Why is this a basis for the subspace

Question

A subspace $\mathcal{U}$ of $\mathbb{R}^3$ defined by the plane with equation $x-2y+z=0$. Letting $y = s$ and $z = t$, we have $x = 2s – t$ and all vectors $(x,y,z)$ in $\mathcal{U}$ can be given by
$$
\left[\begin{array}{r}
x\\
y\\
z
\end{array}\right]
=
s
\left[\begin{array}{r}
2\\
1\\
0
\end{array}\right]
+ t
\left[\begin{array}{r}
-1\\
0\\
1
\end{array}\right]
$$
Hence a basis for $\mathcal{U}$ is
$$
\mathcal{B} = \{\left[\begin{array}{r}
2\\
1\\
0
\end{array}\right], \left[\begin{array}{r}
-1\\
0\\
1
\end{array}\right]\}
$$
Now it is given that an orthogonal basis for $\mathcal{U}$ is
$$
\mathcal{O} =
\{
\left[\begin{array}{r}
2\\
1\\
0
\end{array}\right]
,
\left[\begin{array}{r}
-0.2\\
0.4\\
1
\end{array}\right]
\}
$$
How do I show that $\mathcal{O}$ is also a basis for $\mathcal{U}$? Because I cannot see how to derive that from the plane equation that defines the subspace (because the second component in the vector scaled by $t$ is $0$).

Answer 1

It should be clear that the vectors in $\mathcal O$ are linearly independent since one of the vectors has a null component, whereas the other has not.

Also, both vectors satisfy the equation $x-2y+z=0$, and the subspace defined by this equation has codimension $1$, hence dimension $2$ in a $3$-dimensional space, so a set of $2$ linearly independent vectors is a basis.