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!