Subject: Applied Mathematics 4

Topic: Vector Spaces

Difficulty: Medium

Let u$_1$ = (1,2,0) and u$_2$ = (0,3,1)

v$_1$ = u$_1$ = (1,2,0)

$ v_2 = u_2 - [\frac{\lt u_2,v_1 \gt}{||v_1||^2}]v_1 \\ = (0,3,1,) - [\frac{(0,3,1).(1,2,0)}{(\sqrt{1+4+0})^2}] (1,2,0) \\ = (0,3,1,) - [\frac{6}{5}(1,2,0)] \\ = (0,3,1,) - [\frac{6}{5}, \frac{12}{5},0] \\ = (0-\frac{6}{5},3-\frac{12}{5},1-0) \\ = (-\frac{6}{5},\frac{3}{5},1) \\ ||v_2|| = \sqrt{\frac{36}{25} + \frac{9}{25} + 1} = \sqrt{\frac{45}{25} + 1} = \sqrt{\frac{70}{25}} = \sqrt{\frac{14}{5}} = \sqrt{\frac{14}{5}} \\ q_1 = \frac{v_1}{||v_1||} \frac{(1,2,0,)}{\sqrt{5}} = (\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}},0) \\ q_2 = \frac{v_2}{||v_2||} \frac{(-6/5,3/5,1)}{\sqrt{14/5}} = \frac{\sqrt{5}}{\sqrt{14}}(\frac{-6}{5},\frac{3}{5},1) \\ = (\frac{-6}{\sqrt{70}},\frac{3}{\sqrt{70}},\frac{\sqrt{5}}{\sqrt{14}}) \\ = (\frac{-6}{\sqrt{70}},\frac{3}{\sqrt{70}},\frac{5}{\sqrt{70}})$

Therefore, the orthogonal basis is {$ q_1,q_2 $}