Find the angle $\theta$ (in degrees) between the vectors. (Round your answer to two decimal places.) \[ \begin{array}{l} \mathbf{u}=3 \mathbf{i}-6 \mathbf{j} \\ \mathbf{v}=9 \mathbf{i}+3 \mathbf{j} \end{array} \] \[ \theta= \]
Solution
Step 1 ：We are given the vectors \(\mathbf{u} = 3\mathbf{i} - 6\mathbf{j}\) and \(\mathbf{v} = 9\mathbf{i} + 3\mathbf{j}\).
Step 2 ：We can calculate the dot product of the vectors using the formula \(\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y\), where \(u_x\) and \(u_y\) are the x and y components of \(\mathbf{u}\), and \(v_x\) and \(v_y\) are the x and y components of \(\mathbf{v}\). Plugging in the given components, we get \(\mathbf{u} \cdot \mathbf{v} = 9\).
Step 3 ：We can calculate the magnitudes of the vectors using the formula \(||\mathbf{u}|| = \sqrt{u_x^2 + u_y^2}\) and \(||\mathbf{v}|| = \sqrt{v_x^2 + v_y^2}\). Plugging in the given components, we get \(||\mathbf{u}|| = 6.708203932499369\) and \(||\mathbf{v}|| = 9.486832980505138\).
Step 4 ：We can find the angle between the vectors using the formula \(\theta = \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||}\right)\). Plugging in the values we calculated, we get \(\theta_{rad} = 1.4288992721907328\).
Step 5 ：We convert the angle from radians to degrees to get \(\theta_{deg} = 81.87\).
Step 6 ：Final Answer: The angle \(\theta\) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is \(\boxed{81.87}\) degrees.
