Let 1 4 1 and 5 3 1 SolutionFirst we use GramSchmidt to det

Let $\"x=\"$

$\"\\left.\\vphantom{\\begin{array}{c}\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\end{array}}\$	1	$\"\\left.\\vphantom{\\begin{array}{c}\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\end{array}}\$
4
1

and $\"y=\"$

$\"\\left.\\vphantom{\\begin{array}{c}\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\end{array}}\$	5	$\"\\left.\\vphantom{\\begin{array}{c}\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\!\\strut\\\\\\end{array}}\$
3
1

First, we use Gram-Schmidt to determine an orthogonal basis.

u = x = [ 1, 4, 1 ]

v = y -[ (y.u)/(u.u)]u

= [ 5 , 3 ,1] - [ (5+12+1)/(1+16+1)](1,4,1)

= [5,3,1] - [18/18](1,4,1)

= [5,3,1] - [ 1, 4,1]

= [ 4, -1,0]

Note u and v are orthogonal as their dot product is zero: 4 -4 +0 =0

Now make u and v unit length.

[ 1,4, 1]/sqrt18 , [ 4, -1, 0 ]/sqrt17 determine the orthonormal basis