Given that `vec(A)+vec(B)=vec(C )` and that `vec(C )` is perpendicular to `vec(A)` Further if `|vec(A)|=|vec(C )|`, then what is the angle between `vec(A)` and `vec(B)`

Options

A. `(pi)/4` radian
B. `(pi)/2` radian
C. `(3pi)/(4)` radian
D. `pi` radian

Explanation

To solve the problem, we start with the given information:

1. **Given Equations**:
\[ \vec{A} + \vec{B} = \vec{C} \]
\(\vec{C}\) is perpendicular to \(\vec{A}\), which means:
\[ \vec{A} \cdot \vec{C} = 0 \]
The magnitudes are equal:
\[ |\vec{A}| = |\vec{C}| \]

2. **Using the Perpendicular Condition**:
Since \(\vec{C}\) is perpendicular to \(\vec{A}\), we can express \(\vec{C}\) in terms of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{C} = \vec{A} + \vec{B} \]
Taking the dot product of \(\vec{A}\) with both sides:
\[ \vec{A} \cdot \vec{C} = \vec{A} \cdot (\vec{A} + \vec{B}) = \vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} \]
Since \(\vec{A} \cdot \vec{C} = 0\), we have:
\[ |\vec{A}|^2 + \vec{A} \cdot \vec{B} = 0 \]
This implies:
\[ \vec{A} \cdot \vec{B} = -|\vec{A}|^2 \]

3. **Using Magnitudes**:
We know from the problem statement that:
\[ |\vec{C}| = |\vec{A}| \]
Therefore, we can write:
\[ |\vec{C}|^2 = |\vec{A}|^2 \]
Now, since \(\vec{C} = \vec{A} + \vec{B}\), we can express the magnitude of \(\vec{C}\):
\[ |\vec{C}|^2 = |\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} \]
Setting this equal to \(|\vec{A}|^2\):
\[ |\vec{A}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} \]
Simplifying gives:
\[ 0 = |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} \]

4. **Substituting \(\vec{A} \cdot \vec{B}\)**:
From our earlier result, we substitute \(\vec{A} \cdot \vec{B} = -|\vec{A}|^2\):
\[ 0 = |\vec{B}|^2 - 2 |\vec{A}|^2 \]
Rearranging gives:
\[ |\vec{B}|^2 = 2 |\vec{A}|^2 \]

5. **Finding the Angle**:
Let \(\theta\) be the angle between \(\vec{A}\) and \(\vec{B}\). We know:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Substituting \(\vec{A} \cdot \vec{B} = -|\vec{A}|^2\) and \(|\vec{B}| = \sqrt{2} |\vec{A}|\):
\[ -|\vec{A}|^2 = |\vec{A}| \cdot \sqrt{2} |\vec{A}| \cos \theta \]
Simplifying gives:
\[ -1 = \sqrt{2} \cos \theta \]
Thus:
\[ \cos \theta = -\frac{1}{\sqrt{2}} \implies \theta = 135^\circ \text{ or } \theta = \frac{3\pi}{4} \text{ radians} \]

**Final Answer**:
The angle between \(\vec{A}\) and \(\vec{B}\) is \(135^\circ\) or \(\frac{3\pi}{4}\) radians.

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