question_answer

If \[\vec{a},\,\,\vec{b},\,\,\vec{c}\] are three vectors such that \[\vec{a}+\vec{b}+\vec{c}=0,\] prove that \[\vec{a},\,\,\vec{b}\] and \[\vec{c}\] are coplanar.

Answer:

We have, \[\vec{a}+\vec{b}+\vec{c}=\vec{0}\] \[\Rightarrow \]   \[\vec{a}=-\,(\vec{b}+\vec{c})\] \[\vec{a},\,\,\vec{b},\,\,\vec{c}\] are coplanar of \[\vec{a}\cdot (\vec{b}\times \vec{c})=\vec{0}\] Now,     \[\vec{a}\cdot (\vec{b}\times \vec{c})=-(\vec{b}+\vec{c})\cdot (\vec{b}\times \vec{c})\]             \[=-[\vec{b}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{b}+\vec{c}\times \vec{c}]\]             \[=[0+\vec{b}\times \vec{c}-\vec{b}\times \vec{c}+0]\]             = 0. Hence, \[\vec{a},\,\,\vec{b}\] and \[\vec{c}\] are coplanar.