Question:medium

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

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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.
Updated On: Feb 18, 2026
  • \( [-2, -2, 2]^T \)
  • \( [3, 0, -1]^T \)
  • \( [3, 2, 1]^T \)
  • \( [1, 2, 3]^T \)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Plane Equation.The general vector equation for a plane is \( \vec{n} \cdot \vec{r} = d \), where \( \vec{n} \) represents the normal vector to the plane, \( \vec{r} \) is the position vector of a point on the plane, and \( d \) is a constant related to the plane's distance from the origin.
Step 2: Analyzing the Provided Equation.We are given \( W^T \cdot X = 1 \). Let's express this in vector form.Define \( W = \begin{pmatrix} 1
2
3 \end{pmatrix} \) and \( X = \begin{pmatrix} x
y
z \end{pmatrix} \).Thus, \( W^T = [1, 2, 3] \).The equation \( W^T \cdot X = 1 \) expands to the dot product \( [1, 2, 3] \begin{pmatrix} x
y
z \end{pmatrix} = 1x + 2y + 3z = 1 \), which is a plane equation.
Step 3: Identifying the Normal Vector.Comparing \( W^T \cdot X = 1 \) to the general form \( \vec{n} \cdot \vec{r} = d \), we identify the normal vector \( \vec{n} \) as \( W \).Since \( W^T = [1, 2, 3] \), the column vector \( W \) is \( \begin{pmatrix} 1
2
3 \end{pmatrix} \), or \( [1, 2, 3]^T \).Consequently, the normal vector to the plane is \( [1, 2, 3]^T \).
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