55 10 1 point Let y F VI E Compute the distance d from y to

5.5 10

(1 point) Let y F VI E Compute the distance d from y to the subspace of R4 spanned by Vi and v2. EE

we can write, y = p + (y-p); where p is projection of y in subspace R^4;

we need to find ||y-p||;

Now to find p in given subspace, we need to find normal and ortho-normal vectors in this subspace,

i.e. from v1 and v2 we need to calculate x1 and x2 which are normal and ortho normal vectors in same sub space,

here x1 = v1/|v1|

i.e. x1 = [2,-4,-2,1]/sqrt(25) = [2,-4,-2,1]/5;

and x2 = (v2 - <v2,x1>x1) / |(v2 - <v2,x1>x1)|

Where, where <v2,x1> = inner product of two vectors= (2*2 + (-2*-4)+(-2*4)+1*-4)/5

= 0;

x2 = [2,-2,4,-4]/6

x2 = [1,-1,2,-2]/3

so projection of y over subspace spanned by {x1,x2} is

p = <y,x1>x1 + <y,x2>x2;

where y = [-5,-1,0,1]

<y,x1> = (-5)/5 = -1;

<y,x2> = (-12)/6 = -2

p = [-2,4,2,1]/5+ [-4,4,-8,8]/3

p = [-26,32,-34,43]/15

the distance is d = |y-p|

d = sqrt (4705)/15

d = 4.572 is the answer.