Question
Find the angle between vector bold lower u equals 3 bold lower I plus start root 3 end root bold lower j and vector bold lower v equals negative 2 bold lower I minus 5 bold lower j to the nearest degree.
(1 point)
Responses
82°
82°
38°
38°
142°
142°
98°
(1 point)
Responses
82°
82°
38°
38°
142°
142°
98°
Answers
GPT-4o mini
To find the angle θ between two vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can use the formula:
\[
\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}
\]
where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product of the vectors and \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors.
Let's compute the dot product \(\mathbf{u} \cdot \mathbf{v}\):
\[
\mathbf{u} = 3\mathbf{i} + \sqrt{3}\mathbf{j}
\]
\[
\mathbf{v} = -2\mathbf{i} - 5\mathbf{j}
\]
The dot product is calculated as follows:
\[
\mathbf{u} \cdot \mathbf{v} = (3)(-2) + (\sqrt{3})(-5) = -6 - 5\sqrt{3}
\]
Next, we calculate the magnitudes of the vectors:
\[
|\mathbf{u}| = \sqrt{(3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}
\]
\[
|\mathbf{v}| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
\]
Now we can substitute these into the cosine formula:
\[
\cos(\theta) = \frac{-6 - 5\sqrt{3}}{(2\sqrt{3})(\sqrt{29})}
\]
Calculating the denominator:
\[
(2\sqrt{3})(\sqrt{29}) = 2\sqrt{87}
\]
Now we substitute everything back into the cosine function:
\[
\cos(\theta) = \frac{-6 - 5\sqrt{3}}{2\sqrt{87}}
\]
Now we need to find the numerical value for \(\cos(\theta)\).
Using a calculator:
1. Calculate \(\sqrt{3} \approx 1.732\), then \(5\sqrt{3} \approx 8.660\).
2. Substitute this into the equation:
\[-6 - 5\sqrt{3} \approx -6 - 8.660 \approx -14.660\]
Now, calculate:
\[
\cos(\theta) \approx \frac{-14.660}{2\sqrt{87}} \approx \frac{-14.660}{19.598} \approx -0.747
\]
Next, we find the angle \(\theta\):
\[
\theta \approx \cos^{-1}(-0.747)
\]
Using a calculator, \(\theta \approx 138°\), but this value needs to round to one of the given options.
Thus, since we have:
- \(θ \approx 142°\) as a close match to 138° (since we are given discrete options), we select:
The angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately **142°**.
\[
\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}
\]
where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product of the vectors and \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors.
Let's compute the dot product \(\mathbf{u} \cdot \mathbf{v}\):
\[
\mathbf{u} = 3\mathbf{i} + \sqrt{3}\mathbf{j}
\]
\[
\mathbf{v} = -2\mathbf{i} - 5\mathbf{j}
\]
The dot product is calculated as follows:
\[
\mathbf{u} \cdot \mathbf{v} = (3)(-2) + (\sqrt{3})(-5) = -6 - 5\sqrt{3}
\]
Next, we calculate the magnitudes of the vectors:
\[
|\mathbf{u}| = \sqrt{(3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}
\]
\[
|\mathbf{v}| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
\]
Now we can substitute these into the cosine formula:
\[
\cos(\theta) = \frac{-6 - 5\sqrt{3}}{(2\sqrt{3})(\sqrt{29})}
\]
Calculating the denominator:
\[
(2\sqrt{3})(\sqrt{29}) = 2\sqrt{87}
\]
Now we substitute everything back into the cosine function:
\[
\cos(\theta) = \frac{-6 - 5\sqrt{3}}{2\sqrt{87}}
\]
Now we need to find the numerical value for \(\cos(\theta)\).
Using a calculator:
1. Calculate \(\sqrt{3} \approx 1.732\), then \(5\sqrt{3} \approx 8.660\).
2. Substitute this into the equation:
\[-6 - 5\sqrt{3} \approx -6 - 8.660 \approx -14.660\]
Now, calculate:
\[
\cos(\theta) \approx \frac{-14.660}{2\sqrt{87}} \approx \frac{-14.660}{19.598} \approx -0.747
\]
Next, we find the angle \(\theta\):
\[
\theta \approx \cos^{-1}(-0.747)
\]
Using a calculator, \(\theta \approx 138°\), but this value needs to round to one of the given options.
Thus, since we have:
- \(θ \approx 142°\) as a close match to 138° (since we are given discrete options), we select:
The angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately **142°**.