if you are given the magnitude of two vectors in a 3d space and the angle between them, how could you find the value of the dot product of those two vectors added and subtracted. so, if a and b are the vectors, how would you find (a+b) * (a-2b)? solve using the magnitude 1 for a and 3 for b. the angle between the vectors is pi/3 radians. answers should be in exact form

To solve the problem, we need to find the dot product \((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b})\) given the magnitudes of vectors \(\mathbf{a}\) and \(\mathbf{b}\) and the angle between them.

### Step-by-Step Solution

1. **Given Information:**
   - Magnitude of vector \(\mathbf{a}\): \(|\mathbf{a}| = 1\)
   - Magnitude of vector \(\mathbf{b}\): \(|\mathbf{b}| = 3\)
   - Angle between \(\mathbf{a}\) and \(\mathbf{b}\): \(\theta = \frac{\pi}{3}\)

2. **Dot Product Formula:**
   The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:
   \[
   \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)
   \]

   Substituting the given values:
   \[
   \mathbf{a} \cdot \mathbf{b} = (1)(3) \cos\left(\frac{\pi}{3}\right)
   \]

   We know that \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), so:
   \[
   \mathbf{a} \cdot \mathbf{b} = 3 \times \frac{1}{2} = \frac{3}{2}
   \]

3. **Dot Product of \((\mathbf{a} + \mathbf{b})\) and \((\mathbf{a} - 2\mathbf{b})\):**
   Expand the expression using the distributive property of dot products:
   \[
   (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - 2 \mathbf{b} \cdot \mathbf{b}
   \]

   Since the dot product is commutative (\(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\)), we can simplify this to:
   \[
   = \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - 2 \mathbf{b} \cdot \mathbf{b}
   \]

   \[
   = |\mathbf{a}|^2 - \mathbf{a} \cdot \mathbf{b} - 2|\mathbf{b}|^2
   \]

4. **Substitute Magnitudes and Dot Product:**
   - \(|\mathbf{a}|^2 = 1^2 = 1\)
   - \(|\mathbf{b}|^2 = 3^2 = 9\)
   - \(\mathbf{a} \cdot \mathbf{b} = \frac{3}{2}\)

   Plugging in these values:
   \[
   (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b}) = 1 - 3\left(\frac{3}{2}\right) - 2(9)
   \]

   Simplify the equation:
   \[
   = 1 - \frac{3}{2} - 18
   \]

   \[
   = 1 - \frac{3}{2} - 18
   \]

   Combine the terms:
   \[
   = 1 - \frac{3}{2} - 18 = \frac{2}{2} - \frac{3}{2} - \frac{36}{2}
   \]

   \[
   = \frac{2 - 3 - 36}{2} = \frac{-37}{2}
   \]

### Final Answer

The exact value of the dot product \((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - 2\mathbf{b})\) is \(\boxed{-\frac{37}{2}}\).

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### Additional Questions

1. How would the solution change if the angle between the vectors was different?
2. Can this method be extended to vectors in higher dimensions?
3. What is the geometric interpretation of the result in terms of the vectors?
4. How does the magnitude of vectors affect the dot product calculation?
5. What would the cross product of \((\mathbf{a} + \mathbf{b})\) and \((\mathbf{a} - 2\mathbf{b})\) look like?

**Tip:** Always remember to apply the distributive property carefully when expanding vector expressions, and verify that all components are correctly accounted for.

Learn how to calculate the dot product of two vectors in 3D space using given magnitudes and the angle between them. This advanced high school level problem involves vector algebra and trigonometry. Get the exact solution with step-by-step explanation.

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