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Question
If \[\vec{a,} \vec{b,} \vec{c}\] are three non-coplanar vectors, such that \[\vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = 0,\] then show that \[\vec{d}\] is the null vector.
Solution
Given that: \[\vec{d} \cdot \vec{a} = 0\]
so, either \[\vec{d}\]=0 or \[\vec{d} \perp \vec{a}\]
similarly, \[\vec{d} \cdot \vec{b} = 0\]
so, \[\vec{d}\]=0 or \[\vec{d} \perp \vec{b}\]
Also, \[\vec{d} \cdot \vec{c} = 0\]
so, \[\vec{d}\]=0 or \[\vec{d} \perp \vec{c}\]
But \[\vec{d}\] cannot be perpendicular to \[\vec{a} , \vec{b} , \vec{c}\] as \[\vec{a} , \vec{b} , \vec{c}\] are non-coplanar.
so, \[\vec{d}\]=0. \[\vec{d}\] is a null vector.
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