To solve the problem of finding a unit vector in the same direction as the given vector, we need to normalize the vector. This process involves two main steps:

1. Calculate the magnitude (or norm) of the vector:
   The magnitude of a vector $v = ai + bj$ is given by: $|v| = \sqrt{a^2 + b^2}$
2. Divide the vector by its magnitude:
   A unit vector $u$ in the same direction as $v$ is given by: $u = \frac{v}{|v|}$ This means that each component of the vector $v$ is divided by its magnitude.

Now, let's solve a few of the problems from the image step by step:

Problem 27: $v = 6i + 10j$

1. Magnitude: $|v| = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} = 2\sqrt{34}$
2. Unit Vector: $u = \frac{v}{|v|} = \frac{6i + 10j}{2\sqrt{34}} = \frac{6}{2\sqrt{34}}i + \frac{10}{2\sqrt{34}}j = \frac{3}{\sqrt{34}}i + \frac{5}{\sqrt{34}}j$ So the unit vector is: $u = \frac{3}{\sqrt{34}}i + \frac{5}{\sqrt{34}}j$

Problem 28: $v = 4i - 6j$

1. Magnitude: $|v| = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$
2. Unit Vector: $u = \frac{v}{|v|} = \frac{4i - 6j}{2\sqrt{13}} = \frac{4}{2\sqrt{13}}i - \frac{6}{2\sqrt{13}}j = \frac{2}{\sqrt{13}}i - \frac{3}{\sqrt{13}}j$ So the unit vector is: $u = \frac{2}{\sqrt{13}}i - \frac{3}{\sqrt{13}}j$

Problem 29: $v = i - j$

1. Magnitude: $|v| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$
2. Unit Vector: $u = \frac{v}{|v|} = \frac{i - j}{\sqrt{2}} = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}}j$ So the unit vector is: $u = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}}j$

Would you like to solve more from the list or need further clarifications?

Here are 5 follow-up questions:

1. How do you find the magnitude of a 3D vector?
2. What is the unit vector in the opposite direction of a given vector?
3. Can you apply this method to vectors in higher dimensions?
4. How is this concept used in physics, for example, when working with forces?
5. How does changing the vector's direction affect its unit vector?

Tip: When normalizing vectors, you can simplify calculations by rationalizing denominators if the magnitude has a square root.