Q192 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
For the vector $\vec{A}$ making an angle $\theta$ with the negative X-axis and lying in the third quadrant, the standard resolved form is
- A$-A\sin\theta\,\hat{i} + A\cos\theta\,\hat{j}$
- Correct answerB$-A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$
- C$A\sin\theta\,\hat{i} - A\cos\theta\,\hat{j}$
- D$A\cos\theta\,\hat{i} + A\sin\theta\,\hat{j}$
Solution
The vector $\vec{A}$ lies in the third quadrant and makes an angle $\theta$ with the negative X-axis.
The component of the vector along the negative X-axis is $A\cos\theta$, so $A_x = -A\cos\theta$.
The component of the vector along the negative Y-axis is $A\sin\theta$, so $A_y = -A\sin\theta$.
Therefore, the vector in standard resolved form is $\vec{A} = -A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$.
Answer: $-A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$
The component of the vector along the negative X-axis is $A\cos\theta$, so $A_x = -A\cos\theta$.
The component of the vector along the negative Y-axis is $A\sin\theta$, so $A_y = -A\sin\theta$.
Therefore, the vector in standard resolved form is $\vec{A} = -A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$.
Answer: $-A\cos\theta\,\hat{i} - A\sin\theta\,\hat{j}$
Hence, the correct answer is option: 'B'.