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### Solution for Problems

#### Problem 3: Find \( \mathbf{a} \cdot \mathbf{b} \)
Given:
- \( |\mathbf{a}| = 5 \)
- \( |\mathbf{b}| = \sqrt{14} \)
- The angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( 45^\circ \).

The dot product formula is:
\[
\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)
\]

Substitute the values:
\[
\mathbf{a} \cdot \mathbf{b} = 5 \cdot \sqrt{14} \cdot \cos(45^\circ)
\]

Since \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \), we have:
\[
\mathbf{a} \cdot \mathbf{b} = 5 \cdot \sqrt{14} \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{28}}{2}
\]

Simplify \( \sqrt{28} \) as \( 2\sqrt{7} \):
\[
\mathbf{a} \cdot \mathbf{b} = \frac{5 \cdot 2 \sqrt{7}}{2} = 5\sqrt{7}
\]

Thus, \( \mathbf{a} \cdot \mathbf{b} = 5\sqrt{7} \).

---

#### Problem 4: Determine the relationship between vectors

(a) \( \mathbf{u} = \langle -2, 4, -6 \rangle \), \( \mathbf{v} = \langle 8, 4, -1 \rangle \)

To check:
1. **Orthogonal**: Compute \( \mathbf{u} \cdot \mathbf{v} \):
\[
\mathbf{u} \cdot \mathbf{v} = (-2)(8) + (4)(4) + (-6)(-1) = -16 + 16 + 6 = 6
\]
Since \( \mathbf{u} \cdot \mathbf{v} \neq 0 \), \( \mathbf{u} \) and \( \mathbf{v} \) are **not orthogonal**.

2. **Parallel**: Check if \( \mathbf{v} = k\mathbf{u} \) for some \( k \). The components do not scale proportionally (\(-2 \neq 8k\), etc.), so they are **not parallel**.

Thus, \( \mathbf{u} \) and \( \mathbf{v} \) are **neither**.

---

(b) \( \mathbf{u} = 15\mathbf{i} - 12\mathbf{j} + 9\mathbf{k} \), \( \mathbf{v} = -10\mathbf{i} + 8\mathbf{j} - 6\mathbf{k} \)

Compute \( \mathbf{u} \cdot \mathbf{v} \):
\[
\mathbf{u} \cdot \mathbf{v} = (15)(-10) + (-12)(8) + (9)(-6) = -150 - 96 - 54 = -300
\]
Since \( \mathbf{u} \cdot \mathbf{v} \neq 0 \), \( \mathbf{u} \) and \( \mathbf{v} \) are **not orthogonal**.

For parallelism, the ratios \( \frac{15}{-10}, \frac{-12}{8}, \frac{9}{-6} \) are not equal, so \( \mathbf{u} \) and \( \mathbf{v} \) are **not parallel**.

Thus, \( \mathbf{u} \) and \( \mathbf{v} \) are **neither**.

---

(c) \( \mathbf{u} = \langle c, c, c \rangle \), \( \mathbf{v} = \langle c, 0, -c \rangle \)

Compute \( \mathbf{u} \cdot \mathbf{v} \):
\[
\mathbf{u} \cdot \mathbf{v} = c(c) + c(0) + c(-c) = c^2 - c^2 = 0
\]
Since \( \mathbf{u} \cdot \mathbf{v} = 0 \), \( \mathbf{u} \) and \( \mathbf{v} \) are **orthogonal**.

---

### Summary of Answers:
- **3.** \( 5\sqrt{7} \)
- **4a.** Neither
- **4b.** Neither
- **4c.** Orthogonal

---

Would you like further details on any step?  

### Related Questions:
1. How is the dot product related to the angle between two vectors?
2. Can orthogonality be determined without computing the dot product?
3. How can vectors be tested for parallelism geometrically?
4. What are the properties of unit vectors in terms of dot products?
5. How does the cross product differ from the dot product in vector analysis?

### Tip:
Always simplify square roots and trigonometric functions before substituting in formulas for easier calculations!

Find the dot product of vectors a and b with |a| = 5, |b| = √14, and the angle between them is 45°. Determine if given vectors in parts (a), (b), and (c) are orthogonal, parallel, or neither.

Learn how to find the dot product of vectors with given magnitudes and an angle of 45°. Also, determine whether vectors are orthogonal, parallel, or neither using dot product properties. Suitable for high school students in Grades 11-12.

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