1. **Problem Statement:** Given vectors $v_1 = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$, $v_2 = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$, and $v_3 = \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}$, we want to find an orthogonal set of vectors using the Gram-Schmidt process. 2. **Formula and Rules:** The Gram-Schmidt process constructs an orthogonal set $\{u_1, u_2, u_3\}$ from $\{v_1, v_2, v_3\}$ as follows: $$ u_1 = v_1 $$ $$ u_k = v_k - \sum_{j=1}^{k-1} \mathrm{proj}_{u_j}(v_k) \quad \text{for } k=2,3 $$ where the projection is $$ \mathrm{proj}_{u_j}(v_k) = \frac{v_k \cdot u_j}{u_j \cdot u_j} u_j $$ 3. **Step 1: Compute $u_1$** $$ u_1 = v_1 = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$$ 4. **Step 2: Compute $u_2$** Calculate the projection of $v_2$ onto $u_1$: $$v_2 \cdot u_1 = 1 \times 1 + 0 \times 1 + 1 \times 0 = 1$$ $$u_1 \cdot u_1 = 1^2 + 1^2 + 0^2 = 2$$ $$\mathrm{proj}_{u_1}(v_2) = \frac{1}{2} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix}0.5 \\ 0.5 \\ 0\end{bmatrix}$$ Then, $$ u_2 = v_2 - \mathrm{proj}_{u_1}(v_2) = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} - \begin{bmatrix}0.5 \\ 0.5 \\ 0\end{bmatrix} = \begin{bmatrix}0.5 \\ -0.5 \\ 1\end{bmatrix}$$ 5. **Step 3: Compute $u_3$** Calculate projections of $v_3$ onto $u_1$ and $u_2$: $$v_3 \cdot u_1 = 0 \times 1 + 1 \times 1 + 1 \times 0 = 1$$ $$u_1 \cdot u_1 = 2$$ $$\mathrm{proj}_{u_1}(v_3) = \frac{1}{2} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix}0.5 \\ 0.5 \\ 0\end{bmatrix}$$ Next, $$v_3 \cdot u_2 = 0 \times 0.5 + 1 \times (-0.5) + 1 \times 1 = 0.5$$ $$u_2 \cdot u_2 = 0.5^2 + (-0.5)^2 + 1^2 = 0.25 + 0.25 + 1 = 1.5$$ $$\mathrm{proj}_{u_2}(v_3) = \frac{0.5}{1.5} \begin{bmatrix}0.5 \\ -0.5 \\ 1\end{bmatrix} = \frac{1}{3} \begin{bmatrix}0.5 \\ -0.5 \\ 1\end{bmatrix} = \begin{bmatrix}0.1667 \\ -0.1667 \\ 0.3333\end{bmatrix}$$ Finally, $$\nu_3 = v_3 - \mathrm{proj}_{u_1}(v_3) - \mathrm{proj}_{u_2}(v_3) = \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} - \begin{bmatrix}0.5 \\ 0.5 \\ 0\end{bmatrix} - \begin{bmatrix}0.1667 \\ -0.1667 \\ 0.3333\end{bmatrix} = \begin{bmatrix}-0.6667 \\ 0.6667 \\ 0.6667\end{bmatrix}$$ 6. **Result:** The orthogonal set is $$ u_1 = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \quad \nu_2 = \begin{bmatrix}0.5 \\ -0.5 \\ 1\end{bmatrix}, \quad \nu_3 = \begin{bmatrix}-0.6667 \\ 0.6667 \\ 0.6667\end{bmatrix}$$ These vectors are mutually orthogonal and span the same subspace as the original vectors.