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. \( \mathrm{r}=\langle 7,6\rangle, \mathrm{s}=\langle 2,9\rangle \)
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To find the angle between each pair of vectors using the dot product, we use the formula: \[ \cos(\theta) = \frac{u \cdot v}{\|u\| \|v\|} \] where \( u \cdot v \) is the dot product and \( \|u\|, \|v\| \) are the magnitudes of the vectors. **For \( u=\langle 1,4\rangle \) and \( v=\langle-3,-2\rangle \):** 1. Calculate the dot product: \[ u \cdot v = 1 \cdot (-3) + 4 \cdot (-2) = -3 - 8 = -11 \] 2. Calculate the magnitudes: \[ \|u\| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \] \[ \|v\| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \] 3. Find cos(theta): \[ \cos(\theta) = \frac{-11}{\sqrt{17} \cdot \sqrt{13}} \approx \frac{-11}{14.832} \approx -0.7402 \] \(\theta = \cos^{-1}(-0.7402) \approx 137.1^\circ\) --- **For \( r=\langle 7,6\rangle \) and \( s=\langle 2,9\rangle \):** 1. Calculate the dot product: \[ r \cdot s = 7 \cdot 2 + 6 \cdot 9 = 14 + 54 = 68 \] 2. Calculate the magnitudes: \[ \|r\| = \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85} \] \[ \|s\| = \sqrt{2^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} \] 3. Find cos(theta): \[ \cos(\theta) = \frac{68}{\sqrt{85} \cdot \sqrt{85}} = \frac{68}{85} \approx 0.8 \] \(\theta = \cos^{-1}(0.8) \approx 36.9^\circ\) --- Thus, the angles between the vectors are approximately \(137.1^\circ\) for \(u\) and \(v\), and \(36.9^\circ\) for \(r\) and \(s\).