Question

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

Answer 1

Two non zero vectors are parallel if and only if their cross product is zero vector.
So, we have to prove that cross product of \[\vec{a} - \vec{d}\] and  \[\vec{b}  -  \vec{c}\] is zero vector. \[\left( \vec{a} - \vec{d} \right) \times \left( \vec{b} - \vec{c} \right) = \left( \vec{a} \times \vec{b} \right) - \left( \vec{a} \times \vec{c} \right) - \left( \vec{d} \times \vec{b} \right) + \left( \vec{d} \times \vec{c} \right)\]

Since, it is given that 

\[\vec{a} \times \vec{b} = \vec{c} \times \vec{d} \text { and } \vec{a} \times \vec{c} = \vec{b} \times \vec{d}\]

And,

\[\vec{d} \times \vec{b} = - \vec{b} \times \vec{d} \]

\[ \vec{d} \times \vec{c} = - \vec{c} \times \vec{d}\]

Therefore, 

\[\left( \vec{a} - \vec{d} \right) \times \left( \vec{b} - \vec{c} \right) = \left( \vec{a} \times \vec{b} \right) - \left( \vec{a} \times \vec{c} \right) - \left( \vec{d} \times \vec{b} \right) + \left( \vec{d} \times \vec{c} \right)\]

\[ = \left( \vec{c} \times \vec{d} \right) - \left( \vec{b} \times \vec{d} \right) + \left( \vec{b} \times \vec{d} \right) - \left( \vec{c} \times \vec{d} \right)\]

\[ = \vec{0}\]

Hence,

\[\vec{a} - \vec{d}\] is parallel to 

\[\vec{b} - \vec{c}\] where  \[\vec{a}  \neq  \vec{d}   \text { and }     \vec{b}  \neq  \vec{c}\] .