So we need to find two linearly independent vectors orthogonal to

\[\begin{bmatrix}\frac{1}{2}\\-\frac{1}{2}\\2\end{bmatrix}\]

\[\begin{bmatrix}1\\-1\\4\end{bmatrix}\] (or to avoid fractions). To do so, we can just use the definition of orthogonality:

\(\displaystyle{\left({x},{y},{z}\right)}\cdot{\left({1},-{1},{4}\right)}={x}-{y}+{4}{z}={0}\)

Let's let x=1 and y=0. Then we see that \(\displaystyle{z}=-{\frac{{{1}}}{{{4}}}}\). So

\[\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix}\]

is in W. Now let x=0 and y=1. Then \(\displaystyle{z}={\frac{{{1}}}{{{4}}}}\). So

\[\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\]

is also in W. Thus

\[W=\{\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix},\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\}\]

where

\[\{\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix},\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\}\]

Is a basis for W

\[\begin{bmatrix}\frac{1}{2}\\-\frac{1}{2}\\2\end{bmatrix}\]

\[\begin{bmatrix}1\\-1\\4\end{bmatrix}\] (or to avoid fractions). To do so, we can just use the definition of orthogonality:

\(\displaystyle{\left({x},{y},{z}\right)}\cdot{\left({1},-{1},{4}\right)}={x}-{y}+{4}{z}={0}\)

Let's let x=1 and y=0. Then we see that \(\displaystyle{z}=-{\frac{{{1}}}{{{4}}}}\). So

\[\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix}\]

is in W. Now let x=0 and y=1. Then \(\displaystyle{z}={\frac{{{1}}}{{{4}}}}\). So

\[\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\]

is also in W. Thus

\[W=\{\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix},\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\}\]

where

\[\{\begin{bmatrix}1\\0\\-\frac{1}{4}\end{bmatrix},\begin{bmatrix}0\\1\\\frac{1}{4}\end{bmatrix}\}\]

Is a basis for W