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