Question

For a given vector \( \mathbf{w} = [1 \; 2 \; 3]^T \), the vector normal to the plane defined by \( \mathbf{w}^T \mathbf{x} = 1 \) is:

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

Hint:

The vector multiplying \(\mathbf{x}\) in \(\mathbf{w}^T \mathbf{x} = c\) always represents the normal to the plane.

Answer 1

The Correct Option is D

Step 1: Recall equation of a plane.
The general equation of a plane in 3D is: \[ a x + b y + c z = d \] where the normal vector to the plane is \([a \; b \; c]^T\).

Step 2: Compare with given equation.
Here we are given: \[ \mathbf{w}^T \mathbf{x} = 1 \] with \(\mathbf{w} = [1 \; 2 \; 3]^T\). This expands to: \[ 1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3 = 1 \]

Step 3: Identify normal vector.
Thus the coefficients of \(x_1, x_2, x_3\) directly give the normal vector: \[ \mathbf{n} = [1 \; 2 \; 3]^T \]

Final Answer: \[ \boxed{[1 \; 2 \; 3]^T} \]