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Let's solve the given problem step by step.

### Part (a) - Finding two unit vectors parallel to \( v = 4i + 3j \)

A unit vector is a vector with magnitude 1. To find a unit vector in the direction of \( v = 4i + 3j \), we first find the magnitude of \( v \):

\[
\| v \| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

Now, the unit vector in the same direction as \( v \) is given by:

\[
\hat{v} = \frac{v}{\| v \|} = \frac{4i + 3j}{5} = \left(\frac{4}{5}\right)i + \left(\frac{3}{5}\right)j
\]

Thus, the parallel unit vector with the same direction is:

\[
\boxed{\left(\frac{4}{5}, \frac{3}{5}\right)}
\]

Additionally, there is another unit vector parallel but in the opposite direction, which is:

\[
-\hat{v} = -\left(\frac{4}{5}i + \frac{3}{5}j\right) = \left(\frac{-4}{5}, \frac{-3}{5}\right)
\]

Thus, the two unit vectors parallel to \( v \) are:

\[
\boxed{\left(\frac{4}{5}, \frac{3}{5}\right)} \quad \text{and} \quad \boxed{\left(\frac{-4}{5}, \frac{-3}{5}\right)}
\]

---

### Part (b) - Finding \( b \) if \( v = \frac{1}{8} b \) is a unit vector

Given that \( v = \frac{1}{8} b \) is a unit vector, the magnitude of \( v \) must be 1. Since \( v = \frac{1}{8} b \), we have:

\[
\left\| \frac{1}{8} b \right\| = 1
\]

This implies:

\[
\frac{1}{8} \| b \| = 1
\]

Multiplying both sides by 8, we get:

\[
\| b \| = 8
\]

Therefore, the magnitude of \( b \) is 8.

---

### Part (c) - Finding all values of \( a \) such that \( w = ai - \frac{a}{3} j \) is a unit vector

We are given \( w = ai - \frac{a}{3} j \) and need to find all values of \( a \) such that \( w \) is a unit vector. To do this, we calculate the magnitude of \( w \):

\[
\| w \| = \sqrt{a^2 + \left( \frac{a}{3} \right)^2} = \sqrt{a^2 + \frac{a^2}{9}} = \sqrt{\frac{9a^2}{9} + \frac{a^2}{9}} = \sqrt{\frac{10a^2}{9}} = \frac{\sqrt{10} |a|}{3}
\]

For \( w \) to be a unit vector, \( \| w \| = 1 \), so we set:

\[
\frac{\sqrt{10} |a|}{3} = 1
\]

Multiplying both sides by 3:

\[
\sqrt{10} |a| = 3
\]

Dividing both sides by \( \sqrt{10} \):

\[
|a| = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}
\]

Thus, \( a = \pm \frac{3\sqrt{10}}{10} \).

---

### Summary of Answers:

1. **(a)** The two unit vectors parallel to \( v \) are \( \left( \frac{4}{5}, \frac{3}{5} \right) \) and \( \left( \frac{-4}{5}, \frac{-3}{5} \right) \).
2. **(b)** The magnitude of \( b \) is 8.
3. **(c)** The values of \( a \) such that \( w \) is a unit vector are \( a = \pm \frac{3\sqrt{10}}{10} \).

Let me know if you want further details on any of the steps!

### Follow-up Questions:
1. What is a unit vector, and why is its magnitude always 1?
2. How do you find the direction of a vector if you know its components?
3. What does it mean for two vectors to be parallel?
4. How does scaling a vector affect its magnitude and direction?
5. How can you generalize the process of finding a unit vector for any given vector?

### Tip:
Always verify if a vector is a unit vector by ensuring its magnitude equals 1.

Complete parts (a) through (c) below. a. Find two unit vectors parallel to v = 4i + 3j. b. Find b if v = (1/8)b is a unit vector. c. Find all values of a such that w = ai - (a/3)j is a unit vector.

Learn how to find unit vectors parallel to a given vector, determine a scalar multiplier for a unit vector, and solve vector equations. This problem covers vector algebra concepts including unit vectors and vector magnitude. Suitable for students in Grades 11-12.

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