Let's address the problem step by step. Given:

• $\mathbf{u} = -6\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}$
• $\mathbf{v} = 4\mathbf{j} + 4\mathbf{k}$

Part (a) - Finding the projection of $\mathbf{u}$ onto $\mathbf{v}$:

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: Compute $\mathbf{u} \cdot \mathbf{v}$ (dot product)

$\mathbf{u} \cdot \mathbf{v} = (-6\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}) \cdot (4\mathbf{j} + 4\mathbf{k})$

Since dot products of orthogonal components (like $\mathbf{i} \cdot \mathbf{j}$) are zero, we only focus on the components that match:

$\mathbf{u} \cdot \mathbf{v} = (-4)(4) + (-2)(4) = -16 - 8 = -24$

Step 2: Compute $\mathbf{v} \cdot \mathbf{v}$ (magnitude squared of $\mathbf{v}$)

$\mathbf{v} \cdot \mathbf{v} = (4\mathbf{j} + 4\mathbf{k}) \cdot (4\mathbf{j} + 4\mathbf{k}) = 4^2 + 4^2 = 16 + 16 = 32$

Step 3: Compute the projection

$\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{-24}{32} \mathbf{v} = -\frac{3}{4} \mathbf{v}$

Substitute $\mathbf{v} = 4\mathbf{j} + 4\mathbf{k}$:

$\text{proj}_{\mathbf{v}} \mathbf{u} = -\frac{3}{4} (4\mathbf{j} + 4\mathbf{k}) = -3\mathbf{j} - 3\mathbf{k}$

Thus, the projection of $\mathbf{u}$ onto $\mathbf{v}$ is:

$\text{proj}_{\mathbf{v}} \mathbf{u} = -3\mathbf{j} - 3\mathbf{k}$

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Part (b) - Finding the vector component of $\mathbf{u}$ orthogonal to $\mathbf{v}$:

The vector component of $\mathbf{u}$ orthogonal to $\mathbf{v}$ is given by:

$\mathbf{u}_{\perp \mathbf{v}} = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}$

We already know $\mathbf{u} = -6\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}$ and $\text{proj}_{\mathbf{v}} \mathbf{u} = -3\mathbf{j} - 3\mathbf{k}$.

Now, subtract the projection from $\mathbf{u}$:

$\mathbf{u}_{\perp \mathbf{v}} = (-6\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}) - (-3\mathbf{j} - 3\mathbf{k})$

Simplifying:

$\mathbf{u}_{\perp \mathbf{v}} = -6\mathbf{i} - 4\mathbf{j} - 2\mathbf{k} + 3\mathbf{j} + 3\mathbf{k}$

$\mathbf{u}_{\perp \mathbf{v}} = -6\mathbf{i} - \mathbf{j} + \mathbf{k}$

Thus, the vector component of $\mathbf{u}$ orthogonal to $\mathbf{v}$ is:

$\mathbf{u}_{\perp \mathbf{v}} = -6\mathbf{i} - \mathbf{j} + \mathbf{k}$

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Let me know if you'd like more details or have any questions!

Here are five related questions:

1. How do you calculate the magnitude of a vector in 3D?
2. What is the physical meaning of the projection of one vector onto another?
3. How would the result change if both vectors were unit vectors?
4. Can you derive the vector component parallel to another vector using cross products?
5. What is the significance of orthogonal vectors in applications like physics or computer graphics?

Tip: The projection of a vector gives you the component of one vector that lies in the direction of another. It's especially useful in physics for breaking forces into components!