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. $n\pi - \frac{\pi}{6}, n \in Z$
• B. $2n\pi \pm \frac{11\pi}{6}, n \in Z$
• C. $n\pi \pm \frac{\pi}{6}, n \in Z$
• D. $2n\pi + \frac{11\pi}{6}, n \in Z$

Answer 1

Answer: (D)

$2n\pi + \frac{11\pi}{6}, n \in Z$

Answer 2

The given equation is $(\sqrt{3} \tan \theta + 1) \overrightarrow{a} + (\sqrt{3} \sec \theta - 2) \overrightarrow{b} = 0$.

Since the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are linearly independent, the coefficients of $\overrightarrow{a}$ and $\overrightarrow{b}$ must both be zero. This gives us two equations:

1. $\sqrt{3} \tan \theta + 1 = 0$
2. $\sqrt{3} \sec \theta - 2 = 0$

From equation (1):

$\sqrt{3} \tan \theta = -1$

$\tan \theta = -\frac{1}{\sqrt{3}}$

The general solution for $\tan \theta = -\frac{1}{\sqrt{3}}$ is $\theta = n\pi - \frac{\pi}{6}$, where $n \in Z$.

From equation (2):

$\sqrt{3} \sec \theta = 2$

$\sec \theta = \frac{2}{\sqrt{3}}$

Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\cos \theta = \frac{\sqrt{3}}{2}$.

The general solution for $\cos \theta = \frac{\sqrt{3}}{2}$ is $\theta = 2m\pi \pm \frac{\pi}{6}$, where $m \in Z$.

For $\theta$ to satisfy both equations, the values of $\theta$ must be common to both general solutions.

The first set of solutions is $\theta = n\pi - \frac{\pi}{6}$, $n \in Z$.

The second set of solutions is $\theta = 2m\pi \pm \frac{\pi}{6}$, $m \in Z$.

Let's check which values from the first set are also in the second set.

Case 1: $n$ is even. Let $n = 2k$ for some integer $k$.

Then $\theta = 2k\pi - \frac{\pi}{6}$.

This is of the form $2m\pi - \frac{\pi}{6}$ with $m=k$, which is included in the second set of solutions ($2m\pi \pm \frac{\pi}{6}$ with the negative sign).

Let's verify if this satisfies the condition $\cos \theta = \frac{\sqrt{3}}{2}$.

$\cos(2k\pi - \frac{\pi}{6}) = \cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. This is satisfied.

Let's verify if this satisfies the condition $\tan \theta = -\frac{1}{\sqrt{3}}$.

$\tan(2k\pi - \frac{\pi}{6}) = \tan(-\frac{\pi}{6}) = -\tan(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$. This is satisfied.

So, $\theta = 2k\pi - \frac{\pi}{6}$ for $k \in Z$ is a valid set of solutions.

Case 2: $n$ is odd. Let $n = 2k+1$ for some integer $k$.

Then $\theta = (2k+1)\pi - \frac{\pi}{6} = 2k\pi + \pi - \frac{\pi}{6} = 2k\pi + \frac{5\pi}{6}$.

Let's check if this satisfies the condition $\cos \theta = \frac{\sqrt{3}}{2}$.

$\cos(2k\pi + \frac{5\pi}{6}) = \cos(\frac{5\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$.

This is not equal to $\frac{\sqrt{3}}{2}$. So, values of $\theta$ where $n$ is odd in the first set of solutions are not valid.

Thus, the common solutions are $\theta = 2k\pi - \frac{\pi}{6}$ for $k \in Z$.

We need to match this form with the given options.

Let's rewrite the solution:

$\theta = 2k\pi - \frac{\pi}{6} = 2k\pi + (2\pi - \frac{\pi}{6}) - 2\pi = (2k+2)\pi + (\frac{11\pi}{6} - 2\pi) + 2\pi - 2\pi = (2k+2)\pi - \frac{\pi}{6}$.

Alternatively, $\theta = 2k\pi - \frac{\pi}{6} = 2k\pi + \frac{11\pi}{6} - 2\pi = (2k-2)\pi + \frac{11\pi}{6}$.

Let $m = k-1$. As $k$ takes all integer values, $m$ also takes all integer values.

So the solution set is $\theta = 2m\pi + \frac{11\pi}{6}$, where $m \in Z$.

This matches option (4).

Let's check option (4): $2n\pi + \frac{11\pi}{6}, n \in Z$.

$\theta = 2n\pi + \frac{11\pi}{6}$.

$\tan \theta = \tan(2n\pi + \frac{11\pi}{6}) = \tan(\frac{11\pi}{6}) = \tan(2\pi - \frac{\pi}{6}) = -\tan(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$. This is satisfied.

$\cos \theta = \cos(2n\pi + \frac{11\pi}{6}) = \cos(\frac{11\pi}{6}) = \cos(2\pi - \frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. This is satisfied.

So, option (4) represents the correct general solution.