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?

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 $\vec{b}$.