Find the angle between each of the following Pairs of Vectors. Given you answer Correct to one decimal place.

The given vectors are $\vec{a} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\vec{b} = 4\mathbf{i} - 3\mathbf{j} + 12\mathbf{k}$.

First, calculate the dot product $\vec{a} \cdot \vec{b}$:

$$\vec{a} \cdot \vec{b} = (2)(4) + (1)(-3) + (-2)(12) = 8 - 3 - 24 = -19$$

Next, calculate the magnitudes of the vectors:

$$||\vec{a}|| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ $$||\vec{b}|| = \sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$$

Now, substitute these values into the formula for $\cos \theta$:

$$\cos \theta = \frac{-19}{3 \cdot 13} = \frac{-19}{39}$$

Finally, find the angle $\theta$:

$$\theta = \cos^{-1}\left(\frac{-19}{39}\right) \approx 119.153^\circ$$

Rounding to one decimal place, the angle is $\boxed{\text{119.2}^\circ}$.

Based on your magnitude calculations, the vectors are $\vec{a} = 3\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}$ and $\vec{b} = \mathbf{i} - 6\mathbf{k}$.

First, calculate the dot product $\vec{a} \cdot \vec{b}$:

$$\vec{a} \cdot \vec{b} = (3)(1) + (-5)(0) + (-2)(-6) = 3 + 0 + 12 = 15$$

(Your calculation for the dot product was 10, which led to an incorrect final answer. The correct dot product is 15.)

Next, calculate the magnitudes of the vectors:

$$||\vec{a}|| = \sqrt{3^2 + (-5)^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$$ $$||\vec{b}|| = \sqrt{1^2 + 0^2 + (-6)^2} = \sqrt{1 + 0 + 36} = \sqrt{37}$$

Now, substitute these values into the formula for $\cos \theta$:

$$\cos \theta = \frac{15}{\sqrt{38} \cdot \sqrt{37}} = \frac{15}{\sqrt{1406}}$$

Finally, find the angle $\theta$:

$$\theta = \cos^{-1}\left(\frac{15}{\sqrt{1406}}\right) \approx 66.433^\circ$$

Rounding to one decimal place, the angle is $\boxed{\text{66.4}^\circ}$.