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|`

Explanation

To solve the problem, we start with the given condition:

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

### Step 1: Square both sides

We square both sides of the equation to eliminate the magnitudes:

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

### Step 2: Expand both sides

Using the property of magnitudes, we 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 gives us:

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

### Step 3: Simplify the equation

We can simplify the equation by canceling out the common terms:

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

This leads us to:

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

### Step 4: Combine like terms

Bringing all terms involving \(\vec{V_1} \cdot \vec{V_2}\) to one side gives:

\[ 2 \vec{V_1} \cdot \vec{V_2} + 2 \vec{V_1} \cdot \vec{V_2} = 0 \]

This simplifies to:

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

### Step 5: Conclude about the vectors

Since \(4 \vec{V_1} \cdot \vec{V_2} = 0\), we conclude that:

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

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

### Final Answer

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