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

Explanation

To solve the problem of finding the angle between vectors \(\vec{A}\) and \(\vec{B}\) when their resultant is perpendicular to \(\vec{A}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
We know that the resultant vector \(\vec{R}\) of \(\vec{A}\) and \(\vec{B}\) is perpendicular to \(\vec{A}\). This means that \(\vec{R} \cdot \vec{A} = 0\).

2. **Express the Resultant Vector**:
The resultant vector \(\vec{R}\) can be expressed as:
\[ \vec{R} = \vec{A} + \vec{B} \]

3. **Use the Perpendicular Condition**:
Since \(\vec{R}\) is perpendicular to \(\vec{A}\), we can write:
\[ (\vec{A} + \vec{B}) \cdot \vec{A} = 0 \]
Expanding this using the dot product:
\[ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0 \]

4. **Substitute Magnitudes**:
Let \(A = |\vec{A}|\) and \(B = |\vec{B}|\). The dot product \(\vec{B} \cdot \vec{A}\) can also be expressed as:
\[ \vec{B} \cdot \vec{A} = |\vec{B}| |\vec{A}| \cos(\theta) \]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

5. **Set Up the Equation**:
Substituting this into our equation gives:
\[ A^2 + B A \cos(\theta) = 0 \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ B A \cos(\theta) = -A^2 \]
Dividing both sides by \(A\) (assuming \(A \neq 0\)):
\[ B \cos(\theta) = -A \]

7. **Solving for Cosine**:
From the above equation, we can express \(\cos(\theta)\):
\[ \cos(\theta) = -\frac{A}{B} \]

8. **Finding the Angle**:
Finally, we can find the angle \(\theta\) using the inverse cosine function:
\[ \theta = \cos^{-1}\left(-\frac{A}{B}\right) \]

### Conclusion:
Thus, the angle \(\theta\) between vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \theta = \cos^{-1}\left(-\frac{A}{B}\right) \]