Q102 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
The angle made by vector $\vec{A} = 2\hat{i} + 3\hat{j}$ with x-axis and that with y-axis are respectively.
- A$\tan^{-1} 2\sqrt{13}$ , $\tan^{-1} 3\sqrt{13}$
- Correct answerB$\cos^{-1}\dfrac{2}{\sqrt{13}}$ , $\cos^{-1}\dfrac{3}{\sqrt{13}}$
- C$\cos^{-1}\dfrac{1}{\sqrt{2}}$ , $\cos^{-1}\dfrac{1}{\sqrt{3}}$
- D$\sin^{-1}\dfrac{1}{\sqrt{6}}$ , $\sin^{-1}\dfrac{1}{2\sqrt{3}}$
Solution
Given vector $\vec{A} = 2\hat{i} + 3\hat{j}$
The magnitude of the vector is $|\vec{A}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$
Let $\alpha$ be the angle made by the vector with the x-axis. Then,
$\cos \alpha = \dfrac{A_x}{|\vec{A}|} = \dfrac{2}{\sqrt{13}} \Rightarrow \alpha = \cos^{-1}\left(\dfrac{2}{\sqrt{13}}\right)$
Let $\beta$ be the angle made by the vector with the y-axis. Then,
$\cos \beta = \dfrac{A_y}{|\vec{A}|} = \dfrac{3}{\sqrt{13}} \Rightarrow \beta = \cos^{-1}\left(\dfrac{3}{\sqrt{13}}\right)$
Answer: $\cos^{-1}\dfrac{2}{\sqrt{13}}$ , $\cos^{-1}\dfrac{3}{\sqrt{13}}$
The magnitude of the vector is $|\vec{A}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$
Let $\alpha$ be the angle made by the vector with the x-axis. Then,
$\cos \alpha = \dfrac{A_x}{|\vec{A}|} = \dfrac{2}{\sqrt{13}} \Rightarrow \alpha = \cos^{-1}\left(\dfrac{2}{\sqrt{13}}\right)$
Let $\beta$ be the angle made by the vector with the y-axis. Then,
$\cos \beta = \dfrac{A_y}{|\vec{A}|} = \dfrac{3}{\sqrt{13}} \Rightarrow \beta = \cos^{-1}\left(\dfrac{3}{\sqrt{13}}\right)$
Answer: $\cos^{-1}\dfrac{2}{\sqrt{13}}$ , $\cos^{-1}\dfrac{3}{\sqrt{13}}$
Hence, the correct answer is option: 'B'.