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Exercise 1.2.18

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$∥ v ∥^{2} = 4^{2} + 2^{2} = 20$ and $∥ w ∥^{2} = {(- 1)}^{2} + 2^{2} = 5 .$ Pythagoras is $∥ (3, 4) ∥^{2} = 25 =$ $20 + 5$ for the length of the hypotenuse $v + w = (3, 4)$ .

mathchakra

2021-11-20 14:06

Exercise 1.2.18

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