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Exercise 1.2.16

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$∥ v ∥^{2} = 1 + 1 + \dots + 1 = 9$ so $∥ v ∥ = 3; u = v ∕ 3 = (\frac{1}{3}, \dots, \frac{1}{3})$ is a unit vector in $9 D$ $w = (1, - 1, 0, \dots, 0) ∕ \sqrt{2}$ is a unit vector in the $8 D$ hyperplane perpendicular to $v$

mathchakra

2021-11-20 14:06

Exercise 1.2.16

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