Let's go through each part of the question, which involves finding unit vectors under specified conditions.

(a) Find a unit vector in the same direction as $-3\mathbf{i} + 7\mathbf{j}$.

To find the unit vector in the same direction, follow these steps:

1. First, find the magnitude of the vector $-3\mathbf{i} + 7\mathbf{j}$: $\text{Magnitude} = \sqrt{(-3)^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$
2. The unit vector $\mathbf{u}$ is given by dividing the vector by its magnitude: $\mathbf{u} = \frac{-3\mathbf{i} + 7\mathbf{j}}{\sqrt{58}} = \frac{-3}{\sqrt{58}}\mathbf{i} + \frac{7}{\sqrt{58}}\mathbf{j}$ So, the answer is: $\mathbf{u} = \frac{-3}{\sqrt{58}} \mathbf{i} + \frac{7}{\sqrt{58}} \mathbf{j}$

(b) Find a unit vector oppositely directed to $18\mathbf{i} - 12\mathbf{j} + 12\mathbf{k}$.

To get the unit vector in the opposite direction:

1. First, calculate the magnitude of the vector $18\mathbf{i} - 12\mathbf{j} + 12\mathbf{k}$: $\text{Magnitude} = \sqrt{18^2 + (-12)^2 + 12^2} = \sqrt{324 + 144 + 144} = \sqrt{612}$
2. Simplify the square root of 612: $\sqrt{612} = 6\sqrt{17}$
3. To get the unit vector in the opposite direction, negate the vector and divide by the magnitude: $\mathbf{v} = \frac{-18\mathbf{i} + 12\mathbf{j} - 12\mathbf{k}}{6\sqrt{17}} = \frac{-3}{\sqrt{17}}\mathbf{i} + \frac{2}{\sqrt{17}}\mathbf{j} - \frac{2}{\sqrt{17}}\mathbf{k}$ So, the answer is: $\mathbf{v} = \frac{-3}{\sqrt{17}} \mathbf{i} + \frac{2}{\sqrt{17}} \mathbf{j} - \frac{2}{\sqrt{17}} \mathbf{k}$

(c) Find a unit vector in the same direction as the vector from point $A(-2, 0, 4)$ to point $B(3, 3, 3)$.

The vector from $A$ to $B$ is calculated as: $\mathbf{AB} = (3 - (-2))\mathbf{i} + (3 - 0)\mathbf{j} + (3 - 4)\mathbf{k} = 5\mathbf{i} + 3\mathbf{j} - \mathbf{k}$

1. The magnitude of this vector is: $\text{Magnitude} = \sqrt{5^2 + 3^2 + (-1)^2} = \sqrt{25 + 9 + 1} = \sqrt{35}$
2. The unit vector $\mathbf{w}$ is: $\mathbf{w} = \frac{5\mathbf{i} + 3\mathbf{j} - \mathbf{k}}{\sqrt{35}} = \frac{5}{\sqrt{35}}\mathbf{i} + \frac{3}{\sqrt{35}}\mathbf{j} - \frac{1}{\sqrt{35}}\mathbf{k}$ So, the answer is: $\mathbf{w} = \frac{5}{\sqrt{35}} \mathbf{i} + \frac{3}{\sqrt{35}} \mathbf{j} - \frac{1}{\sqrt{35}} \mathbf{k}$

Do you want more details on any of the steps? Here are some related questions you might find useful:

1. How do you normalize a vector in 3D space?
2. What is the difference between a vector and a unit vector?
3. Why do we need unit vectors, and how are they used in physics?
4. How do you calculate the direction of a vector between two points in space?
5. Can you explain why the magnitude of a vector is always positive?

Tip: When normalizing a vector, always divide by its magnitude to ensure it has a length of 1 while preserving its direction.