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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.$\left[\begin{array}{c}{1 / 3} \\ {1 / 3} \\ {1 / 3}\end{array}\right],\left[\begin{array}{c}{-1 / 2} \\ {0} \\ {1 / 2}\end{array}\right]$

$\left\{\left[\begin{array}{c}{1 / \sqrt{3}} \\ {1 / \sqrt{3}} \\ {1 / \sqrt{3}}\end{array}\right],\left[\begin{array}{c}{-1 / \sqrt{2}} \\ {0} \\ {1 / \sqrt{2}}\end{array}\right]\right\}$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

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Okay, This question wants us to check if this is a North a normal set. And if it's not, we need to normalize the vectors to make it one. So first, let's check to see if they're orthogonal in the first place. So you doubt the where you was the first vector envies. The second will be one third times minus a half plus one third time's positive a half. So those air exactly equal and opposite. So we get zero. So these are orthogonal, but then we have to check if their unit vectors, and if they're not, we'll have to normalize. So the magnitude of you would be equal to the square root of one third squared plus one third squared plus one third square or the square root of 1/9 plus 1/9 plus 1/9. So we actually get one third under the square root, and that would be a one over route three, and that's not equal toe one. So, actually, to get our Ortho normal, you which I'll call you hat. We just have to divide the magnitude of you into our little vector here. So if we do this, we would just get 1/1 over Route three and then times the factor one third, one third, one third or distributing in. We get Route 3/3, Route 3/3, Route 3/3 and that's our first orthogonal vector. Then we'll do the same thing for V. So the magnitude of the would be the square root of, ah, half squared plus zero squared, plus a half squared, giving us the square root of to fourth. So square one half or one over, route to and again that's not equal to one. So we have to normalize by dividing by the magnitude. So the hat would be the over its magnitude so he'd get 1/1 over route to times the vector one half We're sorry, minus one half 01 half. And since one of the route to is just group to, we'd get negative route to over to zero route to over to, and that's our final answer for Ortho Normal set

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