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Question Number 58312 by pete last updated on 21/Apr/19

Show that the angle θ between two unit  vectors a_� ^�  and b_� ^�  is given by cosθ=a_� ^� •b_� ^� .  Hence, given that a_� ^� =i_� cosA+j_� sinA and  b_� ^� =i_� cosB−j_� sinB, prove that cos(A+B)=  cosAcosB−sinAsinB.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angle}\:\theta\:\mathrm{between}\:\mathrm{two}\:\mathrm{unit} \\ $$$$\mathrm{vectors}\:\underset{} {\hat {\mathrm{a}}}\:\mathrm{and}\:\underset{} {\hat {\mathrm{b}}}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{cos}\theta=\underset{} {\hat {\mathrm{a}}}\bullet\underset{} {\hat {\mathrm{b}}}. \\ $$$$\mathrm{Hence},\:\mathrm{given}\:\mathrm{that}\:\underset{} {\hat {\mathrm{a}}}=\underset{} {\mathrm{i}cosA}+\underset{} {\mathrm{j}sinA}\:\mathrm{and} \\ $$$$\underset{} {\hat {\mathrm{b}}}=\underset{} {\mathrm{i}cosB}−\underset{} {\mathrm{j}sinB},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)= \\ $$$$\mathrm{cosAcosB}−\mathrm{sinAsinB}. \\ $$

Commented by tanmay last updated on 21/Apr/19

Commented by tanmay last updated on 21/Apr/19

vector op^→ =a^∧ =icosA+jsinA  vector oQ^→ =b^∧ =icosB−jsinB  angle between op^→  and oQ^→  is<POQ=A+B  a^∧ .b^∧ =∣a^∧ ∣ ∣b^∧ ∣ cos(A+B)=1×1×cos(A+B)=cos(A+B)  a^∧ .b^∧   =(icosA+jsinA).(icosB−jsinB)  =cosAcosB−sinA sinB  hence equating value of a^∧ .b^∧   cos(A+B)=cosAcosB−sinB  formula...  if A^→ =ix_1 +jy_1 +kz_1   B^→ =ix_2 +jy_2 +kz_2   A^→ .B^→ =∣A^→ ∣ ∣B^→ ∣  cosθ  ∣A∣=(√(x_1 ^2 +y_1 ^2 +z_1 ^2 ))   and ∣B^→ ∣=(√(x_2 ^2 +y_2 ^2 +z_2 ^2 ))   A^→ .B^→ =x_1 x_2 +y_1 y_2 +z_1 z_2   cosθ=((A^→ .B^→ )/(∣A^→ ∣ ∣B^→ ∣))=((x_1 x_2 +y_1 y_2 +z_1 z_2 )/((√(x_1 ^2 +y_1 ^2 +z_1 ^2 ))  ×(√(x_2 ^2 +y_2 ^2 +z_2 ^2 )) ))

$${vector}\:{o}\overset{\rightarrow} {{p}}=\overset{\wedge} {{a}}={icosA}+{jsinA} \\ $$$${vector}\:{o}\overset{\rightarrow} {{Q}}=\overset{\wedge} {{b}}={icosB}−{jsinB} \\ $$$${angle}\:{between}\:{o}\overset{\rightarrow} {{p}}\:{and}\:{o}\overset{\rightarrow} {{Q}}\:{is}<{POQ}={A}+{B} \\ $$$$\overset{\wedge} {{a}}.\overset{\wedge} {{b}}=\mid\overset{\wedge} {{a}}\mid\:\mid\overset{\wedge} {{b}}\mid\:{cos}\left({A}+{B}\right)=\mathrm{1}×\mathrm{1}×{cos}\left({A}+{B}\right)={cos}\left({A}+{B}\right) \\ $$$$\overset{\wedge} {{a}}.\overset{\wedge} {{b}} \\ $$$$=\left({icosA}+{jsinA}\right).\left({icosB}−{jsinB}\right) \\ $$$$={cosAcosB}−{sinA}\:{sinB} \\ $$$${hence}\:{equating}\:{value}\:{of}\:\overset{\wedge} {{a}}.\overset{\wedge} {{b}} \\ $$$${cos}\left({A}+{B}\right)={cosAcosB}−{sinB} \\ $$$${formula}... \\ $$$${if}\:\overset{\rightarrow} {{A}}={ix}_{\mathrm{1}} +{jy}_{\mathrm{1}} +{kz}_{\mathrm{1}} \\ $$$$\overset{\rightarrow} {{B}}={ix}_{\mathrm{2}} +{jy}_{\mathrm{2}} +{kz}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{A}}.\overset{\rightarrow} {{B}}=\mid\overset{\rightarrow} {{A}}\mid\:\mid\overset{\rightarrow} {{B}}\mid\:\:{cos}\theta \\ $$$$\mid{A}\mid=\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{1}} ^{\mathrm{2}} }\:\:\:{and}\:\mid\overset{\rightarrow} {{B}}\mid=\sqrt{{x}_{\mathrm{2}} ^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} }\: \\ $$$$\overset{\rightarrow} {{A}}.\overset{\rightarrow} {{B}}={x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} \\ $$$${cos}\theta=\frac{\overset{\rightarrow} {{A}}.\overset{\rightarrow} {{B}}}{\mid\overset{\rightarrow} {{A}}\mid\:\mid\overset{\rightarrow} {{B}}\mid}=\frac{{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{1}} ^{\mathrm{2}} }\:\:×\sqrt{{x}_{\mathrm{2}} ^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} }\:} \\ $$$$ \\ $$

Commented by pete last updated on 21/Apr/19

Thanks very much

$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much} \\ $$

Commented by tanmay last updated on 21/Apr/19

most welcome...

$${most}\:{welcome}... \\ $$

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