A) 1
B) 2
C) 0
D) \[\frac{1}{27}\left( \begin{matrix} -1 & 26 \\ 0 & 27 \\ \end{matrix} \right)\]
Correct Answer: C
Solution :
We have, \[x\cos \theta =y\cos \left( \theta +\frac{2\pi }{3} \right)\] \[z=\cos \left( \theta +\frac{4\pi }{3} \right)=k\] \[\Rightarrow \] \[\cos \theta =\frac{k}{x}\cos \left( \theta +\frac{2\pi }{3} \right)=\frac{k}{y}\] and \[\cos \left( \theta +\frac{4\pi }{3} \right)=\frac{k}{z}\] hence, \[\frac{k}{x}+\frac{k}{y}+\frac{k}{z}=\cos \theta +\cos \left( \theta +\frac{2\pi }{3} \right)\] \[+\cos \left( \theta +\frac{4\pi }{3} \right)\] \[=\cos \theta +\cos \left( \frac{\pi }{3}-\theta \right)-\cos \left( \frac{\pi }{3}+\theta \right)\] \[=\cos \theta -2\cos \frac{\pi }{3}\cos \theta =0\]
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