Use the geometric definition of the dot product in \( \mathbb{R}^{2} \) and \( \mathbb{R}^{3} \) to find the angle in degrees between each pair of vectors. Type as: \#.\#\# 1) \( \mathbf{f}=[-4,3] \) and \( \mathbf{g}=[-8,15] \) \( \square \) degrees Check Show answer 2) \( \mathbf{a}=[2,0,-4] \) and \( \mathbf{b}=[3,-1,1] \) \( \square \) degrees Check Show answer

To find the angle \( \theta \) between each pair of vectors using the geometric definition of the dot product, we can use the formula: \[ \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta) \] Where \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them. We can rearrange the formula to find \( \theta \): \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \quad \Rightarrow \quad \theta = \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\right) \] First, let's calculate for vectors \( \mathbf{f} \) and \( \mathbf{g} \). 1) **Vectors**: \( \mathbf{f} = [-4, 3] \) and \( \mathbf{g} = [-8, 15] \) - **Dot Product**: \[ \mathbf{f} \cdot \mathbf{g} = (-4)(-8) + (3)(15) = 32 + 45 = 77 \] - **Magnitudes**: \[ \|\mathbf{f}\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] \[ \|\mathbf{g}\| = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \] - **Angle**: \[ \cos(\theta) = \frac{77}{5 \times 17} = \frac{77}{85} \] \[ \theta = \cos^{-1}\left(\frac{77}{85}\right) \approx 36.87 \text{ degrees} \] So, 1) \( 36.87 \) degrees --- 2) **Vectors**: \( \mathbf{a} = [2, 0, -4] \) and \( \mathbf{b} = [3, -1, 1] \) - **Dot Product**: \[ \mathbf{a} \cdot \mathbf{b} = (2)(3) + (0)(-1) + (-4)(1) = 6 + 0 - 4 = 2 \] - **Magnitudes**: \[ \|\mathbf{a}\| = \sqrt{(2)^2 + (0)^2 + (-4)^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \] \[ \|\mathbf{b}\| = \sqrt{(3)^2 + (-1)^2 + (1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \] - **Angle**: \[ \cos(\theta) = \frac{2}{(2\sqrt{5})(\sqrt{11})} = \frac{2}{2\sqrt{55}} = \frac{1}{\sqrt{55}} \] \[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{55}}\right) \approx 81.79 \text{ degrees} \] So, 2) \( 81.79 \) degrees