For any vectors `a,b,|axxb|^(2)+(a*b)^(2)` is equal to

Options

A. `|a|^(2)|b|^(2)`
B. `|a+b|`
C. `|a|^(2)-|b|^(2)`
D. 0

Answer

To solve the problem \( | \mathbf{a} \times \mathbf{b} |^2 + ( \mathbf{a} \cdot \mathbf{b} )^2 \), we will follow these steps:

### Step 1: Understand the Cross Product
The magnitude of the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be expressed as:
\[ | \mathbf{a} \times \mathbf{b} | = |\mathbf{a}| |\mathbf{b}| \sin \theta \]
where \(\theta\) is the angle between the vectors \(\mathbf{a}\) and \(\mathbf{b}\).

### Step 2: Square the Cross Product
Now, squaring the magnitude of the cross product gives:
\[ | \mathbf{a} \times \mathbf{b} |^2 = (|\mathbf{a}| |\mathbf{b}| \sin \theta)^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \sin^2 \theta \]

### Step 3: Understand the Dot Product
The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be expressed as:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]

### Step 4: Square the Dot Product
Now, squaring the dot product gives:
\[ ( \mathbf{a} \cdot \mathbf{b} )^2 = (|\mathbf{a}| |\mathbf{b}| \cos \theta)^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \cos^2 \theta \]

### Step 5: Add the Two Results
Now we add the squared cross product and squared dot product:
\[ | \mathbf{a} \times \mathbf{b} |^2 + ( \mathbf{a} \cdot \mathbf{b} )^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \sin^2 \theta + |\mathbf{a}|^2 |\mathbf{b}|^2 \cos^2 \theta \]

### Step 6: Factor Out Common Terms
We can factor out \( |\mathbf{a}|^2 |\mathbf{b}|^2 \):
\[ = |\mathbf{a}|^2 |\mathbf{b}|^2 (\sin^2 \theta + \cos^2 \theta) \]

### Step 7: Use the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = |\mathbf{a}|^2 |\mathbf{b}|^2 \cdot 1 = |\mathbf{a}|^2 |\mathbf{b}|^2 \]

### Conclusion
Thus, we conclude that:
\[ | \mathbf{a} \times \mathbf{b} |^2 + ( \mathbf{a} \cdot \mathbf{b} )^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \]

### Final Answer
The final answer is:
\[ | \mathbf{a} \times \mathbf{b} |^2 + ( \mathbf{a} \cdot \mathbf{b} )^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \]