Question
Compute the dot product of the vectors \( \mathbf{v} \) and \( \mathbf{w} \), and find the angle between the vectors. \( \mathbf{v}=-9 \mathbf{i}-\mathbf{j} \) and \( \mathbf{w}=-\mathbf{i}-9 \mathbf{j} \) \( \mathbf{v} \cdot \mathbf{w}=18 \) (Simplify your answer.) Find the magnitude of the vectors. \( |\mathbf{v}|=\sqrt{82} \) and \( |\mathbf{w}|=\sqrt{82} \) (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 tenth as needed.
Ask by Harris Mejia. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The angle between the vectors is \( 77.4^\circ \).
Solution
To find the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we can use the dot product formula:
\[
\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos(\theta)
\]
Given:
\[
\mathbf{v} \cdot \mathbf{w} = 18
\]
\[
|\mathbf{v}| = \sqrt{82}, \quad |\mathbf{w}| = \sqrt{82}
\]
Plugging these values into the formula:
\[
18 = \sqrt{82} \times \sqrt{82} \times \cos(\theta)
\]
\[
18 = 82 \cos(\theta)
\]
\[
\cos(\theta) = \frac{18}{82} = \frac{9}{41}
\]
Now, solve for \( \theta \):
\[
\theta = \cos^{-1}\left(\frac{9}{41}\right)
\]
Using a calculator to find the inverse cosine:
\[
\theta \approx \cos^{-1}(0.2195) \approx 77.4^\circ
\]
**Answer:** The angle between the vectors is \( \boxed{77.4^\circ} \).
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
To find the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we can use the formula: \[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \] Given that \( \mathbf{v} \cdot \mathbf{w} = 18 \), \( |\mathbf{v}| = \sqrt{82} \), and \( |\mathbf{w}| = \sqrt{82} \), we plug in the values: \[ \cos \theta = \frac{18}{\sqrt{82} \cdot \sqrt{82}} = \frac{18}{82} = \frac{9}{41} \] Now, to find the angle \( \theta \), we take the arccosine: \[ \theta = \arccos\left(\frac{9}{41}\right) \] Calculating this gives us: \[ \theta \approx 69.5^\circ \] So the angle between the vectors is approximately \( 69.5 \) degrees when rounded to the nearest tenth.
