Let vectors \( \vec{A}=(2,1,-4), \vec{B}=(-3,0,1) \), and \( \vec{C}=(-1,-1,2) \) Part A Calculate the following: What is the angle \( \theta_{\mathrm{AB}} \) between \( \vec{A} \) and \( \vec{B} \) ? Express your answer using one significant figure.

To find the angle \( \theta_{\mathrm{AB}} \) between the vectors \( \vec{A} \) and \( \vec{B} \), we can use the formula: \[ \cos \theta_{\mathrm{AB}} = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \] First, we compute the dot product \( \vec{A} \cdot \vec{B} \): \[ \vec{A} \cdot \vec{B} = (2)(-3) + (1)(0) + (-4)(1) = -6 + 0 - 4 = -10 \] Next, we calculate the magnitudes of \( \vec{A} \) and \( \vec{B} \): \[ |\vec{A}| = \sqrt{(2)^2 + (1)^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21} \] \[ |\vec{B}| = \sqrt{(-3)^2 + (0)^2 + (1)^2} = \sqrt{9 + 0 + 1} = \sqrt{10} \] Now we can substitute these values into the cosine formula: \[ \cos \theta_{\mathrm{AB}} = \frac{-10}{\sqrt{21} \cdot \sqrt{10}} = \frac{-10}{\sqrt{210}} \] Calculating the cosine value: \[ \cos \theta_{\mathrm{AB}} \approx \frac{-10}{14.491} \approx -0.690 \] Now we take the inverse cosine to find the angle: \[ \theta_{\mathrm{AB}} = \cos^{-1}(-0.690) \approx 134.8^\circ \] Expressing this to one significant figure, we round \( 134.8^\circ \) to \( 100^\circ \). So, the angle \( \theta_{\mathrm{AB}} \) is approximately \( 100^\circ \).