Let's solve the problem step-by-step.

1. Compute the projection of $\vec{b}$ onto $\vec{a}$

Given: $\vec{b} = \langle 6, -6, -8 \rangle \quad \text{and} \quad \vec{a} = \langle -8, -3, 8 \rangle$

The formula for the projection of $\vec{b}$ onto $\vec{a}$ is: $\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{b} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a}$

Step 1: Compute $\vec{b} \cdot \vec{a}$

The dot product $\vec{b} \cdot \vec{a}$ is calculated as follows: $\vec{b} \cdot \vec{a} = (6)(-8) + (-6)(-3) + (-8)(8)$ Calculating each term: $6 \cdot (-8) = -48$ $-6 \cdot (-3) = 18$ $-8 \cdot 8 = -64$ So, $\vec{b} \cdot \vec{a} = -48 + 18 - 64 = -94$

Step 2: Compute $\vec{a} \cdot \vec{a}$

The dot product $\vec{a} \cdot \vec{a}$ is: $\vec{a} \cdot \vec{a} = (-8)^2 + (-3)^2 + 8^2$ Calculating each term: $(-8)^2 = 64$ $(-3)^2 = 9$ $8^2 = 64$ So, $\vec{a} \cdot \vec{a} = 64 + 9 + 64 = 137$

Step 3: Plug values into the projection formula

$\text{proj}_{\vec{a}} \vec{b} = \frac{-94}{137} \vec{a}$ Substitute $\vec{a} = \langle -8, -3, 8 \rangle$: $\text{proj}_{\vec{a}} \vec{b} = \frac{-94}{137} \langle -8, -3, 8 \rangle$ Distribute $\frac{-94}{137}$ to each component: $\text{proj}_{\vec{a}} \vec{b} = \left\langle \frac{-94 \cdot -8}{137}, \frac{-94 \cdot -3}{137}, \frac{-94 \cdot 8}{137} \right\rangle$ Simplify each term: $\text{proj}_{\vec{a}} \vec{b} = \left\langle \frac{752}{137}, \frac{282}{137}, \frac{-752}{137} \right\rangle$

2. Compute the orthogonal projection of $\vec{b}$ onto $\vec{a}$

The orthogonal projection (or the component of $\vec{b}$ orthogonal to $\vec{a}$) is: $\text{orth}_{\vec{a}} \vec{b} = \vec{b} - \text{proj}_{\vec{a}} \vec{b}$ Substitute $\vec{b} = \langle 6, -6, -8 \rangle$ and $\text{proj}_{\vec{a}} \vec{b} = \left\langle \frac{752}{137}, \frac{282}{137}, \frac{-752}{137} \right\rangle$: $\text{orth}_{\vec{a}} \vec{b} = \left\langle 6, -6, -8 \right\rangle - \left\langle \frac{752}{137}, \frac{282}{137}, \frac{-752}{137} \right\rangle$ This results in: $\text{orth}_{\vec{a}} \vec{b} = \left\langle 6 - \frac{752}{137}, -6 - \frac{282}{137}, -8 + \frac{752}{137} \right\rangle$ Simplify each component: $\text{orth}_{\vec{a}} \vec{b} = \left\langle \frac{30}{137}, -\frac{1104}{137}, \frac{-360}{137} \right\rangle$

Final Answers

1. $\text{proj}_{\vec{a}} \vec{b} = \left\langle \frac{752}{137}, \frac{282}{137}, \frac{-752}{137} \right\rangle$
2. $\text{orth}_{\vec{a}} \vec{b} = \left\langle \frac{30}{137}, -\frac{1104}{137}, \frac{-360}{137} \right\rangle$

Would you like further explanation or details on any step?

Related Questions:

1. How do we interpret the projection of one vector onto another geometrically?
2. How can the dot product help determine if two vectors are orthogonal?
3. What are the properties of an orthogonal projection?
4. How can vector projections be applied in physics?
5. What is the relationship between the projection of $\vec{b}$ onto $\vec{a}$ and the angle between them?

Tip:

Remember that the orthogonal projection of a vector onto another shows the component of the vector perpendicular to the given direction.