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Question Number 39172 by Rio Mike last updated on 03/Jul/18

find the angle between 3i − 4j and i − j

$${find}\:{the}\:{angle}\:{between}\: \\ $$$$\mathrm{3}{i}\:−\:\mathrm{4}{j}\:{and}\:{i}\:−\:{j} \\ $$

Commented by math khazana by abdo last updated on 03/Jul/18

let u^→ =3i−4j ⇒u(3,−4) v^→ =i−j?⇒v^→ (1,−1) cos(u^→ ,v^→ ) = ((u^→ .v^→ )/(∣∣u^→ ∣∣.∣∣v^→ ∣∣)) = ((3×1) +(4))/(5(√2))) =(7/(5(√2))) sin(u^→ ,v^→ ) =((det (u,v))/(∣∣u∣∣.∣∣v∣∣)) =( determinant (((3 1)),((−4 −1)))/(5(√2))) = (1/(5(√2))) ⇒ tan(u,v) = (1/7) ⇒ θ =arctan((1/7)).

$${let}\:\overset{\rightarrow} {{u}}=\mathrm{3}{i}−\mathrm{4}{j}\:\Rightarrow{u}\left(\mathrm{3},−\mathrm{4}\right) \\ $$$$\overset{\rightarrow} {{v}}={i}−{j}?\Rightarrow\overset{\rightarrow} {{v}}\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${cos}\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right)\:=\:\frac{\overset{\rightarrow} {{u}}\:.\overset{\rightarrow} {{v}}}{\mid\mid\overset{\rightarrow} {{u}}\mid\mid.\mid\mid\overset{\rightarrow} {{v}}\mid\mid}\:=\:\frac{\left.\mathrm{3}×\mathrm{1}\right)\:+\left(\mathrm{4}\right)}{\mathrm{5}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{7}}{\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$${sin}\left(\overset{\rightarrow} {{u}}\:,\overset{\rightarrow} {{v}}\right)\:=\frac{{det}\:\left({u},{v}\right)}{\mid\mid{u}\mid\mid.\mid\mid{v}\mid\mid}\:=\frac{\begin{vmatrix}{\mathrm{3}\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{4}\:\:\:−\mathrm{1}}\end{vmatrix}}{\mathrm{5}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${tan}\left({u},{v}\right)\:=\:\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\:\theta\:={arctan}\left(\frac{\mathrm{1}}{\mathrm{7}}\right). \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

vector 3i−4j make angle α with x axis so m_1 =tanα=((−4)/3) m_2 =tanβ=((−1)/1) tanθ=((m_1 ∼m_2 )/(1+m_1 m_2 ))=((−1+(4/3))/(1+(4/3)))=((1/3)/(7/3))=(1/7) θ=tan^(−1) ((1/7)) cosθ=(7/((√(50)) )) another approach... cosθ=((3×1+(−4)×(−1))/((√(3^2 +(−4)^2 )) ×(√(1^2 +(−1)^2 )))) =(7/(√(50)))

$${vector}\:\mathrm{3}{i}−\mathrm{4}{j}\:\:{make}\:{angle}\:\alpha\:{with}\:{x}\:{axis}\: \\ $$$${so}\:{m}_{\mathrm{1}} ={tan}\alpha=\frac{−\mathrm{4}}{\mathrm{3}} \\ $$$${m}_{\mathrm{2}} ={tan}\beta=\frac{−\mathrm{1}}{\mathrm{1}} \\ $$$${tan}\theta=\frac{{m}_{\mathrm{1}} \sim{m}_{\mathrm{2}} }{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} }=\frac{−\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{7}}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$$${cos}\theta=\frac{\mathrm{7}}{\sqrt{\mathrm{50}}\:} \\ $$$$ \\ $$$${another}\:{approach}... \\ $$$${cos}\theta=\frac{\mathrm{3}×\mathrm{1}+\left(−\mathrm{4}\right)×\left(−\mathrm{1}\right)}{\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:×\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{7}}{\sqrt{\mathrm{50}}} \\ $$

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