Subject: Applied Mathematics 4

Topic: Vector Spaces

Difficulty: Medium

Let u_1 = (1,0,4), u_2 = (-1,0,7), u_3 = (2,9,11)

Let v_1 = u_2 = (3,0,4)

$ ||v_1||^2 = (\sqrt{9+0+16})^2 = (\sqrt{25})^2 = 25 \\ \lt u_2,v_1 \gt = (-1,0,7).(3,0,4) = -3+28 = 25 $

Now,

$ v_2 = u_2 - Proj(u_2) \\ = u_2 - [\frac{\lt u_2,v_1 \gt}{||v_1||^2}]v_1 \\ = (-1,0,7) - [\frac{25}{25}(3,0,4)] \\ = (-1,0,7) - (3,0,4) \\ = (-4,0,3) \\ ||v_2|| = \sqrt{(-4)^2 + 0 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5 \\ ||v_2||^2 = 25 \\ v_3 = u_3 - Proj(u_3) \\ = u_3 - [ \frac{\lt u_3,v_1 \gt}{||v_1||^2}]v_1 - [ \frac{\lt u_3,v_2 \gt}{||v_2||^2}]v_2 \\ \lt u_3,v_1 \gt = (2,9,11).(3,0,4) = 6 + 44 = 50 \\ ||v_1||^2 = 25 \\ \lt u_3,v_2 \gt = (2,9,11).(-4,0,3) = -8 + 33 = 25 \\ ||v_2|| = \sqrt{(-4)^2 + 0 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \\ ||v_2||^2 = 25 \\ \therefore v_3 = u_3 - [\frac{50}{25}(3,0,4)] - [\frac{25}{25}(-4,0,3)] \\ = (2,9,11) - [2(3,0,4)] - (-4,0,3) \\ = (2,9,11) - (6,0,8) - (-4,0,3) \\ = (2,9,11) - [(6,0,8) + (-4,0,3)] \\ = (2,9,11) - (2,0,11) \\ = (0,9,0) \\ q_1 = \frac{v_1}{||v_1||} = \frac{(3,0,4)}{\sqrt{9+16}} = \frac{(3,0,4)}{\sqrt{25}} = \frac{(3,0,4)}{5} \\ q_1 = (\frac{3}{5},0,\frac{4}{5}) \\ q_2 = \frac{v_2}{||v_2||} = \frac{(-4,0,3)}{\sqrt{(-4)^2} + 3^2} = \frac{(-4,0,3)}{\sqrt{16+9}} = \frac{(-4,0,3)}{5} \\ q_2 = (\frac{-4}{5},0,\frac{3}{5}) \\ q_3 = \frac{v_3}{||v_3||} = \frac{(0,9,0)}{\sqrt{9^2}} = \frac{(0,9,0)}{9} \\ q_3 = (0,1,0) $

Therefore the required orthogonal basis is {$ (\frac{3}{5},0,\frac{4}{5}), (\frac{-4}{5},0,\frac{3}{5}), (0,1,0) $}