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

Options

A. `|vec( A) + vec( C ) | =| vec( A) + vec( B) |`
B. `|vec( A) + vec( B) | =B`
C. `vec( A) . vec( B) lt 0`
D. `vec(A) .vec(C ) ` may be zero

Answer

To solve the problem, let's break it down step by step.

### Step 1: Understand the Given Information
We have three vectors:
- \( \vec{A} + \vec{B} = \vec{C} \)
- \( |\vec{A}| = 2 |\vec{B}| \)
- \( \vec{B} \cdot \vec{C} = 0 \)

### Step 2: Analyze the Dot Product Condition
The condition \( \vec{B} \cdot \vec{C} = 0 \) implies that vectors \( \vec{B} \) and \( \vec{C} \) are perpendicular to each other.

### Step 3: Express \( \vec{C} \) in Terms of \( \vec{A} \) and \( \vec{B} \)
From the equation \( \vec{A} + \vec{B} = \vec{C} \), we can substitute \( \vec{C} \) in the dot product condition:
\[ \vec{B} \cdot (\vec{A} + \vec{B}) = 0 \]
This expands to:
\[ \vec{B} \cdot \vec{A} + \vec{B} \cdot \vec{B} = 0 \]
Since \( \vec{B} \cdot \vec{B} = |\vec{B}|^2 \), we have:
\[ \vec{B} \cdot \vec{A} + |\vec{B}|^2 = 0 \]
Thus, we can express \( \vec{B} \cdot \vec{A} \) as:
\[ \vec{B} \cdot \vec{A} = -|\vec{B}|^2 \]

### Step 4: Use the Magnitude Relationship
Given \( |\vec{A}| = 2 |\vec{B}| \), let \( |\vec{B}| = b \). Then:
\[ |\vec{A}| = 2b \]
Now, we can express the dot product \( \vec{B} \cdot \vec{A} \) in terms of magnitudes and the angle \( \theta \) between them:
\[ \vec{B} \cdot \vec{A} = |\vec{B}| |\vec{A}| \cos(\theta) = b \cdot 2b \cos(\theta) = 2b^2 \cos(\theta) \]
Setting this equal to our previous expression:
\[ 2b^2 \cos(\theta) = -b^2 \]
Dividing both sides by \( b^2 \) (assuming \( b \neq 0 \)):
\[ 2 \cos(\theta) = -1 \implies \cos(\theta) = -\frac{1}{2} \]
This means:
\[ \theta = \frac{2\pi}{3} \text{ or } 120^\circ \]

### Step 5: Conclusion
We have determined the angle between \( \vec{A} \) and \( \vec{B} \) to be \( 120^\circ \).