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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.)

Ask by Coleman Reyes. in the United States
Feb 03,2025

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The dot product \( \mathbf{v} \cdot \mathbf{w} \) is 8.

Solución

To find the dot product \( \mathbf{v} \cdot \mathbf{w} \) and the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we will follow these steps: 1. **Calculate the dot product** \( \mathbf{v} \cdot \mathbf{w} \). 2. **Find the magnitudes** of \( \mathbf{v} \) and \( \mathbf{w} \). 3. **Use the dot product and magnitudes** to find the angle between the vectors. ### Step 1: Calculate the dot product The dot product of two vectors \( \mathbf{v} = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k} \) and \( \mathbf{w} = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k} \) is given by: \[ \mathbf{v} \cdot \mathbf{w} = a_1 a_2 + b_1 b_2 + c_1 c_2 \] For our vectors: - \( \mathbf{v} = -4 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k} \) (where \( a_1 = -4, b_1 = 2, c_1 = 1 \)) - \( \mathbf{w} = 1 \mathbf{i} + 4 \mathbf{j} + 4 \mathbf{k} \) (where \( a_2 = 1, b_2 = 4, c_2 = 4 \)) Now, we can calculate the dot product: \[ \mathbf{v} \cdot \mathbf{w} = (-4)(1) + (2)(4) + (1)(4) \] Calculating each term: \[ = -4 + 8 + 4 \] Now, summing these values: \[ = -4 + 8 + 4 = 8 \] ### Step 2: Find the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \) The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by: \[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \] Calculating the magnitude of \( \mathbf{v} \): \[ \|\mathbf{v}\| = \sqrt{(-4)^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \] Calculating the magnitude of \( \mathbf{w} \): \[ \|\mathbf{w}\| = \sqrt{(1)^2 + (4)^2 + (4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \] ### Step 3: Find the angle between the vectors The angle \( \theta \) between two vectors can be found using the formula: \[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \] Substituting the values we found: \[ \cos(\theta) = \frac{8}{\sqrt{21} \cdot \sqrt{33}} = \frac{8}{\sqrt{693}} \] Thus, the dot product \( \mathbf{v} \cdot \mathbf{w} \) is: \[ \mathbf{v} \cdot \mathbf{w} = 8 \] And the angle can be found using the inverse cosine function if needed, but the dot product is the primary answer requested. ### Final Answer \[ \mathbf{v} \cdot \mathbf{w} = 8 \]

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The Deep Dive

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 \( \theta = \cos^{-1}\left(\frac{4}{\sqrt{693}}\right) \).

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