The problem requires us to:

1. Find the dot product $(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b})$.
2. Determine the angle between vectors $\mathbf{a}$ and $\mathbf{b}$.

Given vectors: $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ $\mathbf{b} = -\mathbf{i} + 4\mathbf{j} + \mathbf{k}$

Let's solve it step by step.

Step 1: Compute $\mathbf{a} - \mathbf{b}$ and $\mathbf{a} + \mathbf{b}$

$\mathbf{a} - \mathbf{b} = (1 - (-1))\mathbf{i} + (2 - 4)\mathbf{j} + (2 - 1)\mathbf{k}$ $= (1 + 1)\mathbf{i} + (-2)\mathbf{j} + (1)\mathbf{k}$ $= 2\mathbf{i} - 2\mathbf{j} + \mathbf{k}$

$\mathbf{a} + \mathbf{b} = (1 + (-1))\mathbf{i} + (2 + 4)\mathbf{j} + (2 + 1)\mathbf{k}$ $= 0\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}$ $= 6\mathbf{j} + 3\mathbf{k}$

Step 2: Compute the dot product $(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b})$

Using the formula: $\mathbf{v} \cdot \mathbf{w} = v_x w_x + v_y w_y + v_z w_z$

$(2, -2, 1) \cdot (0, 6, 3) = (2 \cdot 0) + (-2 \cdot 6) + (1 \cdot 3)$

$= 0 - 12 + 3 = -9$

So, $(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = -9$.

Step 3: Compute the angle between $\mathbf{a}$ and $\mathbf{b}$

The formula for the angle between two vectors is:

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

First, compute $\mathbf{a} \cdot \mathbf{b}$:

$(1,2,2) \cdot (-1,4,1) = (1 \cdot -1) + (2 \cdot 4) + (2 \cdot 1)$

$= -1 + 8 + 2 = 9$

Next, compute magnitudes:

$|\mathbf{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

$|\mathbf{b}| = \sqrt{(-1)^2 + 4^2 + 1^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$

Now, compute $\cos \theta$:

$\cos \theta = \frac{9}{3 \times 3\sqrt{2}} = \frac{9}{9\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Thus,

$\theta = \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) = 45^\circ$

Final Answers:

1. $(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = -9$
2. The angle between $\mathbf{a}$ and $\mathbf{b}$ is $45^\circ$.

Would you like any additional clarifications?

Related Questions:

1. How do you find the cross product of two vectors in 3D?
2. What is the geometric interpretation of the dot product?
3. How does the dot product help in determining vector projection?
4. What is the significance of the cross product in physics?
5. How can vectors be used in kinematics and physics?

Tip:

The dot product of two perpendicular vectors is always zero. If you find a dot product of zero, the vectors are orthogonal.