If `veca,vecb,vecc and vecd` are unit vectors such that `(vecaxxvecb)*(veccxxvecd)=1 and veca.vecc=1/2,` then

Options

A. a, b, c are non-coplanar
B. a, b, d are non-coplanar
C. b, d are non-parallel
D. a, d are parallel and b, c are parallel

Answer

To solve the problem, we need to analyze the given conditions involving the vectors \( \vec{a}, \vec{b}, \vec{c}, \) and \( \vec{d} \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions:**
- We know that \( \vec{a}, \vec{b}, \vec{c}, \) and \( \vec{d} \) are unit vectors.
- The condition given is \( (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \).
- Additionally, we have \( \vec{a} \cdot \vec{c} = \frac{1}{2} \).

2. **Using the Dot Product of Cross Products:**
- The expression \( (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) \) can be rewritten using the properties of the dot and cross products:
\[ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = |\vec{a} \times \vec{b}| |\vec{c} \times \vec{d}| \cos(\gamma) \]
- Here, \( \gamma \) is the angle between the vectors \( \vec{a} \times \vec{b} \) and \( \vec{c} \times \vec{d} \).

3. **Finding the Magnitudes of the Cross Products:**
- Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\alpha) = 1 \cdot 1 \cdot \sin(\alpha) = \sin(\alpha) \]
- Similarly, for \( \vec{c} \) and \( \vec{d} \):
\[ |\vec{c} \times \vec{d}| = |\vec{c}| |\vec{d}| \sin(\beta) = 1 \cdot 1 \cdot \sin(\beta) = \sin(\beta) \]

4. **Setting Up the Equation:**
- Substituting these into our earlier equation gives:
\[ \sin(\alpha) \sin(\beta) \cos(\gamma) = 1 \]

5. **Analyzing the Conditions:**
- Since \( \sin(\alpha) \) and \( \sin(\beta) \) can each be at most 1, the maximum product \( \sin(\alpha) \sin(\beta) \) can be is 1. This means:
\[ \sin(\alpha) = 1 \quad \text{and} \quad \sin(\beta) = 1 \quad \text{and} \quad \cos(\gamma) = 1 \]
- Thus, we conclude that \( \alpha = 90^\circ \) and \( \beta = 90^\circ \), and \( \gamma = 0^\circ \).

6. **Conclusion About the Vectors:**
- Since \( \gamma = 0^\circ \), this implies that \( \vec{a} \times \vec{b} \) is parallel to \( \vec{c} \times \vec{d} \).
- The angles \( \alpha \) and \( \beta \) being \( 90^\circ \) indicates that \( \vec{a} \) is perpendicular to \( \vec{b} \) and \( \vec{c} \) is perpendicular to \( \vec{d} \).

7. **Finding the Angle Between \( \vec{a} \) and \( \vec{c} \):**
- We know \( \vec{a} \cdot \vec{c} = \frac{1}{2} \).
- Since both are unit vectors, we have:
\[ |\vec{a}| |\vec{c}| \cos(\delta) = \frac{1}{2} \implies 1 \cdot 1 \cdot \cos(\delta) = \frac{1}{2} \]
- Thus, \( \cos(\delta) = \frac{1}{2} \), which gives \( \delta = 60^\circ \).

8. **Final Configuration:**
- We can now visualize the vectors in a plane:
- Let \( \vec{a} \) be along the x-axis.
- \( \vec{b} \) will be along the y-axis (90 degrees to \( \vec{a} \)).
- \( \vec{c} \) will be at a 60-degree angle from \( \vec{a} \).
- \( \vec{d} \) will be perpendicular to \( \vec{c} \).

### Final Conclusion:
- The vectors \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) are all coplanar, and we have established their relationships and angles.