_z w_z \] Here, \( \mathbf{v} = -4\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{w} = \mathbf{i} + 4\mathbf{j} + 4\mathbf{k} \). Identifying the components: - For \( \mathbf{v} \): \( v_x = -4, v_y = 2, v_z = 1 \) - For \( \mathbf{w} \): \( w_x = 1, w_y = 4, w_z = 4 \) Now, we compute the dot product: \[ \mathbf{v} \cdot \mathbf{w} = (-4)(1) + (2)(4) + (1)(4) \] \[ = -4 + 8 + 4 \] \[ = 8 - 4 = 4 \] Thus, \( \mathbf{v} \cdot \mathbf{w} = 4 \). Next, to find the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{w} \), we use the formula: \[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \] First, we calculate the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \): \[ |\mathbf{v}| = \sqrt{(-4)^2 + (2)^2 + (1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \] \[ |\mathbf{w}| = \sqrt{(1)^2 + (4)^2 + (4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \] Now plugging in the values: \[ \cos(\theta) = \frac{4}{\sqrt{21} \sqrt{33}} = \frac{4}{\sqrt{693}} \] To find \( \theta \), we take the arccosine: \[ \theta = \cos^{-1}\left(\frac{4}{\sqrt{693}}\right) \] So the final results are: \[ \mathbf{v} \cdot \mathbf{w} = 4 \] And the angle \( \theta = \cos^{-1}\left(\frac{4}{\sqrt{693}}\right) \).