GATE Electrical 2023 | Question: 1

Question

For a given vector $\mathbf{w}=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{\mathrm{T}}$, the vector normal to the plane defined by $\mathbf{w}^{\mathrm{T}} \mathbf{x}=1$ is 

• A. $\left[\begin{array}{lll}-2 & -2 & 2\end{array}\right]^T$
• B. $\left[\begin{array}{lll}3 & 0 & -1\end{array}\right]^T$
• C. $\left[\begin{array}{lll}3 & 2 & 1\end{array}\right]^T$
• D. $\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^T$

Answer 1

The plane is defined by $w^T x = 1$, where $w = [1, 2, 3]^T$. This can be written in standard form as $x + 2y + 3z = 1$.

 Determine the Normal Vector 
For any plane given by $ax + by + cz = d$, the normal vector is simply the vector of coefficients $[a, b, c]^T$. Comparing this to the plane equation, the normal vector is $[1, 2, 3]^T$.

 Option D is corret.
Shortcut Trick: For any equation of the form $w^T x = k$, the vector $w$ is always the normal to that surface.
Important Rule: The normal to a plane is defined by the coefficients of its linear variables.