Let y 3 3 3 3 and W Span1 0 1 0 1 1 1 0The distance from y
     Let y =[3  3  3  3], and W = Span{[1  0  1  0], [-1  1  1  0]}.The distance from y to W is![Let y =[3 3 3 3], and W = Span{[1 0 1 0], [-1 1 1 0]}.The distance from y to W isSolutionFirst we need to find the vector in the span of W which is closest to   Let y =[3 3 3 3], and W = Span{[1 0 1 0], [-1 1 1 0]}.The distance from y to W isSolutionFirst we need to find the vector in the span of W which is closest to](/WebImages/29/let-y-3-3-3-3-and-w-span1-0-1-0-1-1-1-0the-distance-from-y-1082115-1761568437-0.webp) 
  
  Solution
First we need to find the vector in the span of W which is closest to y
Since the last entry in both the vectors is equal to zero, hence the minimum distance will be sqrt(9) or 3 units
Now we need to set the W vector such that the distance is minimized, so we would try to make atleast one row as 3, but we see that the below combination yields the minimum distance
W = 3[1 0 1 0] + [-1 1 1 0]
=>[2 1 4 0]
Distance = sqrt( (3-2)^2 + (3-1)^2 + (3-4)^2 + (3-0)^2 )
=> sqrt(1+4+9+1)
=>sqrt(15
![Let y =[3 3 3 3], and W = Span{[1 0 1 0], [-1 1 1 0]}.The distance from y to W isSolutionFirst we need to find the vector in the span of W which is closest to   Let y =[3 3 3 3], and W = Span{[1 0 1 0], [-1 1 1 0]}.The distance from y to W isSolutionFirst we need to find the vector in the span of W which is closest to](/WebImages/29/let-y-3-3-3-3-and-w-span1-0-1-0-1-1-1-0the-distance-from-y-1082115-1761568437-0.webp)
