If A = ax + 3az and B = 5ax + 2ay - 6 az , find AB .

To find the angle \(\theta_{AB}\) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\), we can use the dot product formula:

\[
\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta_{AB})
\]

Where:
- \(\mathbf{A} \cdot \mathbf{B}\) is the dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\),
- \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are the magnitudes of vectors \(\mathbf{A}\) and \(\mathbf{B}\), respectively,
- \(\theta_{AB}\) is the angle between the two vectors.

### Step 1: Express the Vectors
Given:
\[
\mathbf{A} = a_x + 3a_z
\]
\[
\mathbf{B} = 5a_x + 2a_y - 6a_z
\]

### Step 2: Calculate the Dot Product \(\mathbf{A} \cdot \mathbf{B}\)
The dot product of \(\mathbf{A}\) and \(\mathbf{B}\) is calculated by multiplying corresponding components and summing them:

\[
\mathbf{A} \cdot \mathbf{B} = (1)(5) + (0)(2) + (3)(-6) = 5 + 0 - 18 = -13
\]

### Step 3: Calculate the Magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\)
The magnitude of a vector \(\mathbf{V} = v_x a_x + v_y a_y + v_z a_z\) is given by:

\[
|\mathbf{V}| = \sqrt{v_x^2 + v_y^2 + v_z^2}
\]

For \(\mathbf{A} = a_x + 3a_z\):
\[
|\mathbf{A}| = \sqrt{1^2 + 0^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
\]

For \(\mathbf{B} = 5a_x + 2a_y - 6a_z\):
\[
|\mathbf{B}| = \sqrt{5^2 + 2^2 + (-6)^2} = \sqrt{25 + 4 + 36} = \sqrt{65}
\]

### Step 4: Calculate the Angle \(\theta_{AB}\)
Using the dot product formula:

\[
\cos(\theta_{AB}) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-13}{\sqrt{10} \times \sqrt{65}} = \frac{-13}{\sqrt{650}}
\]

So, the angle \(\theta_{AB}\) is:

\[
\theta_{AB} = \cos^{-1}\left(\frac{-13}{\sqrt{650}}\right)
\]

Let's calculate this angle:

\[
\theta_{AB} \approx \cos^{-1}\left(\frac{-13}{\sqrt{650}}\right) \approx \cos^{-1}(-0.5085) \approx 120.54^\circ
\]

### Final Answer:
- **Angle \(\theta_{AB}\)**: \( \approx 120.54^\circ \)

Would you like further details or have any questions?

Here are five related questions:
1. How would the angle \(\theta_{AB}\) change if vector \(\mathbf{A}\) had a negative component in \(a_x\)?
2. What is the significance of the dot product in determining the angle between vectors?
3. How can the cross product be used to determine the orientation of two vectors?
4. What is the geometric interpretation of a negative dot product?
5. How does the angle change if vectors \(\mathbf{A}\) and \(\mathbf{B}\) were orthogonal?

**Tip:** When calculating angles between vectors, always ensure that you correctly determine the magnitude of each vector and their dot product.

Learn how to find the angle θ_AB between vectors A = ax + 3az and B = 5ax + 2ay - 6az using the dot product formula and vector magnitudes. This problem involves advanced concepts in vectors and trigonometry, suitable for advanced high school students.

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