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We are given a plane \( T \) defined parametrically by the equation:

\[
\vec{x} = s \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + t \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix}
\]
where \( s, t \in \mathbb{R} \).

To determine which of the vectors are normal to \( T \), we need to find the normal vector to the plane. A normal vector is perpendicular to both direction vectors of the plane, which are:

\[
\vec{d_1} = \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}, \quad \vec{d_2} = \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix}
\]

The cross product of these two vectors will give us a normal vector \( \vec{n} \) to the plane.

First, let's compute the cross product \( \vec{n} = \vec{d_1} \times \vec{d_2} \).

\[
\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -1 \\ 4 & -1 & 0 \end{vmatrix}
\]

This determinant will give us the normal vector.

I will compute this now.The normal vector to the plane \( T \) is:

\[
\vec{n} = \begin{bmatrix} -1 \\ -4 \\ -13 \end{bmatrix}
\]

Now, we need to check if each of the given vectors is normal to the plane by checking if they are parallel to \( \vec{n} \). A vector is parallel to \( \vec{n} \) if it is a scalar multiple of \( \vec{n} \). We will evaluate each vector provided in the options:

1. \( \begin{bmatrix} 3 \\ 12 \\ 39 \end{bmatrix} \)
2. \( \begin{bmatrix} 2 \\ 8 \\ 26 \end{bmatrix} \)
3. \( \begin{bmatrix} -3 \\ -10 \\ -11 \end{bmatrix} \)
4. \( \begin{bmatrix} 4 \\ 11 \\ -18 \end{bmatrix} \)
5. \( \begin{bmatrix} -13 \\ -1 \\ -13 \end{bmatrix} \)

We will now check which vectors are scalar multiples of the normal vector \( \vec{n} \).Here are the results for each vector:

- **a.** \( \begin{bmatrix} 3 \\ 12 \\ 39 \end{bmatrix} \) is normal to \( T \): **True** (it is a scalar multiple of \( \vec{n} \)).
- **b.** \( \begin{bmatrix} 2 \\ 8 \\ 26 \end{bmatrix} \) is normal to \( T \): **True** (it is a scalar multiple of \( \vec{n} \)).
- **c.** \( \begin{bmatrix} -3 \\ -10 \\ -11 \end{bmatrix} \) is normal to \( T \): **False** (not a scalar multiple of \( \vec{n} \)).
- **d.** \( \begin{bmatrix} 4 \\ 11 \\ -18 \end{bmatrix} \) is normal to \( T \): **False** (not a scalar multiple of \( \vec{n} \)).
- **e.** \( \begin{bmatrix} -13 \\ -1 \\ -13 \end{bmatrix} \) is normal to \( T \): **False** (not a scalar multiple of \( \vec{n} \)).

Would you like more details on any part of this solution?

Here are some related questions to further explore this topic:
1. How do you find a normal vector using the cross product in different dimensions?
2. Can you always rely on scalar multiples to determine if vectors are parallel?
3. What are other methods to determine the orientation of a plane in 3D space?
4. How do we define the equation of a plane using points instead of direction vectors?
5. How would you apply this knowledge to solve real-world problems in physics or engineering?

**Tip:** Always verify the direction of your vectors when working with planes, as even small errors in sign can lead to incorrect conclusions!

Given that T is a plane defined by \( \vec{x} = s \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + t \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix}, \) where \( s, t \in \mathbb{R} \), which of the following vectors are normal to T?

Explore how to determine if a vector is normal to a plane in 3D using the cross product of direction vectors. This problem includes a step-by-step solution of finding the normal vector to the plane T.

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