If `vec C=vec A+vec B` then

Options

A. `vec C` is always greater than `|vecA|`
B. `C` is always equal to `A+B`
C. `C` is always equal to `A+B`
D. It is possible to have `|vecC|lt|vecA|` and `|vecC|lt|vecB|`

Answer

To solve the problem given the equation \( \vec{C} = \vec{A} + \vec{B} \), we need to analyze the magnitude of vector \( \vec{C} \) in terms of the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \), and the angle \( \theta \) between them.

### Step 1: Understand the Magnitude of the Resultant Vector
The magnitude of the resultant vector \( \vec{C} \) can be expressed using the law of cosines:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
where \( |\vec{A}| \) and \( |\vec{B}| \) are the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \), respectively, and \( \theta \) is the angle between them.

### Step 2: Determine Maximum and Minimum Values of \( |\vec{C}| \)
To find the maximum and minimum values of \( |\vec{C}| \), we consider the extremes of \( \cos \theta \):
- The maximum value occurs when \( \cos \theta = 1 \):
\[ |\vec{C}|_{\text{max}} = |\vec{A}| + |\vec{B}| \]
- The minimum value occurs when \( \cos \theta = -1 \):
\[ |\vec{C}|_{\text{min}} = ||\vec{A}| - |\vec{B}|| \]

### Step 3: Establish the Range of \( |\vec{C}| \)
From the above calculations, we can conclude that:
\[ ||\vec{A}| - |\vec{B}|| \leq |\vec{C}| \leq |\vec{A}| + |\vec{B}| \]
This means that the magnitude of \( \vec{C} \) can vary between the absolute difference of the magnitudes of \( \vec{A} \) and \( \vec{B} \) and the sum of their magnitudes.

### Step 4: Analyze Specific Cases
Let’s analyze specific values for \( |\vec{A}| \) and \( |\vec{B}| \):
- Assume \( |\vec{A}| = 6 \) and \( |\vec{B}| = 3 \):
- Minimum: \( ||6 - 3|| = 3 \)
- Maximum: \( 6 + 3 = 9 \)
Thus, \( 3 \leq |\vec{C}| \leq 9 \).

### Step 5: Conclusion
Based on the analysis, we can conclude that:
- The magnitude of vector \( \vec{C} \) is always greater than or equal to the absolute difference of the magnitudes of \( \vec{A} \) and \( \vec{B} \) and less than or equal to the sum of their magnitudes. Therefore, \( |\vec{C}| \) can indeed be less than \( |\vec{A}| \) or \( |\vec{B}| \) depending on the angle \( \theta \).