Given \( v=3 i-j \) and \( w=9 i+2 j \), find the angle between \( v \) and \( w \). What is the angle between \( v \) and \( w \) ? \( \square^{\circ} \) (Type your answer in degrees. Do not round until the final answer. Then round to the neares

To find the angle \( \theta \) between the vectors \( v \) and \( w \), we use the formula: \[ \cos(\theta) = \frac{v \cdot w}{\|v\| \|w\|} \] First, we calculate the dot product \( v \cdot w \): \[ v \cdot w = (3)(9) + (-1)(2) = 27 - 2 = 25 \] Next, we find the magnitudes of \( v \) and \( w \): \[ \|v\| = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \] \[ \|w\| = \sqrt{(9)^2 + (2)^2} = \sqrt{81 + 4} = \sqrt{85} \] Now we can plug these values into the cosine formula: \[ \cos(\theta) = \frac{25}{\sqrt{10} \cdot \sqrt{85}} = \frac{25}{\sqrt{850}} \approx \frac{25}{29.154} \approx 0.8563 \] Now, taking the inverse cosine: \[ \theta = \cos^{-1}(0.8563) \approx 31.05^{\circ} \] Rounding to the nearest degree gives us: \(\theta \approx 31^{\circ}\) So, the angle between \( v \) and \( w \) is \( 31^{\circ} \).