Use a dot product to find the angle between each pair of vectors (to the nearest tenth of a degree). 13. \( u=\langle 1,4\rangle, v=\langle-3,-2\rangle \) 14. \( \mathrm{r}=\langle 7,6\rangle, \mathrm{s}=\langle 2,9\rangle \)

To find the angle between two vectors using the **dot product**, you can use the following formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|} \] Where: - \(\mathbf{a} \cdot \mathbf{b}\) is the dot product of vectors \(\mathbf{a}\) and \(\mathbf{b}\). - \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are the magnitudes (lengths) of the vectors. - \(\theta\) is the angle between the vectors. The angle \(\theta\) can then be found by taking the inverse cosine (\(\cos^{-1}\)) of the result. Let's apply this method to the given problems. --- ### Problem 13 **Given Vectors:** \[ \mathbf{u} = \langle 1, 4 \rangle, \quad \mathbf{v} = \langle -3, -2 \rangle \] **Step 1: Compute the Dot Product (\(\mathbf{u} \cdot \mathbf{v}\))** \[ \mathbf{u} \cdot \mathbf{v} = (1)(-3) + (4)(-2) = -3 - 8 = -11 \] **Step 2: Compute the Magnitudes (\(|\mathbf{u}|\) and \(|\mathbf{v}|\))** \[ |\mathbf{u}| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.1231 \] \[ |\mathbf{v}| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.6056 \] **Step 3: Compute \(\cos \theta\)** \[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| \, |\mathbf{v}|} = \frac{-11}{4.1231 \times 3.6056} \approx \frac{-11}{14.837} \approx -0.7415 \] **Step 4: Compute the Angle (\(\theta\))** \[ \theta = \cos^{-1}(-0.7415) \approx 136.3^\circ \] **Answer:** The angle between \(\mathbf{u}\) and \(\mathbf{v}\) is approximately **136.3 degrees**. --- ### Problem 14 **Given Vectors:** \[ \mathbf{r} = \langle 7, 6 \rangle, \quad \mathbf{s} = \langle 2, 9 \rangle \] **Step 1: Compute the Dot Product (\(\mathbf{r} \cdot \mathbf{s}\))** \[ \mathbf{r} \cdot \mathbf{s} = (7)(2) + (6)(9) = 14 + 54 = 68 \] **Step 2: Compute the Magnitudes (\(|\mathbf{r}|\) and \(|\mathbf{s}|\))** \[ |\mathbf{r}| = \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.2195 \] \[ |\mathbf{s}| = \sqrt{2^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} \approx 9.2195 \] **Step 3: Compute \(\cos \theta\)** \[ \cos \theta = \frac{\mathbf{r} \cdot \mathbf{s}}{|\mathbf{r}| \, |\mathbf{s}|} = \frac{68}{9.2195 \times 9.2195} = \frac{68}{85} \approx 0.8 \] **Step 4: Compute the Angle (\(\theta\))** \[ \theta = \cos^{-1}(0.8) \approx 36.9^\circ \] **Answer:** The angle between \(\mathbf{r}\) and \(\mathbf{s}\) is approximately **36.9 degrees**. ---