If `theta` is the abgle between two vectors A and B , then match the following two columns.
`{:(,"Column I ",,"Column II"),((A),A.B=|AxxB|,(P),theta=90^(@)),((B),A.B=B^(2),(Q),theta=0^(@)or 180^(@)),((C),|A+B|=|A-C|,(r),A=B),((D), |AxxB|=AB,(s),"None"):}`

Answer

To solve the problem of matching the conditions given in Column I with the corresponding statements in Column II, we will analyze each condition step by step.

### Step-by-Step Solution:

1. **Condition A: \( A \cdot B = |A \times B| \)**
- We know that the dot product \( A \cdot B \) is given by:
\[ A \cdot B = |A| |B| \cos \theta \]
- The magnitude of the cross product \( |A \times B| \) is given by:
\[ |A \times B| = |A| |B| \sin \theta \]
- Setting these two equal gives:
\[ |A| |B| \cos \theta = |A| |B| \sin \theta \]
- Dividing both sides by \( |A| |B| \) (assuming \( |A|, |B| \neq 0 \)):
\[ \cos \theta = \sin \theta \]
- This implies:
\[ \tan \theta = 1 \implies \theta = 45^\circ \]
- Since there is no option for \( 45^\circ \) in Column II, we match A to (s) "None".

2. **Condition B: \( A \cdot B = B^2 \)**
- The dot product can be expressed as:
\[ A \cdot B = |A| |B| \cos \theta \]
- The right side, \( B^2 \), is:
\[ B^2 = |B| \cdot |B| = |B|^2 \]
- Setting these equal gives:
\[ |A| |B| \cos \theta = |B|^2 \]
- Dividing both sides by \( |B| \) (assuming \( |B| \neq 0 \)):
\[ |A| \cos \theta = |B| \]
- This can only hold true if \( \cos \theta = 1 \) (which implies \( \theta = 0^\circ \)) and \( |A| = |B| \). However, since there is no option for this specific case in Column II, we match B to (s) "None".

3. **Condition C: \( |A + B| = |A - C| \)**
- We start with the equation:
\[ |A + B|^2 = |A - C|^2 \]
- Expanding both sides:
\[ |A|^2 + |B|^2 + 2A \cdot B = |A|^2 + |C|^2 - 2A \cdot C \]
- Canceling \( |A|^2 \) from both sides:
\[ |B|^2 + 2A \cdot B = |C|^2 - 2A \cdot C \]
- This condition can hold true when \( \theta = 90^\circ \) (as the terms involving \( A \cdot B \) and \( A \cdot C \) would cancel out). Thus, we match C to (P) \( \theta = 90^\circ \).

4. **Condition D: \( |A \times B| = AB \)**
- The magnitude of the cross product is:
\[ |A \times B| = |A| |B| \sin \theta \]
- Setting this equal to \( AB \):
\[ |A| |B| \sin \theta = |A| |B| \]
- Dividing both sides by \( |A| |B| \) (assuming \( |A|, |B| \neq 0 \)):
\[ \sin \theta = 1 \]
- This occurs when \( \theta = 90^\circ \). Thus, we match D to (P) \( \theta = 90^\circ \).

### Final Matching:
- A → (s) "None"
- B → (s) "None"
- C → (P) \( \theta = 90^\circ \)
- D → (P) \( \theta = 90^\circ \)

### Summary of Matches:
- A → s
- B → s
- C → P
- D → P