Attempt 1: 10 attempts remaining. Let \( \vec{u}=\left[\begin{array}{c}1 \\ -7 \\ 0 \\ -4\end{array}\right], \vec{v}=\left[\begin{array}{l}0 \\ 0 \\ 1 \\ 7\end{array}\right] \), and let \( W \) the subspace of \( \mathbb{R}^{4} \) spanned by \( \{\vec{u}, \vec{v}\} \). Find a basis for \( W^{\perp} \). Answer: Submit answer

To find a basis for \( W^{\perp} \), the orthogonal complement of the subspace \( W \), we need to find all vectors \( \vec{w} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \in \mathbb{R}^4 \) that are orthogonal to both vectors \( \vec{u} \) and \( \vec{v} \) in \( W \). This means we need to solve the following system of equations derived from the dot product: 1. \( \vec{u} \cdot \vec{w} = 0 \) gives us: \[ 1x_1 - 7x_2 + 0x_3 - 4x_4 = 0 \quad \Rightarrow \quad x_1 - 7x_2 - 4x_4 = 0 \] 2. \( \vec{v} \cdot \vec{w} = 0 \) gives us: \[ 0x_1 + 0x_2 + 1x_3 + 7x_4 = 0 \quad \Rightarrow \quad x_3 + 7x_4 = 0 \quad \Rightarrow \quad x_3 = -7x_4 \] Now, we can express \( x_1 \) and \( x_3 \) in terms of \( x_2 \) and \( x_4 \): - From the first equation: \[ x_1 = 7x_2 + 4x_4 \] - From the second equation, we already have: \[ x_3 = -7x_4 \] Now we can write \( \vec{w} \) as: \[ \vec{w} = \begin{bmatrix} 7x_2 + 4x_4 \\ x_2 \\ -7x_4 \\ x_4 \end{bmatrix} \] This can be expressed as: \[ \vec{w} = x_2 \begin{bmatrix} 7 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 4 \\ 0 \\ -7 \\ 1 \end{bmatrix} \] Thus, a basis for \( W^{\perp} \) is given by the two vectors: \[ \left\{ \begin{bmatrix} 7 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 0 \\ -7 \\ 1 \end{bmatrix} \right\} \]