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 66216 by Rio Michael last updated on 11/Aug/19

∣a ∣ = 3 ,∣b∣= 5 , a.b =−14    ∣a − b∣ = ?

$$\mid{a}\:\mid\:=\:\mathrm{3}\:,\mid{b}\mid=\:\mathrm{5}\:,\:{a}.{b}\:=−\mathrm{14} \\ $$$$\:\:\mid{a}\:−\:{b}\mid\:=\:? \\ $$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

∣a ∣ = 3 ,∣b∣= 5 ⇒a.b ≠−14  a.b=−15  or  a.b=15

$$\mid{a}\:\mid\:=\:\mathrm{3}\:,\mid{b}\mid=\:\mathrm{5}\:\Rightarrow{a}.{b}\:\neq−\mathrm{14} \\ $$$${a}.{b}=−\mathrm{15}\:\:{or}\:\:{a}.{b}=\mathrm{15} \\ $$

Commented by behi83417@gmail.com last updated on 11/Aug/19

∣a−b∣=(√(∣a∣^2 +∣b∣^2 +2∣a∣∣b∣cosθ))=  =(√(3^2 +5^2 +2×(−14)))=(√(9+25−28))=(√6)

$$\mid\mathrm{a}−\mathrm{b}\mid=\sqrt{\mid\mathrm{a}\mid^{\mathrm{2}} +\mid\mathrm{b}\mid^{\mathrm{2}} +\mathrm{2}\mid\mathrm{a}\mid\mid\mathrm{b}\mid\mathrm{cos}\theta}= \\ $$$$=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}×\left(−\mathrm{14}\right)}=\sqrt{\mathrm{9}+\mathrm{25}−\mathrm{28}}=\sqrt{\mathrm{6}} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Aug/19

∣a ∣ = 3 ,∣b∣= 5 , a.b =−15    ∣a − b∣ = ?    ∣a∣ = 3⇒a=±3⇒b=∓5                    [∵a.b=−15]     ∣a − b∣=∣(±3)−(∓5)∣                   =∣(±3)+(±5)∣                   =∣±8∣=8

$$\mid{a}\:\mid\:=\:\mathrm{3}\:,\mid{b}\mid=\:\mathrm{5}\:,\:{a}.{b}\:=−\mathrm{15} \\ $$$$\:\:\mid{a}\:−\:{b}\mid\:=\:? \\ $$$$\:\:\mid{a}\mid\:=\:\mathrm{3}\Rightarrow{a}=\pm\mathrm{3}\Rightarrow{b}=\mp\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because{a}.{b}=−\mathrm{15}\right] \\ $$$$\:\:\:\mid{a}\:−\:{b}\mid=\mid\left(\pm\mathrm{3}\right)−\left(\mp\mathrm{5}\right)\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mid\left(\pm\mathrm{3}\right)+\left(\pm\mathrm{5}\right)\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mid\pm\mathrm{8}\mid=\mathrm{8} \\ $$

Commented by Rio Michael last updated on 11/Aug/19

thanks

$${thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com