You can try other unit vectors f 1 0 0 5 f 0 0 1

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Unformatted text preview: her unit vectors: f (1, 0, 0) = 5, f (0, 0, 1) = 4, f (0, 1/ 2, 1/ 2) = 6/ 2 = 3 2, but you always seem to get something larger than 2. How p we show that f (x, y, z ) can not be lower than 2 for a unit vector? If (x, y, z ) is a unit vector, can then x2 + y 2 + z 2 = 1, or equivalently x2 + y 2 + z 2 = 1. So if (x, y, z ) is a unit vector, then f (x, y, z ) = 5x2 + 2y 2 + 4z 2 = 3x2 + 2z 2 + 2(x2 + y 2 + z 2 ) = 3x2 + 2z 2 + 2, where we are using the fact that x2 + y 2 + z 2 = 1 in the last step. Now squares are always nonnegative, so 3x2 + 2z 2 0, and so f (x, y, z ) is equal to a nonnegative number plus 2. So f (x, y, z ) 2. C EDRIC C HAUVE , FALL 2013 1...
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