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Let $\mathbf{y}=\left[\begin{array}{l}{2} \\ {6}\end{array}\right]$ and $\mathbf{u}=\left[\begin{array}{l}{7} \\ {1}\end{array}\right] .$ Write $\mathbf{y}$ as the sum of a vector in $\operatorname{Span}\{\mathbf{u}\}$ and a vector orthogonal to $\mathbf{u} .$

$\left[\begin{array}{c}{14 / 5} \\ {2 / 5}\end{array}\right]+\left[\begin{array}{c}{-4 / 5} \\ {28 / 5}\end{array}\right]$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

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Harvey Mudd College

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so first to find the part that is in the span of you, we want to take the orthogonal projection of why onto you. So why hat is our projection And we know this is equal to white out you over you dot you times you. So why don't you is 14 plus six? Uh, you dot you is seven times seven is 49 plus one and we want to multiply this coefficient into the vector or seven. So those coefficient comes out to be 20 over 50 which is 2/5 to fist times seven hiss, 14 over five, and then we get to over five on the bottom. So this is the first part. So now when you define, uh, the other part, that is orthogonal. So we take why minus y hat our Why waas to six union. We subtract our 14 5th in 2/5 to get negative 4/5 and 28 5th So therefore we can, right? Why? As a song of 14 over 52 over five, plus negative for over five. 25

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