Question
Given vectors bold lower u equals left angle bracket negative 6 comma 4 right angle bracket and bold lower v equals left angle bracket 3 comma 3 right angle bracket, determine if the vectors are orthogonal. If they are not orthogonal, find the angle between the two vectors.
(1 point)
Responses
The vectors are orthogonal.
The vectors are orthogonal.
The vectors are not orthogonal. The angle between the two vectors is 72.5°.
The vectors are not orthogonal. The angle between the two vectors is 72.5°.
The vectors are not orthogonal. The angle between the two vectors is 101.3°.
The vectors are not orthogonal. The angle between the two vectors is 101.3°.
The vectors are not orthogonal. The angle between the two vectors is 130.6°.
The vectors are not orthogonal. The angle between the two vectors is 130.6°.
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(1 point)
Responses
The vectors are orthogonal.
The vectors are orthogonal.
The vectors are not orthogonal. The angle between the two vectors is 72.5°.
The vectors are not orthogonal. The angle between the two vectors is 72.5°.
The vectors are not orthogonal. The angle between the two vectors is 101.3°.
The vectors are not orthogonal. The angle between the two vectors is 101.3°.
The vectors are not orthogonal. The angle between the two vectors is 130.6°.
The vectors are not orthogonal. The angle between the two vectors is 130.6°.
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Answers
GPT-4o mini
To determine if the vectors \(\mathbf{u} = \langle -6, 4 \rangle\) and \(\mathbf{v} = \langle 3, 3 \rangle\) are orthogonal, we can use the dot product. Two vectors are orthogonal if their dot product equals zero.
The dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as follows:
\[
\mathbf{u} \cdot \mathbf{v} = (-6)(3) + (4)(3) = -18 + 12 = -6
\]
Since the dot product is \(-6\) (not equal to zero), the vectors are not orthogonal.
Next, to find the angle \(\theta\) between the two vectors, we can use the formula:
\[
\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||}
\]
First, we calculate the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\):
\[
||\mathbf{u}|| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]
\[
||\mathbf{v}|| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
Now substituting into the cosine formula:
\[
\cos(\theta) = \frac{-6}{(2\sqrt{13})(3\sqrt{2})} = \frac{-6}{6\sqrt{26}} = \frac{-1}{\sqrt{26}}
\]
Now we find \(\theta\):
\[
\theta = \cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)
\]
Calculating this gives:
\[
\theta \approx 108.2^\circ
\]
Thus, the vectors are not orthogonal, and the angle between the two vectors is approximately \(108.2^\circ\).
Therefore, the correct response is:
**The vectors are not orthogonal. The angle between the two vectors is approximately 108.2°.**
The dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as follows:
\[
\mathbf{u} \cdot \mathbf{v} = (-6)(3) + (4)(3) = -18 + 12 = -6
\]
Since the dot product is \(-6\) (not equal to zero), the vectors are not orthogonal.
Next, to find the angle \(\theta\) between the two vectors, we can use the formula:
\[
\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||}
\]
First, we calculate the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\):
\[
||\mathbf{u}|| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]
\[
||\mathbf{v}|| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
Now substituting into the cosine formula:
\[
\cos(\theta) = \frac{-6}{(2\sqrt{13})(3\sqrt{2})} = \frac{-6}{6\sqrt{26}} = \frac{-1}{\sqrt{26}}
\]
Now we find \(\theta\):
\[
\theta = \cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)
\]
Calculating this gives:
\[
\theta \approx 108.2^\circ
\]
Thus, the vectors are not orthogonal, and the angle between the two vectors is approximately \(108.2^\circ\).
Therefore, the correct response is:
**The vectors are not orthogonal. The angle between the two vectors is approximately 108.2°.**