Let a c o s θ = b c o s ( θ + 2 π 3 ) = c c o s ( θ + 4 π 3 ) = k a\mathrm{cos}\theta =b\mathrm{cos}(\theta +\frac{2\pi }{3})=c\mathrm{cos}(\theta +\frac{4\pi }{3})=k a cos θ = b cos ( θ + 3 2 π ) = c cos ( θ + 3 4 π ) = k
⇒ a = k c o s θ , b = k c o s ( θ + 2 π 3 ) , c = k c o s ( θ + 4 π 3 ) \Rightarrow a=\frac{k}{\mathrm{cos}\theta },b=\frac{k}{\mathrm{cos}(\theta +\frac{2\pi }{3})},c=\frac{k}{\mathrm{cos}(\theta +\frac{4\pi }{3})} ⇒ a = cos θ k , b = cos ( θ + 3 2 π ) k , c = cos ( θ + 3 4 π ) k
⇒ a b + b c + c a = k 2 c o s ( θ + 4 π 3 ) + c o s θ + c o s ( θ + 2 π 3 ) c o s ( θ + 4 π 3 ) c o s θ c o s ( θ + 2 π 3 ) \Rightarrow ab+bc+ca={k}^{2}\frac{\mathrm{cos}(\theta +\frac{4\pi }{3})+\mathrm{cos}\theta +\mathrm{cos}(\theta +\frac{2\pi }{3})}{\mathrm{cos}(\theta +\frac{4\pi }{3})\mathrm{cos}\theta \mathrm{cos}(\theta +\frac{2\pi }{3})} ⇒ ab + b c + c a = k 2 cos ( θ + 3 4 π ) cos θ cos ( θ + 3 2 π ) cos ( θ + 3 4 π ) + cos θ + cos ( θ + 3 2 π )
= k 2 [ c o s θ + 2 c o s ( θ + π ) . c o s ( π 3 ) c o s θ . c o s ( θ + 2 π 3 ) . c o s ( θ + 4 π 3 ) ] ={k}^{2}[\frac{\mathrm{cos}\theta +2\mathrm{cos}(\theta +\pi ).\mathrm{cos}(\frac{\pi }{3})}{\mathrm{cos}\theta .\mathrm{cos}(\theta +\frac{2\pi }{3}).\mathrm{cos}(\theta +\frac{4\pi }{3})}] = k 2 [ cos θ . cos ( θ + 3 2 π ) . cos ( θ + 3 4 π ) cos θ + 2 cos ( θ + π ) . cos ( 3 π ) ]
= k 2 [ c o s θ − 2 c o s θ . 1 2 c o s θ . c o s ( θ + 2 π 3 ) . c o s ( θ + 4 π 3 ) ] = 0 ={k}^{2}[\frac{\mathrm{cos}\theta -2\mathrm{cos}\theta .\frac{1}{2}}{\mathrm{cos}\theta .\mathrm{cos}(\theta +\frac{2\pi }{3}).\mathrm{cos}(\theta +\frac{4\pi }{3})}]=0 = k 2 [ cos θ . cos ( θ + 3 2 π ) . cos ( θ + 3 4 π ) cos θ − 2 cos θ . 2 1 ] = 0
Let ϕ \phi ϕ be the angle between two given vectors.
∴ c o s ϕ = ( a i ^ + b j ^ + c k ^ ) . ( b i ^ + c j ^ + a k ^ ) a 2 + b 2 + c 2 ⋅ b 2 + c 2 + a 2 = a b + b c + c a a 2 + b 2 + c 2 = 0 \therefore \mathrm{cos}\phi =\frac{(a\hat{i}+b\hat{j}+c\hat{k}).(b\hat{i}+c\hat{j}+a\hat{k})}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\cdot \sqrt{{b}^{2}+{c}^{2}+{a}^{2}}}=\frac{ab+bc+ca}{{a}^{2}+{b}^{2}+{c}^{2}}=0 ∴ cos ϕ = a 2 + b 2 + c 2 ⋅ b 2 + c 2 + a 2 ( a i ^ + b j ^ + c k ^ ) . ( b i ^ + c j ^ + a k ^ ) = a 2 + b 2 + c 2 ab + b c + c a = 0
⇒ ϕ = π 2 \Rightarrow \phi =\frac{\pi }{2} ⇒ ϕ = 2 π