1. **Problem:** Use the Gram-Schmidt process to transform the basis \(v_1 = \begin{pmatrix}6 \\ 3 \\ 6\end{pmatrix}, v_2 = \begin{pmatrix}5 \\ 4 \\ 3\end{pmatrix}, v_3 = \begin{pmatrix}-1 \\ 1 \\ 5\end{pmatrix}\) into an orthonormal basis. 2. **Formula and rules:** The Gram-Schmidt process orthogonalizes vectors \(v_1, v_2, v_3\) into \(u_1, u_2, u_3\) by: \[ u_1 = v_1 \] \[ u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = v_2 - \frac{v_2 \cdot u_1}{u_1 \cdot u_1} u_1 \] \[ u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = v_3 - \frac{v_3 \cdot u_1}{u_1 \cdot u_1} u_1 - \frac{v_3 \cdot u_2}{u_2 \cdot u_2} u_2 \] Then normalize each \(u_i\) to get orthonormal vectors \(e_i = \frac{u_i}{\|u_i\|}\). 3. **Step 1: Compute \(u_1\):** \[ u_1 = v_1 = \begin{pmatrix}6 \\ 3 \\ 6\end{pmatrix} \] 4. **Step 2: Compute projection of \(v_2\) on \(u_1\):** \[ v_2 \cdot u_1 = 5 \times 6 + 4 \times 3 + 3 \times 6 = 30 + 12 + 18 = 60 \] \[ u_1 \cdot u_1 = 6^2 + 3^2 + 6^2 = 36 + 9 + 36 = 81 \] \[ \mathrm{proj}_{u_1}(v_2) = \frac{60}{81} u_1 = \frac{20}{27} \begin{pmatrix}6 \\ 3 \\ 6\end{pmatrix} = \begin{pmatrix} \frac{120}{27} \\ \frac{60}{27} \\ \frac{120}{27} \end{pmatrix} = \begin{pmatrix} \frac{40}{9} \\ \frac{20}{9} \\ \frac{40}{9} \end{pmatrix} \] 5. **Step 3: Compute \(u_2\):** \[ u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = \begin{pmatrix}5 \\ 4 \\ 3\end{pmatrix} - \begin{pmatrix} \frac{40}{9} \\ \frac{20}{9} \\ \frac{40}{9} \end{pmatrix} = \begin{pmatrix} 5 - \frac{40}{9} \\ 4 - \frac{20}{9} \\ 3 - \frac{40}{9} \end{pmatrix} = \begin{pmatrix} \frac{45}{9} - \frac{40}{9} \\ \frac{36}{9} - \frac{20}{9} \\ \frac{27}{9} - \frac{40}{9} \end{pmatrix} = \begin{pmatrix} \frac{5}{9} \\ \frac{16}{9} \\ -\frac{13}{9} \end{pmatrix} \] 6. **Step 4: Compute projection of \(v_3\) on \(u_1\):** \[ v_3 \cdot u_1 = (-1) \times 6 + 1 \times 3 + 5 \times 6 = -6 + 3 + 30 = 27 \] \[ \mathrm{proj}_{u_1}(v_3) = \frac{27}{81} u_1 = \frac{1}{3} \begin{pmatrix}6 \\ 3 \\ 6\end{pmatrix} = \begin{pmatrix}2 \\ 1 \\ 2\end{pmatrix} \] 7. **Step 5: Compute projection of \(v_3\) on \(u_2\):** \[ u_2 \cdot u_2 = \left(\frac{5}{9}\right)^2 + \left(\frac{16}{9}\right)^2 + \left(-\frac{13}{9}\right)^2 = \frac{25}{81} + \frac{256}{81} + \frac{169}{81} = \frac{450}{81} = \frac{50}{9} \] \[ v_3 \cdot u_2 = (-1) \times \frac{5}{9} + 1 \times \frac{16}{9} + 5 \times \left(-\frac{13}{9}\right) = -\frac{5}{9} + \frac{16}{9} - \frac{65}{9} = -\frac{54}{9} = -6 \] \[ \mathrm{proj}_{u_2}(v_3) = \frac{-6}{\frac{50}{9}} u_2 = -6 \times \frac{9}{50} u_2 = -\frac{54}{50} u_2 = -\frac{27}{25} u_2 \] \[ = -\frac{27}{25} \begin{pmatrix} \frac{5}{9} \\ \frac{16}{9} \\ -\frac{13}{9} \end{pmatrix} = \begin{pmatrix} -\frac{27}{25} \times \frac{5}{9} \\ -\frac{27}{25} \times \frac{16}{9} \\ -\frac{27}{25} \times -\frac{13}{9} \end{pmatrix} = \begin{pmatrix} -\frac{135}{225} \\ -\frac{432}{225} \\ \frac{351}{225} \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} \\ -\frac{96}{125} \\ \frac{39}{25} \end{pmatrix} \] 8. **Step 6: Compute \(u_3\):** \[ u_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = \begin{pmatrix}-1 \\ 1 \\ 5\end{pmatrix} - \begin{pmatrix}2 \\ 1 \\ 2\end{pmatrix} - \begin{pmatrix} -\frac{3}{5} \\ -\frac{96}{125} \\ \frac{39}{25} \end{pmatrix} = \begin{pmatrix} -1 - 2 + \frac{3}{5} \\ 1 - 1 + \frac{96}{125} \\ 5 - 2 - \frac{39}{25} \end{pmatrix} = \begin{pmatrix} -3 + \frac{3}{5} \\ 0 + \frac{96}{125} \\ 3 - \frac{39}{25} \end{pmatrix} \] \[ = \begin{pmatrix} -\frac{15}{5} + \frac{3}{5} \\ \frac{96}{125} \\ \frac{75}{25} - \frac{39}{25} \end{pmatrix} = \begin{pmatrix} -\frac{12}{5} \\ \frac{96}{125} \\ \frac{36}{25} \end{pmatrix} \] 9. **Step 7: Normalize \(u_1, u_2, u_3\) to get orthonormal basis \(e_1, e_2, e_3\):** \[ \|u_1\| = \sqrt{81} = 9, \quad e_1 = \frac{1}{9} \begin{pmatrix}6 \\ 3 \\ 6\end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{pmatrix} \] \[ \|u_2\| = \sqrt{\frac{50}{9}} = \frac{5 \sqrt{2}}{3}, \quad e_2 = \frac{1}{\|u_2\|} u_2 = \frac{3}{5 \sqrt{2}} \begin{pmatrix} \frac{5}{9} \\ \frac{16}{9} \\ -\frac{13}{9} \end{pmatrix} = \begin{pmatrix} \frac{1}{3 \sqrt{2}} \\ \frac{16}{15 \sqrt{2}} \\ -\frac{13}{15 \sqrt{2}} \end{pmatrix} \] \[ \|u_3\| = \sqrt{ \left(-\frac{12}{5}\right)^2 + \left(\frac{96}{125}\right)^2 + \left(\frac{36}{25}\right)^2 } = \sqrt{ \frac{144}{25} + \frac{9216}{15625} + \frac{1296}{625} } \approx 2.56 \] \[ e_3 = \frac{1}{2.56} \begin{pmatrix} -\frac{12}{5} \\ \frac{96}{125} \\ \frac{36}{25} \end{pmatrix} \approx \begin{pmatrix} -0.94 \\ 0.30 \\ 0.56 \end{pmatrix} \] **Final orthonormal basis:** \[ e_1 = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{pmatrix}, \quad e_2 = \begin{pmatrix} \frac{1}{3 \sqrt{2}} \\ \frac{16}{15 \sqrt{2}} \\ -\frac{13}{15 \sqrt{2}} \end{pmatrix}, \quad e_3 \approx \begin{pmatrix} -0.94 \\ 0.30 \\ 0.56 \end{pmatrix} \] This completes the Gram-Schmidt orthonormalization of the first problem. **Note:** Only the first problem is solved as per instructions.