Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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)) :}

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com