If $\vec{\mathbf{a}}, \vec{\mathbf{b}}, \vec{\mathbf{c}}$ and $\vec{\mathbf{d}}$ are the unit vectors such…

1. $\vec{\mathbf{a}}, \vec{\mathbf{b}}, \vec{\mathbf{c}}$ are non-coplanar
2. $\vec{\mathbf{a}}, \vec{\mathbf{b}}, \vec{\mathbf{d}}$ are non-coplanar
3. $\vec{\mathbf{b}}, \vec{\mathbf{d}}$ are non-parallel
4. $\vec{\mathbf{a}}, \vec{\mathbf{d}}$ are parallel and $\vec{\mathbf{b}}, \vec{\mathbf{c}}$ are parallel

Solution

Let angle between $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$ be $\theta_1, \vec{\mathbf{c}}$ and $\vec{\mathbf{d}}$ be $\theta_2$ and $\vec{\mathbf{a}} \times \vec{\mathbf{b}}$ and $\vec{\mathbf{c}} \times \vec{\mathbf{d}}$ be $\theta$. Since, $(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot(\vec{\mathbf{c}} \times \vec{\mathbf{d}})=1$ $\Rightarrow \quad \sin \theta_1 \cdot \sin \theta_2 \cdot \cos \theta=1$ $\Rightarrow \theta_1=90^{\circ}, \theta_2=90^{\circ}, \theta=0^{\circ}$ $\Rightarrow \vec{\mathbf{a}} \perp \vec{\mathbf{b}}, \vec{\mathbf{c}} \perp \vec{\mathbf{d}},(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \|(\vec{\mathbf{c}} \times \vec{\mathbf{d}})$ So, $\vec{\mathbf{a}} \times \vec{\mathbf{b}}=k(\vec{\mathbf{c}} \times \vec{\mathbf{d}})$ and $\vec{\mathbf{a}} \times \vec{\mathbf{b}}=k(\vec{\mathbf{c}} \times \vec{\mathbf{d}})$ $\Rightarrow \quad(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot \vec{\mathbf{c}}=k(\vec{\mathbf{c}} \times \vec{\mathbf{d}}) \cdot \vec{\mathbf{c}}$ and $(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot \vec{\mathbf{d}}=k(\vec{\mathbf{c}} \times \vec{\mathbf{d}}) \cdot \vec{\mathbf{d}}$ $\Rightarrow\left[\begin{array}{lll}\vec{\mathbf{a}} & \vec{\mathbf{b}} & \vec{\mathbf{c}}\end{array}\right]=0$ and $\left[\begin{array}{lll}\vec{\mathbf{a}} & \vec{\mathbf{b}} & \vec{\mathbf{d}}\end{array}\right]=0$ $\Rightarrow \vec{\mathbf{a}}, \vec{\mathbf{b}}, \vec{\mathbf{c}}$ and $\vec{\mathbf{a}}, \vec{\mathbf{b}}, \vec{\mathbf{d}}$ are coplanar vectors, so options (a) and (b) are incorrect. Let $\quad \vec{\mathbf{b}} \| \vec{\mathbf{d}} \Rightarrow \vec{\mathbf{b}}=\pm \vec{\mathbf{d}}$ As $\quad(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot(\vec{\mathbf{c}} \times \vec{\mathbf{d}})=1$ $(\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot(\vec{\mathbf{c}} \times \vec{\mathbf{b}})=\pm 1$ $\Rightarrow \quad[\vec{\mathbf{a}} \times \vec{\mathbf{b}} \vec{\mathbf{c}} \vec{\mathbf{b}}]=\pm 1$ $\Rightarrow \quad[\vec{\mathbf{c}} \vec{\mathbf{b}} \vec{\mathbf{a}} \times \vec{\mathbf{b}}]=\pm 1$ $\Rightarrow \quad \vec{\mathbf{c}} \cdot[\vec{\mathbf{b}} \times(\vec{\mathbf{a}} \times \vec{\mathbf{b}})]=\pm 1$ $\Rightarrow \quad \vec{\mathbf{c}} \cdot[\vec{\mathbf{a}}-(\vec{\mathbf{b}} \cdot \vec{\mathbf{a}}) \vec{\mathbf{b}}]=\pm 1$ $\Rightarrow \quad \vec{\mathbf{c}} \cdot \vec{\mathbf{a}}=\pm 1 \quad[\because \vec{\mathbf{a}} \cdot \vec{\mathbf{b}}=0]$ which is a contradiction, so option (c) is correct. Let option (d) be correct. $\begin{array}{ll} \Rightarrow & \vec{\mathbf{d}}=\pm \vec{\mathbf{a}} \text { and } \vec{\mathbf{c}}=\pm \vec{\mathbf{b}} \\ \text { As } & (\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot(\vec{\mathbf{c}} \times \vec{\mathbf{d}})=1 \\ \Rightarrow & (\vec{\mathbf{a}} \times \vec{\mathbf{b}}) \cdot(\vec{\mathbf{b}} \times \vec{\mathbf{a}})=\pm 1 \end{array}$ which is a contradiction, so option (d) is incorrect. Alternatively option (c) and (d) may be observed from the given figure.

Asked in: JEE Advanced 2009 (Paper 1)

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