Find the orthogonal projection of \( \mathbf{v}=\left[\begin{array}{c}-11 \\ -13 \\ 1\end{array}\right] \) onto the subspace \( V \) of \( \mathbb{R}^{3} \) spanned by \( \left[\begin{array}{c}3 \\ -4 \\ -1\end{array}\right] \) and \( \left[\begin{array}{c}-2 \\ -4 \\ -2\end{array}\right] \). \( \operatorname{proj}_{v}(\mathbf{v})=\left[\begin{array}{c}\square \\ \hline\end{array}\right] \)

To find the orthogonal projection of \( \mathbf{v} = \begin{bmatrix} -11 \\ -13 \\ 1 \end{bmatrix} \) onto the subspace \( V \) spanned by the vectors \( \mathbf{u_1} = \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix} \) and \( \mathbf{u_2} = \begin{bmatrix} -2 \\ -4 \\ -2 \end{bmatrix} \), we first need to check if these two vectors are linearly independent. 1. **Checking Linear Independence**: To see if \( \mathbf{u_1} \) and \( \mathbf{u_2} \) are independent, we can find a scalar \( k \) such that \( k \mathbf{u_1} = \mathbf{u_2} \): \[ k \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix} = \begin{bmatrix} -2 \\ -4 \\ -2 \end{bmatrix} \] This gives: \[ 3k = -2 \quad \Rightarrow \quad k = -\frac{2}{3} \] Now check the other components: \[ -4k = -4 \quad \Rightarrow \quad k = 1 \quad (\text{which is inconsistent}) \] Hence, \( \mathbf{u_1} \) and \( \mathbf{u_2} \) are linearly independent. 2. **Finding the Orthonormal Basis**: We can use the Gram-Schmidt process: - Let \( \mathbf{e_1} = \mathbf{u_1} = \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix} \). - To find \( \mathbf{e_2} \), we need to make \( \mathbf{u_2} \) orthogonal to \( \mathbf{e_1} \): \[ \mathbf{e_2} = \mathbf{u_2} - \operatorname{proj}_{\mathbf{e_1}}(\mathbf{u_2}) = \mathbf{u_2} - \frac{\mathbf{u_2} \cdot \mathbf{e_1}}{\mathbf{e_1} \cdot \mathbf{e_1}} \mathbf{e_1} \] First calculate the dot products: \[ \mathbf{u_2} \cdot \mathbf{e_1} = (-2)(3) + (-4)(-4) + (-2)(-1) = -6 + 16 + 2 = 12 \] \[ \mathbf{e_1} \cdot \mathbf{e_1} = 3^2 + (-4)^2 + (-1)^2 = 9 + 16 + 1 = 26 \] So, \[ \operatorname{proj}_{\mathbf{e_1}}(\mathbf{u_2}) = \frac{12}{26} \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix} = \frac{6}{13} \begin{bmatrix} 3 \\ -4 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{18}{13} \\ -\frac{24}{13} \\ -\frac{6}{13} \end{bmatrix} \] Then, \[ \mathbf{e_2} = \begin{bmatrix} -2 \\ -4 \\ -2 \end{bmatrix} - \begin{bmatrix} \frac{18}{13} \\ -\frac{24}{13} \\ -\frac{6}{13} \end{bmatrix} = \begin{bmatrix} -2 - \frac{18}{13} \\ -4 + \frac{24}{13} \\ -2 + \frac{6}{13} \end{bmatrix} \] Now compute each component: \[ = \begin{bmatrix} -\frac{26}{13} - \frac{18}{13} \\ -\frac{52}{13} + \frac{24}{13} \\ -\frac{26}{13} + \frac{6}{13} \end{bmatrix} = \begin{bmatrix} -\frac{44}{13} \\ -\frac{28}{13}