If `A+B=C,|A|=2|B|and B.C=0,` then

Options

A. `|A+B|=|A+B|`
B. `|A+C|=B`
C. `A.Blt0`
D. A.C may be zero

Explanation

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

### Step 1: Write down the given equations
We have the following equations:
1. \( \mathbf{A} + \mathbf{B} = \mathbf{C} \)
2. \( |\mathbf{A}| = 2 |\mathbf{B}| \)
3. \( \mathbf{B} \cdot \mathbf{C} = 0 \)

### Step 2: Express \(\mathbf{C}\) in terms of \(\mathbf{A}\) and \(\mathbf{B}\)
From the first equation, we can express \(\mathbf{C}\) as:
\[ \mathbf{C} = \mathbf{A} + \mathbf{B} \]

### Step 3: Substitute \(\mathbf{C}\) into the dot product equation
Substituting \(\mathbf{C}\) into the dot product equation gives:
\[ \mathbf{B} \cdot (\mathbf{A} + \mathbf{B}) = 0 \]
This expands to:
\[ \mathbf{B} \cdot \mathbf{A} + \mathbf{B} \cdot \mathbf{B} = 0 \]

### Step 4: Simplify the dot product equation
We know that \(\mathbf{B} \cdot \mathbf{B} = |\mathbf{B}|^2\). Thus, we can rewrite the equation as:
\[ \mathbf{B} \cdot \mathbf{A} + |\mathbf{B}|^2 = 0 \]
This implies:
\[ \mathbf{B} \cdot \mathbf{A} = -|\mathbf{B}|^2 \]

### Step 5: Use the relationship between magnitudes
Given that \( |\mathbf{A}| = 2 |\mathbf{B}| \), we can substitute this into the dot product. The magnitude of \(\mathbf{A}\) can be expressed as:
\[ |\mathbf{A}|^2 = (2|\mathbf{B}|)^2 = 4|\mathbf{B}|^2 \]

### Step 6: Find the angle between \(\mathbf{A}\) and \(\mathbf{B}\)
Using the dot product formula:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos(\theta) \]
Substituting the magnitudes:
\[ \mathbf{B} \cdot \mathbf{A} = 4|\mathbf{B}|^2 \cos(\theta) \]
Setting this equal to \(-|\mathbf{B}|^2\) from the previous step:
\[ 4|\mathbf{B}|^2 \cos(\theta) = -|\mathbf{B}|^2 \]
Dividing both sides by \(|\mathbf{B}|^2\) (assuming \(|\mathbf{B}| \neq 0\)):
\[ 4 \cos(\theta) = -1 \]
Thus, we find:
\[ \cos(\theta) = -\frac{1}{4} \]

### Step 7: Determine the angle \(\theta\)
To find the angle \(\theta\), we can use the inverse cosine function:
\[ \theta = \cos^{-1}\left(-\frac{1}{4}\right) \]

### Conclusion
The angle between vectors \(\mathbf{A}\) and \(\mathbf{B}\) is \(\theta = \cos^{-1}\left(-\frac{1}{4}\right)\).