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: Recall the normal form of a line.
The normal form of a straight line is \[ x\cos\alpha+y\sin\alpha=p \] where \( p \) is the perpendicular distance of the line from the origin, and \[ \cos\alpha,\ \sin\alpha \] are the direction ratios of the normal divided by its magnitude.

Step 2: Write the given equation clearly.
The given line is \[ x+\sqrt{3}y=2\sqrt{3} \] Comparing this with the general form \[ ax+by+c=0 \] we can rewrite it as \[ x+\sqrt{3}y-2\sqrt{3}=0 \] Here, \[ a=1,\qquad b=\sqrt{3} \]

Step 3: Convert the equation into normal form.
To convert into normal form, divide the whole equation by \[ \sqrt{a^2+b^2} \] Now, \[ \sqrt{a^2+b^2}=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2 \] So dividing the equation \[ x+\sqrt{3}y=2\sqrt{3} \] by \( 2 \), we get \[ \frac{x}{2}+\frac{\sqrt{3}}{2}y=\sqrt{3} \]

Step 4: Compare with the standard normal form.
Now compare \[ \frac{x}{2}+\frac{\sqrt{3}}{2}y=\sqrt{3} \] with \[ x\cos\alpha+y\sin\alpha=p \] Thus, \[ \cos\alpha=\frac{1}{2},\qquad \sin\alpha=\frac{\sqrt{3}}{2},\qquad p=\sqrt{3} \]

Step 5: Find the value of \( \alpha \).
We know that \[ \cos\alpha=\frac{1}{2} \quad \text{and} \quad \sin\alpha=\frac{\sqrt{3}}{2} \] This corresponds to the angle \[ \alpha=\frac{\pi}{3} \]

Step 6: Verify the value of \( p \).
From the comparison above, we directly get \[ p=\sqrt{3} \] Also, since \( p \) is a distance, it must be positive, which is satisfied here.

Step 7: Final conclusion.
Therefore, the required values are \[ \alpha=\frac{\pi}{3},\qquad p=\sqrt{3} \] Hence, the correct option is \[ \boxed{(3)\ \dfrac{\pi}{3}\ \text{and}\ \sqrt{3}} \]