Question

The angle made by vector \(\overset{⃗}{A} = 2\hat{i}+3\hat{j}\) with x-axis and that with y-axis are respectively.

• A. \(tan^{-1}2\sqrt{13}\) , \(tan^{-1}3\sqrt{13}\)
• B. \(cos^{-1}\frac{2}{\sqrt{13}}\) , \(cos^{-1}\frac{3}{\sqrt{13}}\)
• C. \(cos^{-1}\frac{1}{\sqrt{2}}\) , \(cos^{-1}\frac{1}{\sqrt{3}}\)
• D. \(sin^{-1}\frac{1}{\sqrt{6}}\) , \(sin^{-1}\frac{1}{2\sqrt{3}}\)

Hint:

The angle with an axis comes from the direction cosine, which is the component divided by the magnitude.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The angle a vector makes with an axis is found from its direction cosine. For the angle \(\alpha\) with the x-axis, \(\cos\alpha = \frac{A_x}{|\vec A|}\). For the angle \(\beta\) with the y-axis, \(\cos\beta = \frac{A_y}{|\vec A|}\).

Step 2: Key Formula or Approach:
The magnitude is \(|\vec A| = \sqrt{2^2 + 3^2} = \sqrt{13}\).

Step 3: Detailed Explanation:
With the x-axis: \(\cos\alpha = \frac{2}{\sqrt{13}}\), so \(\alpha = \cos^{-1}\frac{2}{\sqrt{13}}\).
With the y-axis: \(\cos\beta = \frac{3}{\sqrt{13}}\), so \(\beta = \cos^{-1}\frac{3}{\sqrt{13}}\).
Option A gives \(\tan^{-1}(2\sqrt{13})\), which is a much larger angle than the vector makes with the x-axis. Option C would be right only for a vector with equal components. Option D uses sine values that do not match \(\frac{3}{\sqrt{13}}\) or \(\frac{2}{\sqrt{13}}\).

Final Answer:
The angles are \(\cos^{-1}\frac{2}{\sqrt{13}}\) and \(\cos^{-1}\frac{3}{\sqrt{13}}\), option (B). \[ \boxed{\cos^{-1}\frac{2}{\sqrt{13}},\ \cos^{-1}\frac{3}{\sqrt{13}}} \]