Let's go through the problem step by step to find both the projections requested.

Given:

$\mathbf{u} = \langle 3, -2, -1 \rangle \quad \text{and} \quad \mathbf{v} = \langle 2, -3, -1 \rangle$

1. Projection of $\mathbf{u}$ along $\mathbf{v}$:

The formula for the projection of vector $\mathbf{u}$ onto vector $\mathbf{v}$ is: $\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}$

Step 1: Dot product $\mathbf{u} \cdot \mathbf{v}$

$\mathbf{u} \cdot \mathbf{v} = (3)(2) + (-2)(-3) + (-1)(-1) = 6 + 6 + 1 = 13$

Step 2: Dot product $\mathbf{v} \cdot \mathbf{v}$

$\mathbf{v} \cdot \mathbf{v} = (2)^2 + (-3)^2 + (-1)^2 = 4 + 9 + 1 = 14$

Step 3: Projection of $\mathbf{u}$ onto $\mathbf{v}$

$\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{13}{14} \mathbf{v} = \frac{13}{14} \langle 2, -3, -1 \rangle$ Multiplying each component by $\frac{13}{14}$: $\text{proj}_{\mathbf{v}} \mathbf{u} = \langle \frac{26}{14}, -\frac{39}{14}, -\frac{13}{14} \rangle = \langle 1.857, -2.786, -0.929 \rangle$

2. Projection of $\mathbf{u}$ orthogonal to $\mathbf{v}$:

This projection is found using: $\mathbf{u}_{\perp \mathbf{v}} = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}$

Subtracting the projection from $\mathbf{u}$: $\mathbf{u}_{\perp \mathbf{v}} = \langle 3, -2, -1 \rangle - \langle 1.857, -2.786, -0.929 \rangle$ $\mathbf{u}_{\perp \mathbf{v}} = \langle 3 - 1.857, -2 + 2.786, -1 + 0.929 \rangle = \langle 1.143, 0.786, -0.071 \rangle$

Final Answers:

1. The projection of $\mathbf{u}$ along $\mathbf{v}$ is approximately: $\langle 1.857, -2.786, -0.929 \rangle$
2. The projection of $\mathbf{u}$ orthogonal to $\mathbf{v}$ is approximately: $\langle 1.143, 0.786, -0.071 \rangle$

Would you like further details or clarification on any steps?

Here are 5 related questions to explore further:

1. How can the projection formula be derived geometrically?
2. What is the significance of the dot product in vector projections?
3. How would the projections change if the vectors $\mathbf{u}$ and $\mathbf{v}$ were in 2D instead of 3D?
4. How do vector projections relate to the concept of orthogonality?
5. Can projections be used to solve systems of linear equations?

Tip: Vector projections are useful in breaking down vectors into components in the direction of another vector and perpendicular to it, often applied in physics and engineering problems.