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Compute the dot product of the vectors \( u \) and \( v \), and find the angle between the vectors. \( u=\langle-12,0,5\rangle \) and \( \mathbf{v}=\langle 1,2,3\rangle \). \( \mathbf{u} \cdot \mathbf{v}=3 \). (Type an integer or a simplified fraction.) Find the magnitude of the vectors. \( |\mathbf{u}|=13 \) and \( |\mathbf{v}|=\sqrt{14} \) (Type exact answers, using radicals as needed.) The angle between the vectors is \( \square^{\circ} \). (Type your answer in degrees. Do not round until the final answer. Then round to the nearest hundredth as needed.)

Ask by Weston Hammond. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The angle between the vectors is \( 86.44^\circ \).

Solution

To find the angle between the vectors \( \mathbf{u} = \langle -12, 0, 5 \rangle \) and \( \mathbf{v} = \langle 1, 2, 3 \rangle \), we follow these steps: 1. **Dot Product**: \[ \mathbf{u} \cdot \mathbf{v} = (-12)(1) + (0)(2) + (5)(3) = -12 + 0 + 15 = 3 \] 2. **Magnitudes**: \[ |\mathbf{u}| = \sqrt{(-12)^2 + 0^2 + 5^2} = \sqrt{144 + 0 + 25} = \sqrt{169} = 13 \] \[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] 3. **Finding the Angle**: The angle \( \theta \) between the vectors can be found using the formula: \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{3}{13 \sqrt{14}} \] Calculating the cosine value: \[ \cos \theta \approx \frac{3}{13 \times 3.7416573867739413} \approx \frac{3}{48.641545} \approx 0.0617 \] To find \( \theta \), take the arccosine of 0.0617: \[ \theta \approx \arccos(0.0617) \approx 86.44^\circ \] (This is calculated using a calculator with sufficient precision to ensure accuracy to the nearest hundredth.) **Final Answer**: The angle between the vectors is \( \boxed{86.44^\circ} \).

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Bonus Knowledge

To compute the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), we use the formula: \[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \] Substituting the values: \[ \mathbf{u} \cdot \mathbf{v} = (-12)(1) + (0)(2) + (5)(3) = -12 + 0 + 15 = 3 \] Now, to find the angle \( \theta \) between the vectors, we can use the formula: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] We already know \( \mathbf{u} \cdot \mathbf{v} = 3 \), \( |\mathbf{u}| = 13 \), and \( |\mathbf{v}| = \sqrt{14} \). Therefore, \[ \cos(\theta) = \frac{3}{13 \cdot \sqrt{14}} \] Now we can find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{3}{13\sqrt{14}}\right) \] Using a calculator to evaluate \( \theta \): First, compute \( 13\sqrt{14} \): \[ 13\sqrt{14} \approx 13 \times 3.741657 = 48.632561 \] So, \[ \cos(\theta) \approx \frac{3}{48.632561} \approx 0.061674 \] Now take the inverse cosine: \[ \theta \approx \cos^{-1}(0.061674) \] Calculating this gives approximately: \[ \theta \approx 86.45^{\circ} \] Thus, the angle between the vectors is \( \boxed{86.45^{\circ}} \).

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