Question

If the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are linearly independent satisfying $(\sqrt{3}\tan{\theta}+1)\overrightarrow{a}+(\sqrt{3}\sec{\theta}-2)\overrightarrow{b}=0$, then the most general values of $\theta$ are

• A. $\theta = n\pi - \frac{\pi}{6}$, where $n \in \mathbb{Z}$
• B. $\theta = 2n\pi \pm \frac{\pi}{6}$, where $n \in \mathbb{Z}$
• C. $\theta = 2n\pi - \frac{\pi}{6}$, where $n \in \mathbb{Z}$
• D. $\theta = n\pi + \frac{\pi}{6}$, where $n \in \mathbb{Z}$

Answer 1

Answer: (C)

$\theta = 2n\pi - \frac{\pi}{6}$, where $n \in \mathbb{Z}$

Answer 2

• For linearly independent vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, if $c_1\overrightarrow{a} + c_2\overrightarrow{b} = \overrightarrow{0}$, then $c_1=0$ and $c_2=0$.
• This leads to solving two equations:
  1. $\sqrt{3}\tan{\theta}+1 = 0 \implies \tan{\theta} = -\frac{1}{\sqrt{3}}$
  2. $\sqrt{3}\sec{\theta}-2 = 0 \implies \sec{\theta} = \frac{2}{\sqrt{3}} \implies \cos{\theta} = \frac{\sqrt{3}}{2}$
• The general solution for $\tan{\theta} = -\frac{1}{\sqrt{3}}$ is $\theta = n\pi - \frac{\pi}{6}$, where $n \in \mathbb{Z}$.
• The general solution for $\cos{\theta} = \frac{\sqrt{3}}{2}$ is $\theta = 2m\pi \pm \frac{\pi}{6}$, where $m \in \mathbb{Z}$.
• To satisfy both conditions, we need to find the intersection of these solution sets.
  • If $\theta = 2m\pi + \frac{\pi}{6}$, then $n\pi - \frac{\pi}{6} = 2m\pi + \frac{\pi}{6} \implies (n-2m)\pi = \frac{\pi}{3} \implies n-2m = \frac{1}{3}$, which has no integer solutions for $n, m$.
  • If $\theta = 2m\pi - \frac{\pi}{6}$, then $n\pi - \frac{\pi}{6} = 2m\pi - \frac{\pi}{6} \implies n\pi = 2m\pi \implies n=2m$. This means $n$ must be an even integer.
• Therefore, the common solution is of the form $\theta = 2k\pi - \frac{\pi}{6}$, where $k \in \mathbb{Z}$.