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

Options

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

Answer

To solve the problem, we need to analyze the given condition mathematically. The condition states that the magnitude of the sum of two vectors \( \vec{V}_1 \) and \( \vec{V}_2 \) is equal to the magnitude of their difference:

\[ |\vec{V}_1 + \vec{V}_2| = |\vec{V}_1 - \vec{V}_2| \]

### Step 1: Express the magnitudes mathematically

The magnitude of the sum of two vectors can be expressed using the formula:

\[ |\vec{V}_1 + \vec{V}_2| = \sqrt{|\vec{V}_1|^2 + |\vec{V}_2|^2 + 2 |\vec{V}_1| |\vec{V}_2| \cos \theta} \]

where \( \theta \) is the angle between \( \vec{V}_1 \) and \( \vec{V}_2 \).

Similarly, the magnitude of the difference of the two vectors is:

\[ |\vec{V}_1 - \vec{V}_2| = \sqrt{|\vec{V}_1|^2 + |\vec{V}_2|^2 - 2 |\vec{V}_1| |\vec{V}_2| \cos \theta} \]

### Step 2: Set the two magnitudes equal

According to the problem, we have:

\[ \sqrt{|\vec{V}_1|^2 + |\vec{V}_2|^2 + 2 |\vec{V}_1| |\vec{V}_2| \cos \theta} = \sqrt{|\vec{V}_1|^2 + |\vec{V}_2|^2 - 2 |\vec{V}_1| |\vec{V}_2| \cos \theta} \]

### Step 3: Square both sides

To eliminate the square roots, we square both sides:

\[ |\vec{V}_1|^2 + |\vec{V}_2|^2 + 2 |\vec{V}_1| |\vec{V}_2| \cos \theta = |\vec{V}_1|^2 + |\vec{V}_2|^2 - 2 |\vec{V}_1| |\vec{V}_2| \cos \theta \]

### Step 4: Simplify the equation

Now, we can simplify the equation:

\[ 2 |\vec{V}_1| |\vec{V}_2| \cos \theta + 2 |\vec{V}_1| |\vec{V}_2| \cos \theta = 0 \]

This simplifies to:

\[ 4 |\vec{V}_1| |\vec{V}_2| \cos \theta = 0 \]

### Step 5: Analyze the result

Since \( |\vec{V}_1| \) and \( |\vec{V}_2| \) are finite and non-zero (as per the problem statement), the only solution to this equation is:

\[ \cos \theta = 0 \]

### Step 6: Conclusion

The condition \( \cos \theta = 0 \) implies that the angle \( \theta \) between the two vectors \( \vec{V}_1 \) and \( \vec{V}_2 \) is \( 90^\circ \). Therefore, the vectors \( \vec{V}_1 \) and \( \vec{V}_2 \) are mutually perpendicular.

**Final Answer:** \( \vec{V}_1 \) and \( \vec{V}_2 \) are mutually perpendicular.

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