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

• A

  $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar
• B

  $\mathbf{b}, \mathbf{d}$ are non-parallel
• Correct answer
  C

  $\mathbf{a}, \mathbf{b}, \mathbf{d}$ are non-coplanar
• D

  a, d are parallel and $\mathbf{b}, \mathbf{c}$ are parallel

Solution

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.

Hence, the correct answer is option: 'C'.