Let v1 12 23 23 01 v2 29 59 49 23 and v3 29 49 59 23 Find

Let v_1 = [1/2 -2/3 -2/3 0/1], v_2 = [-2/9 -5/9 4/9 2/3], and v_3 = [-2/9 4/9 -5/9 2/3]. Find a vector v_4 in R^4 such that {v_1, v_2, v_3, v_4} is an orthonormat set.

Solution

Let v4 = { x, y, w , z}

v1.v4 = x/3 -2y/3 -2w/3 =0

v2.v4 = -2x/9 - 5y/9+ 4w/9 +2z/9 =0

v3.v4 = -2x/9 +4y/9 -5w/9 +2z/3 =0

we get : x-2y -2w =0 -----(1)

-2x -5y +4w +2z =0 -----(2)

-2x +4y -5w +2z =0 ------(3)

Let z = t be a free variable

x = 8z/9 ; y=2z/9 ; w = 2z/9 ;

x = 8t/9 ; y = 2t/9 ; w= 2t/9

Now v4 = { 8t/9 , 2t/9 , 2t/9 , t }

Now v4.v4 = 1

(8t/9)^2 + (2t/9)^2 + (2t/9)^2 + t^2 =1

t^2 [ 64/81 +4/81 + 4/81 +1]=1

t^2(153)/81 =1

t = +/- 9/3sqrt(17)

t = 3/sqrt(17) , -3/sqrt(17)   

There are two choices for v4

v4 = { 8t/9 , 2t/9 , 2t/9 , t }

= { 8/3sqrt17 , 6/sqrt17 , 6 /sqrt17 , 3/sqrt17 }

OR

=  { -8/3sqrt17 , -6/sqrt17 , -6 /sqrt17 , -3/sqrt17 }