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In Exercises 13 and $14,$ find the best approximation to $z$ by vectors of the form $c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}$$$\mathbf{z}=\left[\begin{array}{r}{3} \\ {-7} \\ {2} \\ {3}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{2} \\ {-1} \\ {-3} \\ {1}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{1} \\ {1} \\ {0} \\ {-1}\end{array}\right]$$

$\hat{z}=\left[\begin{array}{c}{-1} \\ {-3} \\ {-2} \\ {3}\end{array}\right]$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 3

Orthogonal Projections

Vectors

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

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so we know that to find the best approximation to see by vectors of the form of a linear combination of you want to be too is to take the orthogonal projection onto the span of the two factors. So to do this, we're gonna find the projections cat, and that's equal to, uh, see ya. Do you want over the one dot product with itself? Times one close si dot Be too. Over 32 dot product with the self times, we too. So let's find these coefficients first by taking all the dot products So z dot product with the one is equal to for plus zero close zero. Clause three gives us seven be Wanda product with itself, we get four plus zero plus one plus nine for 14. So that tells us this first coefficient is seven over 14 or 1/2. Next. Ah, si dot product with me too. We get 10 minus eight plus zero minus two. So to minus two is zero. Eso doesn't even matter what 32 dot product with itself is this coefficient zero. So this term is gone, so we just need to take 1/2 and multiply by view one. So we get half a two is one Hypo 00 minus 1/2 and minus three halfs. So that's our best.

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