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: Use tangents:
The vector points up and right. With the x-axis, $\tan\alpha = \frac{A_y}{A_x} = \frac32$, and with the y-axis, $\tan\beta = \frac{A_x}{A_y} = \frac23$.

Step 2: Convert to cosines:
If $\tan\alpha = \frac32$, the right triangle has sides $2$, $3$ and hypotenuse $\sqrt{13}$, so $\cos\alpha = \frac{2}{\sqrt{13}}$.
In the same way, $\cos\beta = \frac{3}{\sqrt{13}}$. This matches option B.

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