Let \( \vec{a}=\langle 5,-1,0\rangle \) and \( \vec{b}=\langle 2,-2,-4\rangle \). Find the angle between the vector 1) The exact angle is \( \cos ^{-1} \) 2) The approximation in degree is

To find the angle between the vectors \(\vec{a}\) and \(\vec{b}\), we can use the dot product formula: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta. \] First, we calculate the dot product \(\vec{a} \cdot \vec{b}\): \[ \vec{a} \cdot \vec{b} = 5 \cdot 2 + (-1) \cdot (-2) + 0 \cdot (-4) = 10 + 2 + 0 = 12. \] Next, we need the magnitudes of \(\vec{a}\) and \(\vec{b}\): \[ |\vec{a}| = \sqrt{5^2 + (-1)^2 + 0^2} = \sqrt{25 + 1 + 0} = \sqrt{26}, \] \[ |\vec{b}| = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}. \] Now, substituting into the dot product formula gives us: \[ 12 = \sqrt{26} (2\sqrt{6}) \cos \theta \] This simplifies to: \[ 12 = 2\sqrt{156} \cos \theta \] Solving for \(\cos \theta\): \[ \cos \theta = \frac{12}{2\sqrt{156}} = \frac{6}{\sqrt{156}} = \frac{6}{2\sqrt{39}} = \frac{3}{\sqrt{39}}. \] Thus, the exact angle is given by: \[ \theta = \cos^{-1}\left(\frac{3}{\sqrt{39}}\right). \] To find the approximation in degrees, we calculate: \[ \theta \approx \cos^{-1}(0.482) \approx 60.2^\circ. \] So the angles between the vectors are: 1) The exact angle is \( \cos^{-1}\left(\frac{3}{\sqrt{39}}\right) \). 2) The approximation in degrees is approximately \(60.2^\circ\).