Question

If the lines $\left[\begin{array}{ccc}\lambda x& sin\alpha y& cos\alpha \\ x& cos\alpha y& sin\alpha \\ -x& sin\alpha y& -cos\alpha \end{array}\right]$ pass through the same point where $\alpha ϵR$, then $\lambda$ lies is interval

• A. $[-1,1]$
• B. $[-\sqrt{2},\sqrt{2}]$
• C. $[-2,2]$
• D. $[-8,8]$

Answer 1

Answer: (B)

$[-\sqrt{2},\sqrt{2}]$

For non-trivial solution,

$xy\left|\begin{array}{ccc}\lambda & sin\alpha & cos\alpha \\ 1& cos\alpha & sin\alpha \\ -1& sin\alpha & -cos\alpha \end{array}\right|=0$

$\therefore xy=0$ or $\left|\begin{array}{ccc}\lambda & sin\alpha & cos\alpha \\ 1& cos\alpha & sin\alpha \\ -1& sin\alpha & -cos\alpha \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}\lambda & sin\alpha & cos\alpha \\ 1& cos\alpha & sin\alpha \\ -1& sin\alpha & -cos\alpha \end{array}\right|=0$

Expanding along $C_1$

$\Rightarrow \lambda (-{cos}^2\alpha -{sin}^2\alpha )-1(-sin\alpha cos\alpha -sin\alpha cos\alpha )-1({sin}^2\alpha -{cos}^2\alpha )=0$

$\Rightarrow -\lambda +sin2\alpha +cos2\alpha =0$

$\Rightarrow \lambda =sin2\alpha +cos2\alpha =\sqrt{2}sin(\frac{\pi }{4}+2a)$

Since $-1\le sin(\frac{\pi }{4}+2a)\le 1$

$\Rightarrow -\sqrt{2}\le \lambda \le \sqrt{2}$