Question

Let \( S = \{(x, y, z) \in \mathbb{R}^3 \setminus \{(0,0,0)\} : z = -(x + y)\} \). Denote:

\[ S^\perp = \{(p, q, r) \in \mathbb{R}^3 : px + qy + rz = 0 \text{ for all } (x, y, z) \in S\}. \]

Then which one of the following options is correct?

• A. \( S^\perp \) is not a subspace of \( \mathbb{R}^3 \)
• B. \( S^\perp = \{(0,0,0)\} \)
• C. \( \dim(S^\perp) = 1 \)
• D. \( \dim(S^\perp) = 2 \)

Hint:

To find the orthogonal complement \( S^\perp \) of a subspace \( S \), solve the system of equations \( p x + q y + r z = 0 \) for all points in \( S \). The result will give the set of vectors that are orthogonal to every vector in \( S \).

Answer 1

The Correct Option is C

Step 1: The equation of \( S \) is given by \( z = -(x + y) \), so any point \( (x, y, z) \) in \( S \) can be written as \( (x, y, -(x + y)) \). Thus, the set \( S \) is the plane given by the equation \( z = -(x + y) \) in \( \mathbb{R}^3 \).

Step 2: To find \( S^\perp \), we need to find all vectors \( (p, q, r) \) such that \( px + qy + rz = 0 \) for all points \( (x, y, z) \in S \). Using \( z = -(x + y) \), we substitute into the equation:

\[ px + qy + r (-(x + y)) = 0 \]
This simplifies to:
\[ px + qy - r x - r y = 0 \quad \Rightarrow \quad (p - r)x + (q - r)y = 0 \]
For this to hold for all values of \( x \) and \( y \), we must have:
\[ p - r = 0 \quad \text{and} \quad q - r = 0 \]
Thus, \( p = r \) and \( q = r \), so the vector \( (p, q, r) \) must be of the form \( (r, r, r) \) for some scalar \( r \).

Step 3: Therefore, \[ S^\perp = \{(r, r, r) : r \in \mathbb{R}\} \] which is a one-dimensional subspace of \( \mathbb{R}^3 \).
Thus, \( \dim(S^\perp) = 1 \).