When two vectors `vecA` and `vecB` of magnitudes `a` and `b` respectively are added, the magnitude of resultant vector is always

Options

A. Equal to `(a+b)`
B. Less than `(a+b)`
C. Greater than `(a+b)`
D. Not greater than `(a+b)`

Answer

To solve the problem of finding the magnitude of the resultant vector when two vectors \(\vec{A}\) and \(\vec{B}\) of magnitudes \(a\) and \(b\) are added, we can follow these steps:

### Step 1: Understand Vector Addition
When two vectors are added, the magnitude of the resultant vector depends on the angle between them. The resultant vector \(\vec{R}\) can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta} \]
where \(\theta\) is the angle between the two vectors.

### Step 2: Identify the Magnitudes
Given:
- Magnitude of vector \(\vec{A} = a\)
- Magnitude of vector \(\vec{B} = b\)

### Step 3: Substitute the Magnitudes into the Formula
Substituting the magnitudes into the formula gives:
\[ |\vec{R}| = \sqrt{a^2 + b^2 + 2ab\cos\theta} \]

### Step 4: Analyze the Cosine Term
The value of \(\cos\theta\) can vary from -1 to 1:
- When \(\theta = 0^\circ\) (vectors are in the same direction), \(\cos\theta = 1\):
\[ |\vec{R}|_{\text{max}} = \sqrt{a^2 + b^2 + 2ab} = \sqrt{(a+b)^2} = a + b \]
- When \(\theta = 180^\circ\) (vectors are in opposite directions), \(\cos\theta = -1\):
\[ |\vec{R}|_{\text{min}} = \sqrt{a^2 + b^2 - 2ab} = \sqrt{(a-b)^2} = |a - b| \]

### Step 5: Conclusion
Thus, the magnitude of the resultant vector \(|\vec{R}|\) can vary between \(|a - b|\) and \(a + b\):
\[ |a - b| \leq |\vec{R}| \leq a + b \]
This means the magnitude of the resultant vector is always less than or equal to \(a + b\) and greater than or equal to \(|a - b|\).

### Final Answer
The magnitude of the resultant vector is always:
\[ |a - b| \leq |\vec{R}| \leq a + b \]

---