Let y vector 1 1 and u vector 3 2 Write y vector as the su
     Let y vector = [1 -1] and u vector = [3 -2]. Write y vector as the sum of two orthogonal vectors, x vector_1 in Span {u} and x vector_2 orthogonal to u vector.  x vector_1 = [], x vector_2 = []![Let y vector = [1 -1] and u vector = [3 -2]. Write y vector as the sum of two orthogonal vectors, x vector_1 in Span {u} and x vector_2 orthogonal to u vector.  Let y vector = [1 -1] and u vector = [3 -2]. Write y vector as the sum of two orthogonal vectors, x vector_1 in Span {u} and x vector_2 orthogonal to u vector.](/WebImages/43/let-y-vector-1-1-and-u-vector-3-2-write-y-vector-as-the-su-1134525-1761606904-0.webp) 
  
  Solution
let x1 = proju(y) = [(y.u)/(u.u)]u = [(3+2)/(9+4)]u = (5/13)( 3,-2)T = (15/13,-10/13)T and x2= y-x1 = (1,-1)T-(15/13,-10/13)T= ( -2/13, -3/13)T. Further x2.u = ( -2/13, -3/13)T. ( 3,-2)T= -6/13 +6/13 = 0. Hence x2 is orthogonal to u. Also, x1, being a scalar multiple of u is in span(u}.
![Let y vector = [1 -1] and u vector = [3 -2]. Write y vector as the sum of two orthogonal vectors, x vector_1 in Span {u} and x vector_2 orthogonal to u vector.  Let y vector = [1 -1] and u vector = [3 -2]. Write y vector as the sum of two orthogonal vectors, x vector_1 in Span {u} and x vector_2 orthogonal to u vector.](/WebImages/43/let-y-vector-1-1-and-u-vector-3-2-write-y-vector-as-the-su-1134525-1761606904-0.webp)
