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Question Number 92465 by Rio Michael last updated on 07/May/20

for a 2d  vectors if ∣a + b∣ = ∣a−b∣ what relationship does a and b have?

$$\mathrm{for}\:\mathrm{a}\:\mathrm{2d}\:\:\mathrm{vectors}\:\mathrm{if}\:\mid{a}\:+\:{b}\mid\:=\:\mid{a}−{b}\mid\:\mathrm{what}\:\mathrm{relationship}\:\mathrm{does}\:{a}\:\mathrm{and}\:{b}\:\mathrm{have}? \\ $$$$ \\ $$

Commented by mr W last updated on 07/May/20

a⊥b

$$\boldsymbol{{a}}\bot\boldsymbol{{b}} \\ $$

Commented by Rio Michael last updated on 07/May/20

thanks sir, any prove for that sir?

$$\mathrm{thanks}\:\mathrm{sir},\:\mathrm{any}\:\mathrm{prove}\:\mathrm{for}\:\mathrm{that}\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 07/May/20

∣a+b∣ and ∣a−b∣ are the both diagonals  of the paralleogram with a and b as  sides. when both diagonals should be  equal, the parallelogram must be a  rectangle, i.e. a and b must be ⊥ to  each other.

$$\mid{a}+{b}\mid\:{and}\:\mid{a}−{b}\mid\:{are}\:{the}\:{both}\:{diagonals} \\ $$$${of}\:{the}\:{paralleogram}\:{with}\:{a}\:{and}\:{b}\:{as} \\ $$$${sides}.\:{when}\:{both}\:{diagonals}\:{should}\:{be} \\ $$$${equal},\:{the}\:{parallelogram}\:{must}\:{be}\:{a} \\ $$$${rectangle},\:{i}.{e}.\:{a}\:{and}\:{b}\:{must}\:{be}\:\bot\:{to} \\ $$$${each}\:{other}. \\ $$

Commented by mr W last updated on 13/May/20

Commented by Rio Michael last updated on 07/May/20

following your explanation i checked that  thank you sir.

$$\mathrm{following}\:\mathrm{your}\:\mathrm{explanation}\:\mathrm{i}\:\mathrm{checked}\:\mathrm{that} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Prithwish Sen 1 last updated on 07/May/20

∵  a+b and a−b are the two diagonals of the   parallalogram formed by the vectors a and b  as the two adjacent sides then if ∣a+b∣=∣a−b∣  only implies that the diagonals are same and it  can only happen when the parallalogram so   formed by a and b is a rectangle. And that is why  the two adjacent sides must be ⊥ and i.e   a⊥b.

$$\because\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\:\mathrm{and}\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\:\mathrm{are}\:\mathrm{the}\:\mathrm{two}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{parallalogram}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{two}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{then}\:\mathrm{if}\:\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$$$\mathrm{only}\:\mathrm{implies}\:\mathrm{that}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{are}\:\mathrm{same}\:\mathrm{and}\:\mathrm{it} \\ $$$$\mathrm{can}\:\mathrm{only}\:\mathrm{happen}\:\mathrm{when}\:\mathrm{the}\:\mathrm{parallalogram}\:\mathrm{so}\: \\ $$$$\mathrm{formed}\:\mathrm{by}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{is}\:\mathrm{a}\:\boldsymbol{\mathrm{rectangle}}.\:\mathrm{And}\:\mathrm{that}\:\mathrm{is}\:\mathrm{why} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{must}\:\mathrm{be}\:\bot\:\mathrm{and}\:\mathrm{i}.\mathrm{e}\: \\ $$$$\boldsymbol{\mathrm{a}}\bot\boldsymbol{\mathrm{b}}. \\ $$

Commented by Rio Michael last updated on 07/May/20

another good explanation thanks

$$\mathrm{another}\:\mathrm{good}\:\mathrm{explanation}\:\mathrm{thanks} \\ $$

Answered by behi83417@gmail.com last updated on 07/May/20

∣a+b∣=(√(a^2 +b^2 −2abcosθ))  ∣a−b∣=(√(a^2 +b^2 +2abcosθ))  [θ=angle between a and b]  ∣a+b∣=∣a−b∣⇒  a^2 +b^2 −2abcosθ=a^2 +b^2 +2abcosθ  ⇒4abcosθ=0⇒ { ((a=0  (or))),((b=0  (or))),((cosθ=0⇒θ=(π/2)⇒a⊥b)) :}

$$\mid\mathrm{a}+\mathrm{b}\mid=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcos}\theta} \\ $$$$\mid\mathrm{a}−\mathrm{b}\mid=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2abcos}\theta} \\ $$$$\left[\theta=\mathrm{angle}\:\mathrm{between}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\right] \\ $$$$\mid\mathrm{a}+\mathrm{b}\mid=\mid\mathrm{a}−\mathrm{b}\mid\Rightarrow \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcos}\theta=\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2abcos}\theta \\ $$$$\Rightarrow\mathrm{4abcos}\theta=\mathrm{0}\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{0}\:\:\left(\mathrm{or}\right)}\\{\mathrm{b}=\mathrm{0}\:\:\left(\mathrm{or}\right)}\\{\mathrm{cos}\theta=\mathrm{0}\Rightarrow\theta=\frac{\pi}{\mathrm{2}}\Rightarrow\mathrm{a}\bot\mathrm{b}}\end{cases} \\ $$

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