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Question Number 118286 by 1549442205PVT last updated on 16/Oct/20

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣a−b∣ ;(b)∣ a+b∣>∣ a−b∣;(c)∣a+ b∣< ∣a−b∣

$$\mathrm{What}\:\mathrm{condition}\:\mathrm{should}\:\mathrm{be}\:\mathrm{satisfied}\:\mathrm{by} \\ $$ $$\mathrm{the}\:\mathrm{vectors}\:\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\: \\ $$ $$\mathrm{relations}\:\mathrm{to}\:\mathrm{hold}\:\mathrm{true}\::\left(\mathrm{a}\right)\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$ $$;\left(\mathrm{b}\right)\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid>\mid\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid;\left(\mathrm{c}\right)\mid\boldsymbol{\mathrm{a}}+\:\boldsymbol{\mathrm{b}}\mid<\:\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$

Answered by TANMAY PANACEA last updated on 16/Oct/20

a)∣a+b∣=∣a−b∣ ∣a+b∣^2 =∣a−b∣^2 (a+b).(a+b)=(a−b).(a−b) a^2 +b^2 +2abcosθ=a^2 +b^2 −2abcosθ [a.b=abcosθ a.a=a^2 b.b=b^2 ] 4abcosθ=0 θ=(π/2) so a⊥b

$$\left.{a}\right)\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid=\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$ $$\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid^{\mathrm{2}} =\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} \\ $$ $$\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)=\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right) \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta \\ $$ $$\left[\boldsymbol{{a}}.\boldsymbol{{b}}={abcos}\theta\:\:\:\:\:\:\:\boldsymbol{{a}}.\boldsymbol{{a}}={a}^{\mathrm{2}} \:\:\:\:\boldsymbol{{b}}.\boldsymbol{{b}}={b}^{\mathrm{2}} \right] \\ $$ $$\mathrm{4}{abcos}\theta=\mathrm{0}\:\:\:\theta=\frac{\pi}{\mathrm{2}}\:\:\:{so}\:\boldsymbol{{a}}\bot\boldsymbol{{b}} \\ $$

Answered by TANMAY PANACEA last updated on 16/Oct/20

b)similarly ∣a+b∣>∣a−b∣ (√(a^2 +b^2 +2abcosθ)) >(√(a^2 +b^2 −2abcosθ)) cosθ>0 so θ=acute angle (π/2)>θ>0 c)a+b∣<∣a−b∣ (√(a^2 +b^2 +2abcosθ)) <(√(a^2 +b^2 −2abcosθ)) cosθ=−ve cosθ<0 π>θ>(π/2) correction done plz check

$$\left.{b}\right){similarly} \\ $$ $$\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid>\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$ $$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta}\:>\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta}\: \\ $$ $${cos}\theta>\mathrm{0}\:\:{so}\:\theta={acute}\:{angle} \\ $$ $$\frac{\pi}{\mathrm{2}}>\theta>\mathrm{0} \\ $$ $$\left.{c}\right)\boldsymbol{{a}}+\boldsymbol{{b}}\mid<\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$ $$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta}\:<\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta}\: \\ $$ $${cos}\theta=−{ve} \\ $$ $${cos}\theta<\mathrm{0} \\ $$ $$\pi>\theta>\frac{\pi}{\mathrm{2}}\:\:{correction}\:{done} \\ $$ $${plz}\:{check} \\ $$

Commented byTANMAY PANACEA last updated on 16/Oct/20

most welcome sir

$${most}\:{welcome}\:{sir} \\ $$

Commented byTANMAY PANACEA last updated on 16/Oct/20

yes sir...

$${yes}\:{sir}... \\ $$

Commented by1549442205PVT last updated on 16/Oct/20

Thank sir.You are welcome. You corrected exactly.When θ=π then there two possibilities ∣a+b∣=∣a−b∣ =0 o r ∣a+b∣<∣a−b∣ Question means the inequalities are really true.

$$\mathrm{Thank}\:\mathrm{sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$ $$\mathrm{You}\:\mathrm{corrected}\:\mathrm{exactly}.\mathrm{When}\: \\ $$ $$\theta=\pi\:\mathrm{then}\:\mathrm{there}\:\mathrm{two}\:\mathrm{possibilities} \\ $$ $$\mid\mathrm{a}+\mathrm{b}\mid=\mid\mathrm{a}−\mathrm{b}\mid\:=\mathrm{0}\:\mathrm{o}\:\mathrm{r}\:\mid\mathrm{a}+\mathrm{b}\mid<\mid\mathrm{a}−\mathrm{b}\mid \\ $$ $$\mathrm{Question}\:\mathrm{means}\:\mathrm{the}\:\mathrm{inequalities}\:\mathrm{are} \\ $$ $$\mathrm{really}\:\mathrm{true}. \\ $$

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