Why is this a basis for the subspace? 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 $0$).
 A: 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.
A: Note the following things: 


*

*Vectors in $O$ are linearly independent. 

*It's dimension is $2$. 

*And all vectors in $U$ can be formed using a linear combination of the vectors given in $O$.
A: Solve the system
$$\begin{pmatrix}-\frac15\\\frac25\\1\end{pmatrix}=s\begin{pmatrix}2\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\0\\1\end{pmatrix}=\begin{pmatrix}2s-t\\s\\t\end{pmatrix}$$
Check the above is soluble and that's all...
A: Let $u=(2,1,0)$ and $v=(-1,0,1)$. 
If you use "Gram-Schmidt" to get an orthogonal vector to $u$ using the vector $v$, you get 
$$w=v-\frac{v\cdot u}{u\cdot u} u=(-1,0,1)-\frac{-2}{5}(2,1,0)=(-\frac{1}{5},\frac{2}{5},1)=(-0.2,0.4,1)$$
So, you can see, that since $w$ is a (non-trivial) linear combination of $u,v$, then $\{u,w\}$ is also a basis of your space. Even more, the way we constructed $w$ gives you that this is an orthogonal basis, of course you can also see this just by checking $u\cdot w=0$.
