Two vector A and B have rqual magnitude x .Angle between tjem is `60^(@)` Then match the following two columns.
`{:(,"Column I ",,"Column II"),((A),"|A+B|",(P),(sqrt(3))/(2)x^(2)),((B),|A-B|,(Q),x),((C), A.B,(r),sqrt(3)x),((D), |AxxB|,(s),"None"):}`

Explanation

To solve the problem, we need to calculate the following for two vectors \( \mathbf{A} \) and \( \mathbf{B} \) with equal magnitude \( x \) and an angle of \( 60^\circ \) between them:

1. Magnitude of \( \mathbf{A} + \mathbf{B} \)
2. Magnitude of \( \mathbf{A} - \mathbf{B} \)
3. Dot product \( \mathbf{A} \cdot \mathbf{B} \)
4. Magnitude of the cross product \( \mathbf{A} \times \mathbf{B} \)

### Step 1: Calculate \( |\mathbf{A} + \mathbf{B}| \)

Using the formula for the magnitude of the sum of two vectors:

\[ |\mathbf{A} + \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 |\mathbf{A}| |\mathbf{B}| \cos \theta} \]

Substituting \( |\mathbf{A}| = |\mathbf{B}| = x \) and \( \theta = 60^\circ \):

\[ |\mathbf{A} + \mathbf{B}| = \sqrt{x^2 + x^2 + 2 \cdot x \cdot x \cdot \cos(60^\circ)} \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ |\mathbf{A} + \mathbf{B}| = \sqrt{x^2 + x^2 + 2 \cdot x^2 \cdot \frac{1}{2}} = \sqrt{x^2 + x^2 + x^2} = \sqrt{3x^2} = \sqrt{3} x \]

### Step 2: Calculate \( |\mathbf{A} - \mathbf{B}| \)

Using the formula for the magnitude of the difference of two vectors:

\[ |\mathbf{A} - \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 - 2 |\mathbf{A}| |\mathbf{B}| \cos \theta} \]

Substituting the same values:

\[ |\mathbf{A} - \mathbf{B}| = \sqrt{x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(60^\circ)} \]

\[ |\mathbf{A} - \mathbf{B}| = \sqrt{x^2 + x^2 - x^2} = \sqrt{x^2} = x \]

### Step 3: Calculate \( \mathbf{A} \cdot \mathbf{B} \)

Using the dot product formula:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]

Substituting the values:

\[ \mathbf{A} \cdot \mathbf{B} = x \cdot x \cdot \cos(60^\circ) = x^2 \cdot \frac{1}{2} = \frac{x^2}{2} \]

### Step 4: Calculate \( |\mathbf{A} \times \mathbf{B}| \)

Using the magnitude of the cross product formula:

\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \]

Substituting the values:

\[ |\mathbf{A} \times \mathbf{B}| = x \cdot x \cdot \sin(60^\circ) = x^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} x^2 \]

### Summary of Results

1. \( |\mathbf{A} + \mathbf{B}| = \sqrt{3} x \) (matches with \( r \))
2. \( |\mathbf{A} - \mathbf{B}| = x \) (matches with \( Q \))
3. \( \mathbf{A} \cdot \mathbf{B} = \frac{x^2}{2} \) (matches with \( S \), which is "None" since it is not listed)
4. \( |\mathbf{A} \times \mathbf{B}| = \frac{\sqrt{3}}{2} x^2 \) (matches with \( P \))

### Final Matching

- \( A \) matches with \( r \)
- \( B \) matches with \( Q \)
- \( C \) matches with \( S \) (None)
- \( D \) matches with \( P \)