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Question
If \[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\], prove that \[xy + yz + zx = 0\]
Solution
\[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\]
\[ \Rightarrow \frac{\cos\theta}{\frac{1}{x}} = \frac{\cos\left( \theta + \frac{2\pi}{3} \right)}{\frac{1}{y}} = \frac{\cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{z}}\]
\[ \Rightarrow \frac{\cos\theta}{\frac{1}{x}} = \frac{\cos\left( \theta + \frac{2\pi}{3} \right)}{\frac{1}{y}} = \frac{\cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{z}} = \frac{\cos\theta + \cos\left( \theta + \frac{2\pi}{3} \right) + \cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \left( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = . . . = \frac{a + c + e + . . .}{b + d + f + . . .} \right)\]
\[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\]
\[ \Rightarrow \frac{\cos\theta}{\frac{1}{x}} = \frac{\cos\left( \theta + \frac{2\pi}{3} \right)}{\frac{1}{y}} = \frac{\cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{z}}\]
\[ \Rightarrow \frac{\cos\theta}{\frac{1}{x}} = \frac{\cos\left( \theta + \frac{2\pi}{3} \right)}{\frac{1}{y}} = \frac{\cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{z}} = \frac{\cos\theta + \cos\left( \theta + \frac{2\pi}{3} \right) + \cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \left( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = . . . = \frac{a + c + e + . . .}{b + d + f + . . .} \right)\]
\[\Rightarrow \frac{\cos\theta}{\frac{1}{x}} = \frac{\cos\left( \theta + \frac{2\pi}{3} \right)}{\frac{1}{y}} = \frac{\cos\left( \theta + \frac{4\pi}{3} \right)}{\frac{1}{z}} = \frac{0}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}\]
\[ \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\]
\[ \Rightarrow \frac{yz + zx + xy}{xyz} = 0\]
\[ \Rightarrow xy + yz + zx = 0\]
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