Question

If the normal form of the equation of a straight line \( x+\sqrt{3}y=2\sqrt{3} \) is \( x\cos\alpha+y\sin\alpha=p \), then the values of \( \alpha \) and \( p \) are respectively

• A. \( \dfrac{\pi}{6} \) and \( \sqrt{6} \)
• B. \( \dfrac{\pi}{3} \) and \( \sqrt{6} \)
• C. \( \dfrac{\pi}{3} \) and \( \sqrt{3} \)
• D. \( \dfrac{\pi}{6} \) and \( \sqrt{3} \)
• E. \( \dfrac{\pi}{4} \) and \( \sqrt{6} \)

Hint:

To convert a line into normal form, divide the equation by \( \sqrt{a^2+b^2} \). Then compare the coefficients directly with \( \cos\alpha \) and \( \sin\alpha \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The normal form of a line is \(x\cos\alpha + y\sin\alpha = p\), where \(p\) is the length of the perpendicular from the origin to the line, and \(\alpha\) is the angle this perpendicular makes with the positive x-axis. To convert a general equation \(Ax+By=C\) (with \(C>0\)) to normal form, we divide the entire equation by \(\sqrt{A^2+B^2}\).
Step 2: Key Formula or Approach:
Given the equation \(Ax+By=C\).
1. Ensure \(C\) is positive. If not, multiply the equation by -1. 2. Calculate the normalizing factor \( \sqrt{A^2+B^2} \). 3. Divide the equation by this factor: \( \frac{A}{\sqrt{A^2+B^2}}x + \frac{B}{\sqrt{A^2+B^2}}y = \frac{C}{\sqrt{A^2+B^2}} \). 4. Compare this with \(x\cos\alpha + y\sin\alpha = p\) to find \(\cos\alpha, \sin\alpha,\) and \(p\).
Step 3: Detailed Explanation:
The given equation is \(x + \sqrt{3}y = 2\sqrt{3}\).
Here, \(A=1\), \(B=\sqrt{3}\), and \(C=2\sqrt{3}\). Since C is positive, we can proceed.
Calculate the normalizing factor:
\[ \sqrt{A^2+B^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \] Divide the entire equation by 2:
\[ \frac{1}{2}x + \frac{\sqrt{3}}{2}y = \frac{2\sqrt{3}}{2} \] \[ \frac{1}{2}x + \frac{\sqrt{3}}{2}y = \sqrt{3} \] Now, compare this with the normal form \(x\cos\alpha + y\sin\alpha = p\).
We have:
\[ p = \sqrt{3} \] \[ \cos\alpha = \frac{1}{2} \] \[ \sin\alpha = \frac{\sqrt{3}}{2} \] Since both \(\cos\alpha\) and \(\sin\alpha\) are positive, the angle \(\alpha\) lies in the first quadrant. The angle for which these conditions are met is \(\alpha = \frac{\pi}{3}\) (or 60\(^\circ\)).
Step 4: Final Answer:
The values are \(\alpha = \frac{\pi}{3}\) and \(p = \sqrt{3}\). This corresponds to option (C).