question_answer

If \[\vec{a}\times \vec{b}=\vec{c}\times \vec{d}\] and \[\vec{a}\times \vec{c}=\vec{b}\times \vec{d},\] show that \[\vec{a}-\vec{d}\] is parallel to \[\vec{b}-\vec{c},\] where \[\vec{a}\ne \vec{d}\] and \[\vec{b}\ne \vec{c}.\]

Answer:

Recall that two non-zero vectors are parallel iff cross-product is zero vector. Therefore, to prove that \[\vec{a}-\vec{d}\] is parallel to \[\vec{b}-\vec{c},\] it is sufficient to show that \[(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=0.\] Now,     \[(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=\vec{a}\times (\vec{b}-\vec{c})-\vec{d}\times (\vec{b}-\vec{c})\] \[=\vec{a}\times \vec{b}-\vec{a}\times \vec{c}-\vec{d}\times \vec{b}+\vec{d}\times \vec{c}\]                                     [using distributive law] \[=\vec{c}\times \vec{d}-\vec{b}\times \vec{d}-\vec{d}\times \vec{b}+\vec{d}\times \vec{c}\] \[[\because \,\,\,\vec{a}\times \vec{b}=\vec{c}\times \vec{d},\,\,\vec{a}\times \vec{c}=\vec{b}\times \vec{d}]\] \[=\vec{c}\times \vec{d}-\vec{b}\times \vec{d}+\vec{b}\times \vec{d}-\vec{c}\times \vec{d}\] \[=\vec{0}\]  Hence, \[(\vec{a}-\vec{d})\] is parallel to \[(\vec{b}-\vec{c}).\]