1. Diketahui basis \(\vec{u}_1 = (1,1,0), \vec{u}_2 = (0,1,1), \vec{u}_3 = (1,0,1)\) di \(\mathbb{R}^3\) dengan hasil kali dalam Euclides. Tugas kita adalah menggunakan proses Gram-Schmidt untuk mengubah basis ini menjadi basis orthonormal. 2. Proses Gram-Schmidt digunakan untuk mengubah basis menjadi basis orthogonal, kemudian dinormalisasi menjadi orthonormal. Rumusnya adalah: \[ \vec{v}_1 = \vec{u}_1 \] \[ \vec{v}_2 = \vec{u}_2 - \mathrm{proj}_{\vec{v}_1}(\vec{u}_2) = \vec{u}_2 - \frac{\langle \vec{u}_2, \vec{v}_1 \rangle}{\langle \vec{v}_1, \vec{v}_1 \rangle} \vec{v}_1 \] \[ \vec{v}_3 = \vec{u}_3 - \mathrm{proj}_{\vec{v}_1}(\vec{u}_3) - \mathrm{proj}_{\vec{v}_2}(\vec{u}_3) = \vec{u}_3 - \frac{\langle \vec{u}_3, \vec{v}_1 \rangle}{\langle \vec{v}_1, \vec{v}_1 \rangle} \vec{v}_1 - \frac{\langle \vec{u}_3, \vec{v}_2 \rangle}{\langle \vec{v}_2, \vec{v}_2 \rangle} \vec{v}_2 \] 3. Hitung \(\vec{v}_1\): \[ \vec{v}_1 = (1,1,0) \] 4. Hitung \(\vec{v}_2\): \[ \langle \vec{u}_2, \vec{v}_1 \rangle = 0 \cdot 1 + 1 \cdot 1 + 1 \cdot 0 = 1 \] \[ \langle \vec{v}_1, \vec{v}_1 \rangle = 1^2 + 1^2 + 0^2 = 2 \] \[ \mathrm{proj}_{\vec{v}_1}(\vec{u}_2) = \frac{1}{2} (1,1,0) = \left(\frac{1}{2}, \frac{1}{2}, 0\right) \] \[ \vec{v}_2 = (0,1,1) - \left(\frac{1}{2}, \frac{1}{2}, 0\right) = \left(-\frac{1}{2}, \frac{1}{2}, 1\right) \] 5. Hitung \(\vec{v}_3\): \[ \langle \vec{u}_3, \vec{v}_1 \rangle = 1 \cdot 1 + 0 \cdot 1 + 1 \cdot 0 = 1 \] \[ \langle \vec{u}_3, \vec{v}_2 \rangle = 1 \cdot \left(-\frac{1}{2}\right) + 0 \cdot \frac{1}{2} + 1 \cdot 1 = -\frac{1}{2} + 0 + 1 = \frac{1}{2} \] \[ \langle \vec{v}_2, \vec{v}_2 \rangle = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 1^2 = \frac{1}{4} + \frac{1}{4} + 1 = \frac{3}{2} \] \[ \mathrm{proj}_{\vec{v}_1}(\vec{u}_3) = \frac{1}{2} (1,1,0) = \left(\frac{1}{2}, \frac{1}{2}, 0\right) \] \[ \mathrm{proj}_{\vec{v}_2}(\vec{u}_3) = \frac{\frac{1}{2}}{\frac{3}{2}} \left(-\frac{1}{2}, \frac{1}{2}, 1\right) = \frac{1}{3} \left(-\frac{1}{2}, \frac{1}{2}, 1\right) = \left(-\frac{1}{6}, \frac{1}{6}, \frac{1}{3}\right) \] \[ \vec{v}_3 = (1,0,1) - \left(\frac{1}{2}, \frac{1}{2}, 0\right) - \left(-\frac{1}{6}, \frac{1}{6}, \frac{1}{3}\right) = \left(1 - \frac{1}{2} + \frac{1}{6}, 0 - \frac{1}{2} - \frac{1}{6}, 1 - 0 - \frac{1}{3}\right) = \left(\frac{2}{3}, -\frac{2}{3}, \frac{2}{3}\right) \] 6. Normalisasi setiap \(\vec{v}_i\) untuk mendapatkan basis orthonormal \(\vec{e}_i\): \[ \|\vec{v}_1\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \] \[ \vec{e}_1 = \frac{1}{\sqrt{2}} (1,1,0) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \] \[ \|\vec{v}_2\| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4} + \frac{1}{4} + 1} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2} \] \[ \vec{e}_2 = \frac{1}{\|\vec{v}_2\|} \vec{v}_2 = \frac{2}{\sqrt{6}} \left(-\frac{1}{2}, \frac{1}{2}, 1\right) = \left(-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right) \] \[ \|\vec{v}_3\| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{3 \times \frac{4}{9}} = \sqrt{\frac{12}{9}} = \frac{2}{\sqrt{3}} \] \[ \vec{e}_3 = \frac{1}{\|\vec{v}_3\|} \vec{v}_3 = \frac{\sqrt{3}}{2} \left(\frac{2}{3}, -\frac{2}{3}, \frac{2}{3}\right) = \left(\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right) \] Jadi, basis orthonormalnya adalah: \[ \vec{e}_1 = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right), \quad \vec{e}_2 = \left(-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right), \quad \vec{e}_3 = \left(\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right) \]