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Exercise 1.2.20

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We know that $(v - w) \cdot (v - w) = v \cdot v - 2 v \cdot w + w \cdot w .$ The Law of Cosines writes $∥ v ∥ ∥ w ∥ \cos 𝜃$ for $v \cdot w .$ Here $𝜃$ is the angle between $v$ and $w .$ When $𝜃 < 9 0^{∘}$ this $v \cdot w$ is positive, so in this case $v \cdot v + w \cdot w$ is larger than $∥ v - w ∥^{2}$ .

Pythagoras changes from equality $a^{2} + b^{2} = c^{2}$ to inequality when $𝜃 < 9 0^{∘}$ or $𝜃 > 9 0^{∘}$ .

mathchakra

2021-11-20 14:07

Exercise 1.2.20

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