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Given \( \mathbf{v}=-5 \mathbf{i}-\mathbf{j} \) and \( \mathbf{w}=-\mathbf{i}-5 \mathbf{j} \) (a) find the dot product \( \mathbf{v} \cdot \mathbf{w} \); (b) find the angle between \( \mathbf{v} \) and \( \mathbf{w} \). (a) \( \mathbf{v} \cdot \mathbf{w}=10 \) (Simplify your answer.) (b) What is the angle between \( \mathbf{v} \) and \( \mathbf{w} \) ? \( \square^{\circ} \) (Do not round until the final answer. Then round to the nearest tenth as needed.)

Ask by Tran Moran. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The angle between \( \mathbf{v} \) and \( \mathbf{w} \) is approximately \( 67.4^\circ \).

Solution

To find the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we'll use the dot product formula: \[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| \, |\mathbf{w}|} \] **Given:** \[ \mathbf{v} = -5\mathbf{i} - \mathbf{j} \\ \mathbf{w} = -\mathbf{i} - 5\mathbf{j} \] **Step 1: Compute the Dot Product \( \mathbf{v} \cdot \mathbf{w} \)** \[ \mathbf{v} \cdot \mathbf{w} = (-5)(-1) + (-1)(-5) = 5 + 5 = 10 \] **Step 2: Compute the Magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \)** \[ |\mathbf{v}| = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \] \[ |\mathbf{w}| = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \] **Step 3: Compute \( \cos \theta \)** \[ \cos \theta = \frac{10}{\sqrt{26} \times \sqrt{26}} = \frac{10}{26} = \frac{5}{13} \approx 0.384615385 \] **Step 4: Find the Angle \( \theta \)** \[ \theta = \cos^{-1}\left(\frac{5}{13}\right) \approx \cos^{-1}(0.384615385) \approx 67.38^\circ \] **Final Answer:** \[ \theta \approx 67.4^\circ \] **Answer:** Problem (b) Answer: The angle between \( \mathbf{v} \) and \( \mathbf{w} \) is approximately \( 67.4^\circ \).

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To solve the problem, we first find the dot product \( \mathbf{v} \cdot \mathbf{w} \). ### (a) Given: \[ \mathbf{v} = -5 \mathbf{i} - \mathbf{j} \quad \text{and} \quad \mathbf{w} = -\mathbf{i} - 5 \mathbf{j} \] The dot product is calculated as follows: \[ \mathbf{v} \cdot \mathbf{w} = (-5) \cdot (-1) + (-1) \cdot (-5) = 5 + 5 = 10 \] ### (b) Now, we'll find the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{w} \) using the formula: \[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \] First, we need to calculate the magnitudes \( \|\mathbf{v}\| \) and \( \|\mathbf{w}\| \): \[ \|\mathbf{v}\| = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \] \[ \|\mathbf{w}\| = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \] Now substituting the dot product and magnitudes into the cosine formula: \[ \cos(\theta) = \frac{10}{\sqrt{26} \cdot \sqrt{26}} = \frac{10}{26} = \frac{5}{13} \] To find the angle \( \theta \), we take the arccosine: \[ \theta = \cos^{-1}\left(\frac{5}{13}\right) \] Calculating this gives: \[ \theta \approx 62.3^\circ \] So, your final answers are: (a) \( \mathbf{v} \cdot \mathbf{w} = 10 \) (b) \( 62.3^\circ \)

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