The problem asks us to determine if each of the given vectors lies in the same plane as the vectors $\mathbf{u} = \langle 3, -3, -2 \rangle$ and $\mathbf{v} = \langle -2, 2, -2 \rangle$.

Approach:

1. The vectors $\mathbf{u}$ and $\mathbf{v}$ define a plane, and any vector that lies in this plane must be a linear combination of $\mathbf{u}$ and $\mathbf{v}$. This means that for any vector $\mathbf{w} = \langle a, b, c \rangle$, there must exist scalars $\lambda_1$ and $\lambda_2$ such that: $\mathbf{w} = \lambda_1 \mathbf{u} + \lambda_2 \mathbf{v}.$ If such scalars exist, then the vector $\mathbf{w}$ lies in the plane.

2. We can check this by setting up a system of equations for each vector, using the components of the given vectors.

Let's proceed:

1. Vector $\mathbf{w_1} = \langle 6, -6, 6 \rangle$:

We want to check if $\mathbf{w_1}$ can be expressed as $\lambda_1 \mathbf{u} + \lambda_2 \mathbf{v}$. This gives us the system: $\lambda_1 \cdot 3 + \lambda_2 \cdot (-2) = 6,$ $\lambda_1 \cdot (-3) + \lambda_2 \cdot 2 = -6,$ $\lambda_1 \cdot (-2) + \lambda_2 \cdot (-2) = 6.$ Solving this system, we find that no solution exists for $\lambda_1$ and $\lambda_2$. Hence, $\mathbf{w_1}$ does not lie in the plane.

2. Vector $\mathbf{w_2} = \langle -3, 3, -1 \rangle$:

Setting up the system: $\lambda_1 \cdot 3 + \lambda_2 \cdot (-2) = -3,$ $\lambda_1 \cdot (-3) + \lambda_2 \cdot 2 = 3,$ $\lambda_1 \cdot (-2) + \lambda_2 \cdot (-2) = -1.$ Solving this, we also find that there is no solution for $\lambda_1$ and $\lambda_2$. Thus, $\mathbf{w_2}$ does not lie in the plane.

3. Vector $\mathbf{w_3} = \langle 4, -4, -6 \rangle$:

For this vector, we set up the system: $\lambda_1 \cdot 3 + \lambda_2 \cdot (-2) = 4,$ $\lambda_1 \cdot (-3) + \lambda_2 \cdot 2 = -4,$ $\lambda_1 \cdot (-2) + \lambda_2 \cdot (-2) = -6.$ Solving this system, we find that it does have a solution for $\lambda_1$ and $\lambda_2$. Therefore, $\mathbf{w_3}$ lies in the plane.

4. Vector $\mathbf{w_4} = \langle -1, 2, 3 \rangle$:

Finally, we set up the system: $\lambda_1 \cdot 3 + \lambda_2 \cdot (-2) = -1,$ $\lambda_1 \cdot (-3) + \lambda_2 \cdot 2 = 2,$ $\lambda_1 \cdot (-2) + \lambda_2 \cdot (-2) = 3.$ After solving this, we find that no solution exists for $\lambda_1$ and $\lambda_2$. Therefore, $\mathbf{w_4}$ does not lie in the plane.

Summary:

• Vector 1: F
• Vector 2: F
• Vector 3: T
• Vector 4: F

Would you like more details on the steps used to solve this, or do you have any questions?

Additional Questions:

1. How do you solve a system of linear equations in three dimensions?
2. What is the geometric interpretation of a vector lying in a plane?
3. Can you explain the concept of linear combinations in more depth?
4. How is the cross product used to define a plane in vector space?
5. What does it mean when vectors are linearly independent?

Tip: If two vectors define a plane, any vector in the plane can be written as a linear combination of the two.