Successfully reported this slideshow.

# a- Write y as the sum of a vector in span {u} and a vector orthogonal.docx

a. Write y as the sum of a vector in span {u} and a vector orthogonal to u.
b. Compute the distance from y to the line through u and the origin.
Solution
y=(1,2) and u=(4,-3) Let a vector orthogonal to u be w = (x,y) =>u.w = 0 =>4x-3y = 0 => 4x = 3y If y = 4 => x = 3 w = (3,4) Let y = au + bw where a and b are some constants. => (1,2) = a(4,-3) + b(3,4) = (4a+3b, -3a+4b) => 4a + 3b = 1, -3a + 4b = 2 Solving for a, b we get a = -2/25 and b = 11/25 Therefore y = (-2/25)*u + (11/25)*w Line passing through u, origin is: y+3 = (-3/4)(x-4) =>3x+4y = 0 Distance of y from this line is: Distance = (3*1 + 4*2)/5 = 11/5
.

a. Write y as the sum of a vector in span {u} and a vector orthogonal to u.
b. Compute the distance from y to the line through u and the origin.
Solution
y=(1,2) and u=(4,-3) Let a vector orthogonal to u be w = (x,y) =>u.w = 0 =>4x-3y = 0 => 4x = 3y If y = 4 => x = 3 w = (3,4) Let y = au + bw where a and b are some constants. => (1,2) = a(4,-3) + b(3,4) = (4a+3b, -3a+4b) => 4a + 3b = 1, -3a + 4b = 2 Solving for a, b we get a = -2/25 and b = 11/25 Therefore y = (-2/25)*u + (11/25)*w Line passing through u, origin is: y+3 = (-3/4)(x-4) =>3x+4y = 0 Distance of y from this line is: Distance = (3*1 + 4*2)/5 = 11/5
.