If `|vecV_(1)+vecV_(2)|=|vecV_(1)-vecV_(2)|` and `V_(2)` is finite, then

Options

A. `V_(1)` is parallel to `V_(2)`
B. `vecV_(1)=vecV_(2)`
C. `V_(1)` and `V_(2)` are mutually perpendicular
D. `|vecV_(1)|=|vecV_(2)|`

Answer

To solve the problem, we need to analyze the condition given:

\[ |\vec{V_1} + \vec{V_2}| = |\vec{V_1} - \vec{V_2}| \]

### Step 1: Square both sides
To eliminate the absolute values, we square both sides of the equation:

\[ |\vec{V_1} + \vec{V_2}|^2 = |\vec{V_1} - \vec{V_2}|^2 \]

### Step 2: Expand both sides
Using the property of magnitudes, we can expand both sides:

\[ (\vec{V_1} + \vec{V_2}) \cdot (\vec{V_1} + \vec{V_2}) = (\vec{V_1} - \vec{V_2}) \cdot (\vec{V_1} - \vec{V_2}) \]

This leads to:

\[ \vec{V_1} \cdot \vec{V_1} + 2 \vec{V_1} \cdot \vec{V_2} + \vec{V_2} \cdot \vec{V_2} = \vec{V_1} \cdot \vec{V_1} - 2 \vec{V_1} \cdot \vec{V_2} + \vec{V_2} \cdot \vec{V_2} \]

### Step 3: Simplify the equation
Now, we can simplify the equation:

\[ |\vec{V_1}|^2 + 2 \vec{V_1} \cdot \vec{V_2} + |\vec{V_2}|^2 = |\vec{V_1}|^2 - 2 \vec{V_1} \cdot \vec{V_2} + |\vec{V_2}|^2 \]

### Step 4: Cancel common terms
We can cancel \( |\vec{V_1}|^2 \) and \( |\vec{V_2}|^2 \) from both sides:

\[ 2 \vec{V_1} \cdot \vec{V_2} = -2 \vec{V_1} \cdot \vec{V_2} \]

### Step 5: Combine like terms
This simplifies to:

\[ 4 \vec{V_1} \cdot \vec{V_2} = 0 \]

### Step 6: Conclusion
From this equation, we conclude:

\[ \vec{V_1} \cdot \vec{V_2} = 0 \]

This implies that the vectors \( \vec{V_1} \) and \( \vec{V_2} \) are perpendicular to each other.

Thus, the final answer is that \( \vec{V_1} \) and \( \vec{V_2} \) are mutually perpendicular.