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.

Answer 1

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.