The resultant of `vecA` and `vecB` is perpendicular to `vecA`. What is the angle between `vecA` and `vecB` ?

Options

A. `cos ^(-1)((A)/(B))`
B. `cos^(-1)(-(A)/(B))`
C. `sin^(-1)((A)/(B))`
D. `sin^(-1)(-(A)/(B))`

Explanation

To determine the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that their resultant is perpendicular to \(\vec{A}\), we can follow these steps:

1. **Understand the given condition**:
- The resultant vector \(\vec{R}\) of \(\vec{A}\) and \(\vec{B}\) is perpendicular to \(\vec{A}\).
- This implies that the dot product of \(\vec{R}\) and \(\vec{A}\) is zero: \(\vec{R} \cdot \vec{A} = 0\).

2. **Express the resultant vector**:
- The resultant vector \(\vec{R}\) is given by \(\vec{R} = \vec{A} + \vec{B}\).

3. **Apply the dot product condition**:
- Since \(\vec{R}\) is perpendicular to \(\vec{A}\), we have:
\[ (\vec{A} + \vec{B}) \cdot \vec{A} = 0 \]
- Expanding this dot product:
\[ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0 \]

4. **Simplify the dot product**:
- Let \(A\) and \(B\) be the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\), respectively.
- Let \(\theta\) be the angle between \(\vec{A}\) and \(\vec{B}\).
- The dot product \(\vec{A} \cdot \vec{A} = A^2\).
- The dot product \(\vec{B} \cdot \vec{A} = B \cdot A \cdot \cos(\theta)\).
- So, the equation becomes:
\[ A^2 + AB \cos(\theta) = 0 \]

5. **Solve for \(\cos(\theta)\)**:
- Rearrange the equation to solve for \(\cos(\theta)\):
\[ A^2 + AB \cos(\theta) = 0 \]
\[ AB \cos(\theta) = -A^2 \]
\[ \cos(\theta) = -\frac{A}{B} \]

6. **Find the angle \(\theta\)**:
- The angle \(\theta\) can be found using the inverse cosine function:
\[ \theta = \cos^{-1}\left(-\frac{A}{B}\right) \]

Therefore, the angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta = \cos^{-1}\left(-\frac{A}{B}\right)\).