Question

For a given vector \( W^T = [1, 2, 3] \), the vector is normal to the plane defined by \( W^T \cdot X = 1 \).

• A. \( [-2, -2, 2]^T \)
• B. \( [3, 0, -1]^T \)
• C. \( [3, 2, 1]^T \)
• D. \( [1, 2, 3]^T \)

Hint:

In any plane equation of the form \(ax + by + cz = d\), the vector of coefficients \( \langle a, b, c \rangle \) is always the normal vector to the plane.

Answer 1

The Correct Option is D

Step 1: Understand the equation of a plane. The general vector equation of a plane is given by \( \vec{n} \cdot \vec{r} = d \), where \( \vec{n} \) is a vector normal (perpendicular) to the plane, \( \vec{r} \) is the position vector of any point on the plane, and \( d \) is a constant related to the distance of the plane from the origin.
Step 2: Analyze the given equation. The given equation is \( W^T \cdot X = 1 \). Let's write this in vector notation. Let \( W = \begin{pmatrix} 1
2
3 \end{pmatrix} \) and \( X = \begin{pmatrix} x
y
z \end{pmatrix} \). Then \( W^T = [1, 2, 3] \). The equation \( W^T \cdot X = 1 \) is the dot product \( [1, 2, 3] \begin{pmatrix} x
y
z \end{pmatrix} = 1x + 2y + 3z = 1 \). This is the standard form of a plane equation.
Step 3: Identify the normal vector. By comparing the given equation \( W^T \cdot X = 1 \) with the general form \( \vec{n} \cdot \vec{r} = d \), we can directly identify the normal vector \( \vec{n} \) as the vector \( W \). Given \( W^T = [1, 2, 3] \), the column vector \( W \) is \( \begin{pmatrix} 1
2
3 \end{pmatrix} \), or \( [1, 2, 3]^T \). Therefore, the vector normal to the plane is \( [1, 2, 3]^T \).