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This problem is about expressing the vector \(\vec{CH}\) in terms of vectors \(\vec{a}\) and \(\vec{b}\) for a triangle \(ABC\), where:

- \(\vec{CA} = \vec{a}\) and \(\vec{CB} = \vec{b}\)
- \(|\vec{a}| = 3\), \(|\vec{b}| = \sqrt{2}\), and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\)
- \(CH\) is the height of the triangle, and \(H \in AB\) (meaning \(H\) lies on line segment \(AB\))

### Goal:
We need to express the vector \(\vec{CH}\) in terms of vectors \(\vec{a}\) and \(\vec{b}\).

### Step-by-step approach:

1. **Identify the properties of the height**:
   The vector \(\vec{CH}\) is perpendicular to \(AB\), meaning \(\vec{CH} \perp \vec{AB}\). We can find vector \(\vec{AB}\) as:
   \[
   \vec{AB} = \vec{CB} - \vec{CA} = \vec{b} - \vec{a}
   \]
   Since \(\vec{CH}\) is perpendicular to \(\vec{AB}\), the dot product between \(\vec{CH}\) and \(\vec{AB}\) is zero:
   \[
   \vec{CH} \cdot \vec{AB} = 0
   \]
   This will give us a relationship between the components of \(\vec{CH}\), which we can express as a linear combination of \(\vec{a}\) and \(\vec{b}\).

2. **Express \(\vec{CH}\) as a linear combination**:
   We assume that \(\vec{CH} = \alpha \vec{a} + \beta \vec{b}\), where \(\alpha\) and \(\beta\) are scalars. Substituting this into the dot product condition:
   \[
   (\alpha \vec{a} + \beta \vec{b}) \cdot (\vec{b} - \vec{a}) = 0
   \]
   Expanding this expression:
   \[
   \alpha \vec{a} \cdot \vec{b} - \alpha \vec{a} \cdot \vec{a} + \beta \vec{b} \cdot \vec{b} - \beta \vec{b} \cdot \vec{a} = 0
   \]
   Using the given magnitudes and dot products:
   - \(\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 9\)
   - \(\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 2\)
   - \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{4}\right) = 3 \times \sqrt{2} \times \frac{\sqrt{2}}{2} = 3\)

   Substituting these into the equation:
   \[
   \alpha(3 - 9) + \beta(2 - 3) = 0
   \]
   Simplifying:
   \[
   -6\alpha - \beta = 0
   \]
   Therefore, \(\beta = -6\alpha\).

3. **Express \(\vec{CH}\)**:
   Substituting \(\beta = -6\alpha\) into the expression for \(\vec{CH}\):
   \[
   \vec{CH} = \alpha \vec{a} - 6\alpha \vec{b}
   \]
   Factoring out \(\alpha\):
   \[
   \vec{CH} = \alpha (\vec{a} - 6 \vec{b})
   \]
   The scalar \(\alpha\) can be determined based on additional information about the geometry of the triangle or the length of \(CH\), but the vector form of \(\vec{CH}\) in terms of \(\vec{a}\) and \(\vec{b}\) is:
   \[
   \boxed{\vec{CH} = \alpha (\vec{a} - 6 \vec{b})}
   \]

Would you like to go over any particular step in more detail? Here are 5 related questions to explore further:

1. How can we determine the value of \(\alpha\)?
2. What is the significance of the dot product condition \(\vec{CH} \cdot \vec{AB} = 0\)?
3. Can the length of \(\vec{CH}\) be found geometrically from the triangle's properties?
4. How does the angle between \(\vec{a}\) and \(\vec{b}\) affect the solution?
5. How would the solution change if the angle between \(\vec{a}\) and \(\vec{b}\) were different?

**Tip:** When solving geometry problems involving vectors, identifying perpendicular relationships (e.g., using dot products) can simplify your approach.

Given a triangle ABC, express vector CH in terms of vectors a and b, where CA = a, CB = b. Given that |a| = 3, |b| = √2, and the angle between a and b is π/4, CH is the height of the triangle with H on line AB.

Learn how to express the vector CH in a triangle ABC in terms of vectors a and b. This problem covers key concepts in vector algebra, including dot products, and triangle geometry. Suitable for advanced high school or college students.

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