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

Explanation

To solve the problem step by step, we start with the given information and apply vector concepts.

### Step 1: Understand the Given Information
We have three vectors:
- \(\vec{A}\)
- \(\vec{B}\)
- \(\vec{C}\)

The relationships given are:
1. \(\vec{A} + \vec{B} = \vec{C}\)
2. \(\vec{C}\) is perpendicular to \(\vec{A}\) (\(\vec{A} \cdot \vec{C} = 0\))
3. \(|\vec{A}| = |\vec{C}|\)

### Step 2: Use the Perpendicular Condition
Since \(\vec{C}\) is perpendicular to \(\vec{A}\), we can express this condition mathematically:
\[ \vec{A} \cdot \vec{C} = 0 \]
Substituting \(\vec{C} = \vec{A} + \vec{B}\):
\[ \vec{A} \cdot (\vec{A} + \vec{B}) = 0 \]
This expands to:
\[ \vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} = 0 \]
Since \(\vec{A} \cdot \vec{A} = |\vec{A}|^2\), we have:
\[ |\vec{A}|^2 + \vec{A} \cdot \vec{B} = 0 \]
Thus,
\[ \vec{A} \cdot \vec{B} = -|\vec{A}|^2 \]

### Step 3: Use the Magnitude Condition
Given that \(|\vec{A}| = |\vec{C}|\), we substitute \(|\vec{C}|\) into the equation:
\[ |\vec{C}| = |\vec{A} + \vec{B}| \]
Using the property of magnitudes:
\[ |\vec{C}|^2 = |\vec{A} + \vec{B}|^2 \]
This expands to:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A} \cdot \vec{B} \]
Since \(|\vec{C}| = |\vec{A}|\), we can write:
\[ |\vec{A}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A} \cdot \vec{B} \]
Cancelling \(|\vec{A}|^2\) from both sides gives:
\[ 0 = |\vec{B}|^2 + 2\vec{A} \cdot \vec{B} \]

### Step 4: Substitute \(\vec{A} \cdot \vec{B}\)
We already found that \(\vec{A} \cdot \vec{B} = -|\vec{A}|^2\). Substituting this in:
\[ 0 = |\vec{B}|^2 - 2|\vec{A}|^2 \]
This implies:
\[ |\vec{B}|^2 = 2|\vec{A}|^2 \]
Thus, we can express \(|\vec{B}|\) in terms of \(|\vec{A}|\):
\[ |\vec{B}| = |\vec{A}| \sqrt{2} \]

### Step 5: Find the Angle Between \(\vec{A}\) and \(\vec{B}\)
Using the dot product formula:
\[ \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta \]
Substituting the values we have:
\[ -|\vec{A}|^2 = |\vec{A}|(|\vec{A}|\sqrt{2})\cos\theta \]
This simplifies to:
\[ -|\vec{A}|^2 = |\vec{A}|^2\sqrt{2}\cos\theta \]
Dividing both sides by \(|\vec{A}|^2\) (assuming \(|\vec{A}| \neq 0\)):
\[ -1 = \sqrt{2}\cos\theta \]
Thus,
\[ \cos\theta = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]
This corresponds to an angle of:
\[ \theta = 135^\circ \quad \text{(or } \frac{3\pi}{4} \text{ radians)} \]

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

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