For any four vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$, the expression $(\vec{b} \times \vec{c}) . (\vec{a} \times \vec{d})+(\vec{c} \times \vec{a}) . (\vec{b} \times \vec{d})+(\vec{a} \times \vec{b}) . (\vec{c} \times \vec{d})$ is always equal to:
Answer & explanation
Correct answer: option 4
$(\vec{b} \times \vec{c}) . (\vec{a} \times \vec{d})=(\vec{b} . \vec{a})(\vec{c} . \vec{d})-(\vec{b} . \vec{d})(\vec{c} . \vec{a})$
$(\vec{c} \times \vec{a}) . (\vec{b} \times \vec{d})=(\vec{c} . \vec{b})(\vec{a} . \vec{d})-(\vec{c} . \vec{d})(\vec{a} . \vec{b})$
$(\vec{a} \times \vec{b}) . (\vec{c} \times \vec{d})=(\vec{a} . \vec{c})(\vec{b} . \vec{d})-(\vec{a} . \vec{d})(\vec{b} . \vec{c})$
$\Rightarrow(\vec{b} \times \vec{c}) . (\vec{a} \times \vec{d})+(\vec{c} \times \vec{d}) . (\vec{b} \times \vec{d})+(\vec{a} \times \vec{b}) . (\vec{c} \times \vec{d})=0$
Hence (4) is correct answer.