Question Number 118286 by 1549442205PVT last updated on 16/Oct/20
$$\mathrm{What}\:\mathrm{condition}\:\mathrm{should}\:\mathrm{be}\:\mathrm{satisfied}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{vectors}\:\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{relations}\:\mathrm{to}\:\mathrm{hold}\:\mathrm{true}\::\left(\mathrm{a}\right)\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$$$;\left(\mathrm{b}\right)\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid>\mid\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid;\left(\mathrm{c}\right)\mid\boldsymbol{\mathrm{a}}+\:\boldsymbol{\mathrm{b}}\mid<\:\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$
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](https://www.tinkutara.com/question/Q118290.png)
$$\left.{a}\right)\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid=\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$$$\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid^{\mathrm{2}} =\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid^{\mathrm{2}} \\ $$$$\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)=\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right).\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta \\ $$$$\left[\boldsymbol{{a}}.\boldsymbol{{b}}={abcos}\theta\:\:\:\:\:\:\:\boldsymbol{{a}}.\boldsymbol{{a}}={a}^{\mathrm{2}} \:\:\:\:\boldsymbol{{b}}.\boldsymbol{{b}}={b}^{\mathrm{2}} \right] \\ $$$$\mathrm{4}{abcos}\theta=\mathrm{0}\:\:\:\theta=\frac{\pi}{\mathrm{2}}\:\:\:{so}\:\boldsymbol{{a}}\bot\boldsymbol{{b}} \\ $$
Answered by TANMAY PANACEA last updated on 16/Oct/20
$$\left.{b}\right){similarly} \\ $$$$\mid\boldsymbol{{a}}+\boldsymbol{{b}}\mid>\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta}\:>\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta}\: \\ $$$${cos}\theta>\mathrm{0}\:\:{so}\:\theta={acute}\:{angle} \\ $$$$\frac{\pi}{\mathrm{2}}>\theta>\mathrm{0} \\ $$$$\left.{c}\right)\boldsymbol{{a}}+\boldsymbol{{b}}\mid<\mid\boldsymbol{{a}}−\boldsymbol{{b}}\mid \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{abcos}\theta}\:<\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta}\: \\ $$$${cos}\theta=−{ve} \\ $$$${cos}\theta<\mathrm{0} \\ $$$$\pi>\theta>\frac{\pi}{\mathrm{2}}\:\:{correction}\:{done} \\ $$$${plz}\:{check} \\ $$
Commented by TANMAY PANACEA last updated on 16/Oct/20
$${most}\:{welcome}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 16/Oct/20
$${yes}\:{sir}… \\ $$
Commented by 1549442205PVT last updated on 16/Oct/20
$$\mathrm{Thank}\:\mathrm{sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$$$\mathrm{You}\:\mathrm{corrected}\:\mathrm{exactly}.\mathrm{When}\: \\ $$$$\theta=\pi\:\mathrm{then}\:\mathrm{there}\:\mathrm{two}\:\mathrm{possibilities} \\ $$$$\mid\mathrm{a}+\mathrm{b}\mid=\mid\mathrm{a}−\mathrm{b}\mid\:=\mathrm{0}\:\mathrm{o}\:\mathrm{r}\:\mid\mathrm{a}+\mathrm{b}\mid<\mid\mathrm{a}−\mathrm{b}\mid \\ $$$$\mathrm{Question}\:\mathrm{means}\:\mathrm{the}\:\mathrm{inequalities}\:\mathrm{are} \\ $$$$\mathrm{really}\:\mathrm{true}. \\ $$