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The number of all possible triplets (x, y, z) such that (x+y)+(y+2 z) \cos 2 \theta+(z-x) \sin ^2 \theta=0\: f\! or \: all \: \theta is

Option: 1

0


Option: 2

1


Option: 3

3


Option: 4

infinite


Answers (1)

best_answer

We have

(x+y)+(y+2 k) \cos 2 \theta+(z-x)\left(\frac{1-\cos 2 \theta}{2}\right)=0

\begin{gathered} \Rightarrow(2 x+2 y)+(2 y+4 z) \cos 2 \theta+(z-x) \\ -(z-x) \cos 2 \theta=0 \\ \Rightarrow(2 x+2 y+z-x)+(2 y+4 z-z+x) \cos 2 \theta=0 \\ (x+2 y+z)+(x+2 y+3 z) \cos 2 \theta=0 \end{gathered}
Which is zero for all values of \theta

\Rightarrow x+2 y+z=0 \\ x+2 y+3 z=0

\begin{aligned} & \frac{x}{6-2}=\frac{y}{1-3}=\frac{z}{2-2}=\lambda \\\\ & x=2 \lambda, y=-\lambda, z=0 \\ &(2 \lambda,-\lambda, 0) \end{aligned}

For any \lambda \in R

Hence, there are infinite number of triplets.

 

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mansi

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