If `vec(a) and vec(b)` are unit vectors, then what is the value of `|vec(a) xx vec(b)|^(2) + (vec(a).vec(b))^(2)`?

Options

A. A) 0
B. B) 2
C. C) 1
D. D) `1//2`

Answer

To solve the problem, we need to evaluate the expression \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 \) given that \(\vec{a}\) and \(\vec{b}\) are unit vectors.

### Step-by-Step Solution:

1. **Understand the properties of unit vectors**:
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Calculate the magnitude of the cross product**:
The magnitude of the cross product of two vectors can be expressed as:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
where \(\theta\) is the angle between the two vectors. Since both vectors are unit vectors, this simplifies to:
\[ |\vec{a} \times \vec{b}| = 1 \cdot 1 \cdot \sin \theta = \sin \theta \]

3. **Square the magnitude of the cross product**:
Now, squaring the magnitude gives:
\[ |\vec{a} \times \vec{b}|^2 = (\sin \theta)^2 \]

4. **Calculate the dot product**:
The dot product of two vectors can be expressed as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Again, since both vectors are unit vectors, this simplifies to:
\[ \vec{a} \cdot \vec{b} = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]

5. **Square the dot product**:
Now, squaring the dot product gives:
\[ (\vec{a} \cdot \vec{b})^2 = (\cos \theta)^2 \]

6. **Combine the results**:
Now, substituting back into the original expression:
\[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = \sin^2 \theta + \cos^2 \theta \]

7. **Use the Pythagorean identity**:
From trigonometry, we know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

8. **Conclusion**:
Therefore, the value of the expression is:
\[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = 1 \]

### Final Answer:
The value of \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 \) is \( 1 \).

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