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To solve this problem, we need to compute the **projection of** \(\vec{a}\) **onto** \(\vec{b}\) and the **orthogonal complement** of \(\vec{a}\) with respect to \(\vec{b}\). The projection formula and the orthogonal component can be found as follows:

### Given:
\[
\vec{a} = \langle 7, 3, 6 \rangle
\]
\[
\vec{b} = \langle 3, 1, -5 \rangle
\]

### Step 1: Compute the Projection of \(\vec{a}\) onto \(\vec{b}\)
The formula for the projection of \(\vec{a}\) onto \(\vec{b}\) is:
\[
\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}
\]

1. **Dot Product** \(\vec{a} \cdot \vec{b}\):
   \[
   \vec{a} \cdot \vec{b} = (7)(3) + (3)(1) + (6)(-5) = 21 + 3 - 30 = -6
   \]

2. **Dot Product** \(\vec{b} \cdot \vec{b}\):
   \[
   \vec{b} \cdot \vec{b} = (3)^2 + (1)^2 + (-5)^2 = 9 + 1 + 25 = 35
   \]

3. **Compute** \(\text{proj}_{\vec{b}} \vec{a}\):
   \[
   \text{proj}_{\vec{b}} \vec{a} = \frac{-6}{35} \vec{b} = \frac{-6}{35} \langle 3, 1, -5 \rangle
   \]
   \[
   = \left\langle \frac{-18}{35}, \frac{-6}{35}, \frac{30}{35} \right\rangle
   \]
   Simplifying the fractions, we get:
   \[
   \text{proj}_{\vec{b}} \vec{a} = \left\langle -\frac{18}{35}, -\frac{6}{35}, \frac{6}{7} \right\rangle
   \]

### Step 2: Compute the Orthogonal Complement of \(\vec{a}\) with respect to \(\vec{b}\)
The orthogonal complement (or component) of \(\vec{a}\) with respect to \(\vec{b}\) is:
\[
\text{Orthogonal complement} = \vec{a} - \text{proj}_{\vec{b}} \vec{a}
\]

Substitute \(\vec{a} = \langle 7, 3, 6 \rangle\) and \(\text{proj}_{\vec{b}} \vec{a} = \left\langle -\frac{18}{35}, -\frac{6}{35}, \frac{6}{7} \right\rangle\):
\[
\text{Orthogonal complement} = \left\langle 7, 3, 6 \right\rangle - \left\langle -\frac{18}{35}, -\frac{6}{35}, \frac{6}{7} \right\rangle
\]
\[
= \left\langle 7 + \frac{18}{35}, 3 + \frac{6}{35}, 6 - \frac{6}{7} \right\rangle
\]

Simplifying each component:
1. First component: \(7 + \frac{18}{35} = \frac{245 + 18}{35} = \frac{263}{35}\)
2. Second component: \(3 + \frac{6}{35} = \frac{105 + 6}{35} = \frac{111}{35}\)
3. Third component: \(6 - \frac{6}{7} = \frac{42 - 6}{7} = \frac{36}{7}\)

So, the orthogonal complement is:
\[
\text{Orthogonal complement} = \left\langle \frac{263}{35}, \frac{111}{35}, \frac{36}{7} \right\rangle
\]

### Final Answer:
1. **Projection of** \(\vec{a}\) **onto** \(\vec{b}\):
   \[
   \text{proj}_{\vec{b}} \vec{a} = \left\langle -\frac{18}{35}, -\frac{6}{35}, \frac{6}{7} \right\rangle
   \]

2. **Orthogonal complement**:
   \[
   \text{Orthogonal complement} = \left\langle \frac{263}{35}, \frac{111}{35}, \frac{36}{7} \right\rangle
   \]

Would you like further details on any specific step?

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Here are some additional questions to deepen your understanding:
1. How would the process change if the vector \(\vec{b}\) were a unit vector?
2. Can the projection of \(\vec{a}\) onto \(\vec{b}\) ever equal \(\vec{a}\) itself? Under what conditions?
3. What does the orthogonal complement represent geometrically?
4. How would you find the angle between \(\vec{a}\) and \(\vec{b}\) using the dot product?
5. What would the result be if \(\vec{a}\) and \(\vec{b}\) were perpendicular?

**Tip:** The orthogonal complement of \(\vec{a}\) with respect to \(\vec{b}\) is always perpendicular to \(\vec{b}\).

Let \( \vec{a} = \langle 7, 3, 6 \rangle \) and \( \vec{b} = \langle 3, 1, -5 \rangle \). Compute the projection of \( \vec{a} \) onto \( \vec{b} \) and the vector component of \( \vec{a} \) orthogonal to \( \vec{b} \).

Learn how to compute the projection of vector \( \vec{a} = \langle 7, 3, 6 \rangle \) onto vector \( \vec{b} = \langle 3, 1, -5 \rangle \) and determine the orthogonal complement. This solution uses vector projection formulas and the concept of orthogonality, ideal for students in Grades 10-12.

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