Mathematics - Vector Question with Solution | TestHub

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, $(\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{d})=1$ $\Rightarrow \sin {\theta }_1\cdot \sin {\theta }_2\cdot \cos \theta =1$ $\Rightarrow {\theta }_1=90^{\circ },{\theta }_2=90^{\circ },\theta =0^{\circ }$ $\Rightarrow \mathbf{a}\perp \mathbf{b},\mathbf{c}\perp \mathbf{d},(\mathbf{a}\times \mathbf{b})\Vert (\mathbf{c}\times \mathbf{d})$ So, $\mathbf{a}\times \mathbf{b}=k(\mathbf{c}\times \mathbf{d})$ and $\mathbf{a}\times \mathbf{b}=k(\mathbf{c}\times \mathbf{d})$ $\Rightarrow (\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}=k(\mathbf{c}\times \mathbf{d})\cdot \mathbf{c}$ 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 $\mathbf{b}\Vert \mathbf{d}\Rightarrow \mathbf{b}=\pm \mathbf{d}$ As $(\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{d})=1$ $(\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{b})=\pm 1$ $\Rightarrow [\mathbf{a}\times \mathbf{bcb}]=\pm 1$ $\Rightarrow [\mathbf{cba}\times \mathbf{b}]=\pm 1$ $\Rightarrow \mathbf{c}\cdot [\mathbf{b}\times (\mathbf{a}\times \mathbf{b})]=\pm 1$ $\Rightarrow \mathbf{c}\cdot [\mathbf{a}-(\mathbf{b}\cdot \mathbf{a})\mathbf{b}]=\pm 1$ $\Rightarrow \mathbf{c}\cdot \mathbf{a}=\pm 1[\because \mathbf{a}\cdot \mathbf{b}=0]$ 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.