सिद्ध कीजिए कि `|vecA xx vecB|^(2) + |vecA * vecB|^(2) = (AB)^(2).`

Options

A. zero
B. `A^(2)B^(2)`
C. AB
D. `sqrt(AB)`

Explanation

To prove the equation \( |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = (AB)^2 \), we will follow these steps:

### Step 1: Write the expressions for \(|\vec{A} \times \vec{B}|\) and \(|\vec{A} \cdot \vec{B}|\)

We know that:
- The magnitude of the cross product is given by:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
- The magnitude of the dot product is given by:
\[ |\vec{A} \cdot \vec{B}| = |\vec{A}| |\vec{B}| \cos \theta \]

### Step 2: Square both expressions

Now, we will square both expressions:
1. For the cross product:
\[ |\vec{A} \times \vec{B}|^2 = (|\vec{A}| |\vec{B}| \sin \theta)^2 = |\vec{A}|^2 |\vec{B}|^2 \sin^2 \theta \]
2. For the dot product:
\[ |\vec{A} \cdot \vec{B}|^2 = (|\vec{A}| |\vec{B}| \cos \theta)^2 = |\vec{A}|^2 |\vec{B}|^2 \cos^2 \theta \]

### Step 3: Add the squared expressions

Now, we will add these two squared expressions:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = |\vec{A}|^2 |\vec{B}|^2 \sin^2 \theta + |\vec{A}|^2 |\vec{B}|^2 \cos^2 \theta \]

### Step 4: Factor out the common term

We can factor out \( |\vec{A}|^2 |\vec{B}|^2 \) from the left-hand side:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = |\vec{A}|^2 |\vec{B}|^2 (\sin^2 \theta + \cos^2 \theta) \]

### Step 5: Use the Pythagorean identity

Using the Pythagorean identity, we know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Thus, we can simplify:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = |\vec{A}|^2 |\vec{B}|^2 \cdot 1 = |\vec{A}|^2 |\vec{B}|^2 \]

### Step 6: Write the final result

Finally, we can express this as:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = (|\vec{A}| |\vec{B}|)^2 \]
This proves that:
\[ |\vec{A} \times \vec{B}|^2 + |\vec{A} \cdot \vec{B}|^2 = (AB)^2 \]