Question-187496

Question Number 187496 by mathlove last updated on 18/Feb/23

Commented by mr W last updated on 18/Feb/23

things like ∣(A^→ /B^→ )∣ don′t exist! if you mean ((∣A^→ ∣)/(∣B^→ ∣)), then you should also write that what you mean.

$${things}\:{like}\:\mid\frac{\overset{\rightarrow} {{A}}}{\overset{\rightarrow} {{B}}}\mid\:{don}'{t}\:{exist}!\: \\ $$$${if}\:{you}\:{mean}\:\frac{\mid\overset{\rightarrow} {{A}}\mid}{\mid\overset{\rightarrow} {{B}}\mid},\:{then}\:{you}\:{should} \\ $$$${also}\:{write}\:{that}\:{what}\:{you}\:{mean}. \\ $$

Answered by manxsol last updated on 18/Feb/23

((∣p−q∣^2 )/(∣p+q∣^2 ))=((∣p∣^2 −2p•q+∣q∣^2 )/(∣p∣^2 +2p•q+∣q∣^2 )) =((2−2cosθ)/(2+2cosθ)) p•q=(1)(1)cosθ ∣((p−q)/(p+q))∣^2 =((1−cosθ)/(1+cosθ))=((1−cos(θ/2))/(1+cos(θ/2))) =((1−(1−2sin^2 (θ/2)))/(1+(2cos^2 (θ/2)−1)))=((2sin^2 (θ/2))/(2cos^2 (θ/2))) ∣((p−q)/(p+q))∣=tan(θ/2)

$$\frac{\mid{p}−{q}\mid^{\mathrm{2}} }{\mid{p}+{q}\mid^{\mathrm{2}} }=\frac{\mid{p}\mid^{\mathrm{2}} −\mathrm{2}{p}\bullet{q}+\mid{q}\mid^{\mathrm{2}} }{\mid{p}\mid^{\mathrm{2}} +\mathrm{2}{p}\bullet{q}+\mid{q}\mid^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}−\mathrm{2}{cos}\theta}{\mathrm{2}+\mathrm{2}{cos}\theta}\:\:\:\:\:\:\:{p}\bullet{q}=\left(\mathrm{1}\right)\left(\mathrm{1}\right){cos}\theta \\ $$$$\mid\frac{{p}−{q}}{{p}+{q}}\mid^{\mathrm{2}} =\frac{\mathrm{1}−{cos}\theta}{\mathrm{1}+{cos}\theta}=\frac{\mathrm{1}−{cos}\frac{\theta}{\mathrm{2}}}{\mathrm{1}+{cos}\frac{\theta}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}+\left(\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}−\mathrm{1}\right)}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$$\mid\frac{{p}−{q}}{{p}+{q}}\mid={tan}\frac{\theta}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Commented by manxsol last updated on 18/Feb/23

$${true},{no}\:{existe}\:{division} \\ $$$$\:{de}\:{vectores} \\ $$

Commented by mr W last updated on 18/Feb/23

$$¡{De}\:{nada}! \\ $$

Commented by manxsol last updated on 18/Feb/23

exercise debe ser asi to prove ((∣p^(→) −q^(→) ∣^2 )/(∣p^(→) +q^→ ∣^2 ))=tg(θ/2) p and q vectors unitary gracias por la atencion Sir W

$${exercise}\:{debe}\:{ser}\:{asi} \\ $$$${to}\:{prove}\:\:\:\frac{\mid\overset{\rightarrow} {{p}}−\overset{\rightarrow} {{q}}\mid^{\mathrm{2}} }{\mid\overset{\rightarrow} {{p}}+\overset{\rightarrow} {{q}}\mid^{\mathrm{2}} }={tg}\frac{\theta}{\mathrm{2}} \\ $$$${p}\:{and}\:{q}\:{vectors}\:{unitary} \\ $$$$ \\ $$$${gracias}\:{por}\:{la}\:{atencion}\: \\ $$$$\:{Sir}\:{W} \\ $$

Commented by mr W last updated on 18/Feb/23

$${what}\:{is}\:\mid\frac{{p}−{q}}{{p}+{q}}\mid? \\ $$

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