If \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{d}\) are the unit vectors such that \((\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1\) and \(\mathbf{a} \cdot \mathbf{c}=\frac{1}{2}\), then

Let angle between \(\mathbf{a}\) and \(\mathbf{b}\) be \(\theta_1, \mathbf{c}\) and \(\mathbf{d}\) be \(\theta_2\) and \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{c} \times \mathbf{d}\) be \(\theta\).
Since, \((\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1\)
\(\Rightarrow \quad \sin \theta_1 \cdot \sin \theta_2 \cdot \cos \theta=1\)
\(\Rightarrow \theta_1=90^{\circ}, \theta_2=90^{\circ}, \theta=0^{\circ}\)
\(\Rightarrow \mathbf{a} \perp \mathbf{b}, \mathbf{c} \perp \mathbf{d},(\mathbf{a} \times \mathbf{b}) \|(\mathbf{c} \times \mathbf{d})\)
So, \(\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})\)
and \(\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})\)
\(\Rightarrow \quad(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{c}\)
and \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{d}\)
\(\Rightarrow\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=0\) and \(\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{d}\end{array}\right]=0\)

\(\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{a}, \mathbf{b}, \mathbf{d}\) are coplanar vectors, so options (a) and (b) are incorrect.
Let \(\quad \mathbf{b} \| \mathbf{d} \Rightarrow \mathbf{b}=\pm \mathbf{d}\)
As \(\quad(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1\)
\((\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{b})=\pm 1\)
\(\Rightarrow \quad[\mathbf{a} \times \mathbf{b} \mathbf{c} \mathbf{b}]=\pm 1\)
\(\Rightarrow \quad[\mathbf{c} \mathbf{b} \mathbf{a} \times \mathbf{b}]=\pm 1\)
\(\Rightarrow \quad \mathbf{c} \cdot[\mathbf{b} \times(\mathbf{a} \times \mathbf{b})]=\pm 1\)
\(\Rightarrow \quad \mathbf{c} \cdot[\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b}]=\pm 1\)
\(\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=\pm 1 \quad[\because \mathbf{a} \cdot \mathbf{b}=0]\)
which is a contradiction, so option (c) is correct.


Let option (d) be correct.
\[
\begin{array}{ll}
\Rightarrow & \mathbf{d}=\pm \mathbf{a} \text { and } \mathbf{c}=\pm \mathbf{b} \\
\text { As } & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1 \\
\Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{a})=\pm 1
\end{array}
\]
which is a contradiction, so option (d) is incorrect.
Alternatively option (c) and (d) may be observed from the given figure.