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Question Number 128830 by bemath last updated on 10/Jan/21

Answered by liberty last updated on 10/Jan/21

let ∣a^→ ∣ = ∣b^→ ∣ = 1 ; also ∣ a^→ +b^→ ∣ = 1 we want to compute ∣a^→ −b^→ ∣ . by using equality (x+y)^2 +(x−y)^2 =2(x^2 +y^2 ) ⇔ ∣a^→ +b^→ ∣^2 +∣a^→ −b^→ ∣^2 = 2(∣a^→ ∣^2 +∣b^→ ∣^2 ) ⇔ 1+ ∣a^→ −b^→ ∣^2 =2(1+1) ⇔ ∣a^→ −b^→ ∣ = (√3) .

let∣a→∣=∣b→∣=1;also∣a→+b→∣=1wewanttocompute∣a→−b→∣.byusingequality(x+y)2+(x−y)2=2(x2+y2)⇔∣a→+b→∣2+∣a→−b→∣2=2(∣a→∣2+∣b→∣2)⇔1+∣a→−b→∣2=2(1+1)⇔∣a→−b→∣=3.

Answered by mr W last updated on 10/Jan/21

Commented by mr W last updated on 10/Jan/21

∣a∣=∣b∣=1 ∣a+b∣=1 ⇒∣a−b∣=(√3)

∣a∣=∣b∣=1∣a+b∣=1⇒∣a−b∣=3

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