# Every Plane Through the Origin in the Three Dimensional Space is a Subspace ## Problem 294

Prove that every plane in the $3$-dimensional space $\R^3$ that passes through the origin is a subspace of $\R^3$. Add to solve later

## Proof.

Each plane $P$ in $\R^3$ through the origin is given by the equation
$ax+by+cz=0$ for some real numbers $a, b, c$.
That is, the plane $P$ is a set of vectors $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ satisfying the equation $ax+by+cz=0$:
$P=\left\{\, \begin{bmatrix} x \\ y \\ z \end{bmatrix}\in \R^3 \quad \middle| \quad ax+by+cz=0 \,\right \}.$

Now the equation can be written as the matrix equation
$A\mathbf{x}=\mathbf{0},$ where $A$ is the $1\times 3$ matrix $A$, $\mathbf{x}\in \R^3$, and $\mathbf{0}$ is the $1$-dimensional zero vector given by
$A=\begin{bmatrix} a & b & c \\ \end{bmatrix}, \mathbf{x}=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, \mathbf{0}=.$

Thus, the plane can be written as
\begin{align*}
P&=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\}\\
\end{align*}
and this is the definition of the nullspace $\calN(A)$ of $A$. That is
$P=\calN(A).$

Therefore, the plane $P$ is the nullspace of the $1\times 3$ matrix $A$.
Since the nullspace of a matrix is always a subspace, we conclude that the plane $P$ is a subspace of $\R^3$.
Therefore, every plane in $\R^3$ through the origin is a subspace of $\R^3$. Add to solve later

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