If `overset(rarr)A` and `overset(rarr)B` are two vectors of non-zero magnitude, which of the following relations cannot be possible ?

Options

A. `overset(rarr)A +overset(rarr)B=overset(rarr)A-overset(rarr)B`
B. `overset(rarr)A.overset(rarr)B=|overset(rarr)A xxoverset(rarr)B|`
C. `|overset(rarr)A+overset(rarr)B|=|overset(rarr)A-overset(rarr)B|`
D. `|overset(rarr)A+overset(rarr)B| gt |overset(rarr)A|+|overset(rarr)B|`

Explanation

To solve the question regarding which relation cannot be possible between two non-zero vectors \( \overset{\rarr}{A} \) and \( \overset{\rarr}{B} \), we will analyze each option provided.

### Step-by-Step Solution:

1. **Option A: \( \overset{\rarr}{A} + \overset{\rarr}{B} = \overset{\rarr}{A} - \overset{\rarr}{B} \)**

Rearranging gives:
\[ \overset{\rarr}{A} + \overset{\rarr}{B} - \overset{\rarr}{A} + \overset{\rarr}{B} = 0 \]
This simplifies to:
\[ 2\overset{\rarr}{B} = 0 \]
Since \( \overset{\rarr}{B} \) is a non-zero vector, this equation is not possible.

**Conclusion**: This relation cannot be possible.

2. **Option B: \( \overset{\rarr}{A} \cdot \overset{\rarr}{B} = |\overset{\rarr}{A} \times \overset{\rarr}{B}| \)**

The dot product is given by:
\[ \overset{\rarr}{A} \cdot \overset{\rarr}{B} = |\overset{\rarr}{A}| |\overset{\rarr}{B}| \cos \theta \]
The magnitude of the cross product is:
\[ |\overset{\rarr}{A} \times \overset{\rarr}{B}| = |\overset{\rarr}{A}| |\overset{\rarr}{B}| \sin \theta \]
Setting them equal gives:
\[ |\overset{\rarr}{A}| |\overset{\rarr}{B}| \cos \theta = |\overset{\rarr}{A}| |\overset{\rarr}{B}| \sin \theta \]
Dividing both sides by \( |\overset{\rarr}{A}| |\overset{\rarr}{B}| \) (non-zero), we get:
\[ \cos \theta = \sin \theta \]
This occurs when \( \theta = 45^\circ \). Thus, this relation is possible.

3. **Option C: \( |\overset{\rarr}{A} + \overset{\rarr}{B}| = |\overset{\rarr}{A}| \)**

Squaring both sides gives:
\[ |\overset{\rarr}{A} + \overset{\rarr}{B}|^2 = |\overset{\rarr}{A}|^2 \]
Expanding the left side:
\[ |\overset{\rarr}{A}|^2 + |\overset{\rarr}{B}|^2 + 2 \overset{\rarr}{A} \cdot \overset{\rarr}{B} = |\overset{\rarr}{A}|^2 \]
This simplifies to:
\[ |\overset{\rarr}{B}|^2 + 2 \overset{\rarr}{A} \cdot \overset{\rarr}{B} = 0 \]
Since \( |\overset{\rarr}{B}|^2 \) is non-negative, this implies \( \overset{\rarr}{A} \cdot \overset{\rarr}{B} \) must be negative. This is possible if \( \overset{\rarr}{B} \) is in the opposite direction to \( \overset{\rarr}{A} \).

4. **Option D: \( |\overset{\rarr}{A} + \overset{\rarr}{B}| > |\overset{\rarr}{A}| + |\overset{\rarr}{B}| \)**

By the triangle inequality, we know:
\[ |\overset{\rarr}{A} + \overset{\rarr}{B}| \leq |\overset{\rarr}{A}| + |\overset{\rarr}{B}| \]
Therefore, this relation cannot be true, as it contradicts the triangle inequality.

### Final Conclusion:
The relations that cannot be possible are:
- **Option A**: \( \overset{\rarr}{A} + \overset{\rarr}{B} = \overset{\rarr}{A} - \overset{\rarr}{B} \)
- **Option D**: \( |\overset{\rarr}{A} + \overset{\rarr}{B}| > |\overset{\rarr}{A}| + |\overset{\rarr}{B}| \)