Question

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. \(-A\sin \theta \hat{i}+A\cos \theta \hat{j}\)
• B. \(-A\cos \theta \hat{i}-A\sin \theta \hat{j}\)
• C. \(A\sin \theta \hat{i}-A\cos \theta \hat{j}\)
• D. \(A\cos \theta \hat{i}+A\sin \theta \hat{j}\)

Hint:

In the third quadrant both the x and y components are negative. The angle is measured from the negative X-axis, so the x part uses cosine and the y part uses sine.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A vector in the third quadrant points to the left and downward, so both of its components are negative.

Step 2: Resolve along the axes
The angle \(\theta\) is taken from the negative X-axis. The component along that axis is \(A\cos\theta\) and the component along the negative Y-axis is \(A\sin\theta\).

Step 3: Attach the signs
Both components are negative, so \(\vec{A}=-A\cos\theta\,\hat{i}-A\sin\theta\,\hat{j}\).

Step 4: Check
The magnitude is \(\sqrt{A^{2}\cos^{2}\theta+A^{2}\sin^{2}\theta}=A\), as it should be.

Final Answer:
The vector is resolved as minus A cos theta along i and minus A sin theta along j. \[ \boxed{\text{(B) }-A\cos \theta \hat{i}-A\sin \theta \hat{j}} \]