Let $\vec{u}=\left[\begin{array}{c}0 \\ 1 \\ 0 \\ -6\end{array}\right], \vec{v}=\left[\begin{array}{c}0 \\ 0 \\ 1 \\ -9\end{array}\right]$, and let $W$ the subspace of $\mathbb{R}^{4}$ spanned by $\{\vec{u}, \vec{v}\}$. Find a basis for $W^{\perp}$. Answer:

Step 1 ：Let \(\vec{u}=\left[\begin{array}{c}0 \ 1 \ 0 \ -6\end{array}\right], \vec{v}=\left[\begin{array}{c}0 \ 0 \ 1 \ -9\end{array}\right]\), and let \(W\) be the subspace of \(\mathbb{R}^{4}\) spanned by \(\{\vec{u}, \vec{v}\}\). We are to find a basis for \(W^{\perp}\).

Step 2 ：The subspace \(W^{\perp}\) is the orthogonal complement of \(W\), which means it consists of all vectors in \(\mathbb{R}^{4}\) that are orthogonal to every vector in \(W\). Since \(W\) is spanned by \(\vec{u}\) and \(\vec{v}\), we need to find vectors that are orthogonal to both \(\vec{u}\) and \(\vec{v}\). This means that the dot product of the vectors we are looking for with both \(\vec{u}\) and \(\vec{v}\) should be zero.

Step 3 ：We can set up a system of linear equations to find such vectors. The equations are \(x2 - 6*x4 = 0\) and \(x3 - 9*x4 = 0\).

Step 4 ：The solution to the system of equations is \(\{x2: 6*x4, x3: 9*x4\}\). This means that any vector in \(W^{\perp}\) can be written in the form \([x1, 6*x4, 9*x4, x4]\) for some values of \(x1\) and \(x4\).

Step 5 ：We can choose \(x1\) and \(x4\) to be 1 to get two basis vectors for \(W^{\perp}\). The basis vectors are \(\left[\begin{array}{c}1 \ 6 \ 9 \ 1\end{array}\right]\) and \(\left[\begin{array}{c}0 \ 0 \ 0 \ 1\end{array}\right]\).

Step 6 ：\(\boxed{\text{Final Answer: The basis for } W^{\perp} \text{ is } \left\{\left[\begin{array}{c}1 \ 6 \ 9 \ 1\end{array}\right], \left[\begin{array}{c}0 \ 0 \ 0 \ 1\end{array}\right]\right\}}\)