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 solve the problem, we need to find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that their resultant \(\vec{R} = \vec{A} + \vec{B}\) is perpendicular to \(\vec{A}\).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the resultant vector \(\vec{R}\) is given by:
\[ \vec{R} = \vec{A} + \vec{B} \]
Since \(\vec{R}\) is perpendicular to \(\vec{A}\), we can use the property of dot products.

2. **Using the Dot Product**:
If two vectors are perpendicular, their dot product is zero. Therefore, we can write:
\[ \vec{R} \cdot \vec{A} = 0 \]
Substituting \(\vec{R}\) into this equation gives:
\[ (\vec{A} + \vec{B}) \cdot \vec{A} = 0 \]

3. **Expanding the Dot Product**:
Using the distributive property of the dot product, we can expand the left-hand side:
\[ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0 \]
Here, \(\vec{A} \cdot \vec{A}\) is the magnitude of \(\vec{A}\) squared, denoted as \(|\vec{A}|^2\), and \(\vec{B} \cdot \vec{A}\) can be expressed as \(|\vec{B}||\vec{A}|\cos\theta\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

4. **Setting Up the Equation**:
We can rewrite the equation as:
\[ |\vec{A}|^2 + |\vec{B}||\vec{A}|\cos\theta = 0 \]

5. **Isolating \(\cos\theta\)**:
Rearranging the equation gives:
\[ |\vec{B}||\vec{A}|\cos\theta = -|\vec{A}|^2 \]
Dividing both sides by \(|\vec{B}||\vec{A}|\) (assuming \(|\vec{B}|\) and \(|\vec{A}|\) are not zero):
\[ \cos\theta = -\frac{|\vec{A}|}{|\vec{B}|} \]

6. **Finding the Angle \(\theta\)**:
To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{|\vec{A}|}{|\vec{B}|}\right) \]

### Final Answer:
The angle between \(\vec{A}\) and \(\vec{B}\) is:
\[ \theta = \cos^{-1}\left(-\frac{|\vec{A}|}{|\vec{B}|}\right) \]