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Determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set.$\left[\begin{array}{c}{1 / \sqrt{18}} \\ {4 / \sqrt{18}} \\ {1 / \sqrt{18}}\end{array}\right],\left[\begin{array}{c}{1 / \sqrt{2}} \\ {0} \\ {-1 / \sqrt{2}}\end{array}\right],\left[\begin{array}{r}{-2 / 3} \\ {1 / 3} \\ {-2 / 3}\end{array}\right]$

Thus, the set $\left\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\}$ is an orthonormal set.

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

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So we're determined if the set of factors is Ortho Norman's so first we need to see if there are thousands of each other. So we need to take, um, not products of all possible pairs of factors. So if we take the one dot be too, we give one over ah, screw to 36 plus zero minus one over Square 36 which is zero. So those two are thought more of each other. Let's take 31 dot b three. So we get negative two. Over there, you square roots of 18 plus for over the square roots of 18 minus two over three squirts of 18 which equals zero. Lastly, we'll take the other combination. B two dot b three. So we get negative two over three squares, a two plus zero plus two over Yuri Square roots to which again is zero. So the set is orthogonal, so check if it's orphan normal. You need to check, uh, the norms of a defector. So if we calculate the normal fee one. So we take the square root of the some of the components squared, so one over square to 18 squared is one over 18 and then we have plus four square to 16. Hurry. 18 on again, plus one over 18. So we get square root of 18 over 18 which is one so next normal V two, we have square root. Uh, 1/2 well zero plus 1/2 which is squared a one which is one. And lastly, the normal B three. We have the square root of negative 2/3 squared, which is for over nine 1/3 squared won't open nine and 2/3 where you can. It's for over nine. So we have a square root of nine over nine, which is what? So the set is orphan often.

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