To determine if the vectors u and v are orthogonal, we first calculate their dot product. Two vectors are orthogonal if their dot product is zero.
Given: u = \(\begin{pmatrix} -6 \ 4 \end{pmatrix}\) v = \(\begin{pmatrix} 3 \ 3 \end{pmatrix}\)
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 not zero (\(-6 \neq 0\)), the vectors are not orthogonal.
Next, to find the angle \( \theta \) between the two vectors, we use the formula:
\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \]
First, we need to find the magnitudes of both vectors:
\[ |\mathbf{u}| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]
\[ |\mathbf{v}| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]
Now substituting the values:
\[ \cos(\theta) = \frac{-6}{(2\sqrt{13})(3\sqrt{2})} \]
Calculating the denominator:
\[ (2\sqrt{13})(3\sqrt{2}) = 6\sqrt{26} \]
So,
\[ \cos(\theta) = \frac{-6}{6\sqrt{26}} = \frac{-1}{\sqrt{26}} \]
Now, to find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{-1}{\sqrt{26}}\right) \]
Calculating this value using a calculator gives us approximately:
\[ \theta \approx 108.2^\circ \]
However, this value does not appear in your responses. Therefore, based on the provided options and the computed angle being roughly \( 108^\circ \):
The best match is:
- The vectors are not orthogonal. The angle between the two vectors is 101.3°.
(Although this is not the exact angle computed, it is the closest option provided.)