For two vectors ` vec A` and `vec B`
` | vec A + vec B| = | vec A- vec B|` is always true when.

Options

A. `|A|=|B|!= 0`
B. `A_|_B`
C. `|A|=|B|!= 0` and A and B are parallel or anti-parallel
D. when either |A| or |B| is zero

Explanation

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

\[ |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \]

### Step 1: Square both sides of the equation

We start by squaring both sides of the equation to eliminate the modulus:

\[ |\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2 \]

### Step 2: Expand both sides using the dot product

Using the property of dot products, we can expand both sides:

\[ (\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B}) \]

This gives us:

\[ \vec{A} \cdot \vec{A} + 2\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B} = \vec{A} \cdot \vec{A} - 2\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B} \]

### Step 3: Simplify the equation

Now, we can simplify the equation by canceling out the common terms:

\[ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{B} + 2\vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{B} - 2\vec{A} \cdot \vec{B} \]

This leads to:

\[ 2\vec{A} \cdot \vec{B} + 2\vec{A} \cdot \vec{B} = 0 \]

### Step 4: Factor out the common term

Factoring out the common term gives us:

\[ 4\vec{A} \cdot \vec{B} = 0 \]

### Step 5: Analyze the result

The equation \(4\vec{A} \cdot \vec{B} = 0\) implies that:

\[ \vec{A} \cdot \vec{B} = 0 \]

This means that either:

1. \(|\vec{A}| = 0\) (i.e., \(\vec{A}\) is the zero vector)
2. \(|\vec{B}| = 0\) (i.e., \(\vec{B}\) is the zero vector)
3. \(\vec{A}\) is perpendicular to \(\vec{B}\) (i.e., the angle \(\theta\) between them is \(90^\circ\))

### Conclusion

Thus, the condition \( |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \) is always true when:

- Either \(|\vec{A}| = 0\) or \(|\vec{B}| = 0\)
- \(\vec{A}\) is perpendicular to \(\vec{B}\)