derivation-or-proof-or-full-explanation-of-R-R-1-2-R-2-2-2R-1-R-2-cos-1-2-

Question Number 47639 by azharkhan250963@gmail.com last updated on 12/Nov/18

derivation or proof or full explanation

of R = {(R_{1}^{2} + R_{2}^{2} + {2 R}_{1} . R_{2} \cos θ)}^{1 / 2}

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

\vec{R} = {\vec{R}}_{1} + {\vec{R}}_{2}

\vec{R} . \vec{R} = ({\vec{R}}_{1} + {\vec{R}}_{2}) . ({\vec{R}}_{1} + {\vec{R}}_{2}) \dots . . (1)

[n o w \vec{A} . \vec{B} =∣ \vec{A} ∣∣ \vec{B} ∣ c o s α = a b c o s α

∣ \vec{A} ∣= a ∣ \vec{B} ∣= b

w h e r e α i s a n g l e b e t w e e n \vec{A} a n d \vec{B}

a g a i n \vec{P} . \vec{P} =∣ \vec{P} ∣∣ \vec{P} ∣ c o s 0^{o} =∣ \vec{P} ∣^{2} = p^{2}]

s o f r o m 1 w e g e t

\vec{R} . \vec{R} = {\vec{R}}_{1} . {\vec{R}}_{1} + 2 {\vec{R}}_{1} . {\vec{R}}_{2} + {\vec{R}}_{2} . {\vec{R}}_{2}

R^{2} = R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} c o s θ

R = \sqrt{R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} c o s θ}

w h e r e ∣ \vec{R} ∣= R ∣ {\vec{R}}_{1} ∣= R_{1} ∣ {\vec{R}}_{2} ∣= R_{2}