First, we compute $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$:

$ 2 \mathbf{a} = -8 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} + 8 \hat{\mathbf{k}} $

$ \frac{\mathbf{b}}{2} = \frac{1}{\sqrt{2}} \hat{\mathbf{i}} - \frac{1}{\sqrt{2}} \hat{\mathbf{j}} $

To find the angle between $2 \mathbf{a}$ and $\frac{\mathbf{b}}{2}$, we use the dot product formula:

$ \cos \theta = \frac{2 \mathbf{a} \cdot \frac{\mathbf{b}}{2}}{|2 \mathbf{a}|\left|\frac{\mathbf{b}}{2}\right|} $

This simplifies to:

$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} $

Calculating the dot product $\mathbf{a} \cdot \mathbf{b}$:

$ \begin{aligned} \mathbf{a} \cdot \mathbf{b} &= (-4)(\sqrt{2}) + (2)(-\sqrt{2}) + (4)(0) \\ &= -4\sqrt{2} - 2\sqrt{2} \\ &= -6\sqrt{2} \end{aligned} $

Next, we find the magnitudes of $\mathbf{a}$ and $\mathbf{b}$:

$ |\mathbf{a}| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 $

$ |\mathbf{b}| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 $

Substitute these into the cosine formula:

$ \cos \theta = \frac{-6\sqrt{2}}{6 \times 2} = \frac{-6\sqrt{2}}{12} = -\frac{1}{\sqrt{2}} $

The angle $\theta$ is calculated as:

$ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) $

Since $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ corresponds to $\pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$:

$ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} = 135^{\circ} $

Vector Algebra

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