Question

Show that the four points A, B, C and D with position vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\], \[\vec{d}\] respectively are coplanar if and only if \[3 \vec{a} - 2 \vec{b} + \vec{c} - 2 \vec{d} = \vec{0} .\]

Answer 1

Necessary Condition: Firstly , let \[\vec{a} , \vec{b} , \vec{c}\] are coplanar vectors.  Then, one of them is expressible as a linear combination of the other two.
Let \[\vec{c} = x \vec{a} + y \vec{b}\] for some scalars x, y .
 Then,
\[\vec{c} = x \vec{a} + y \vec{b}\] for some scalars x, y
\[\Rightarrow l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} ,\]  where \[l = x, m = y, n = - 1\] 
Thus, if \[\vec{a} , \vec{b} , \vec{c}\]  are coplanar vectors, then there exists a scalars \[l, m, n\] not all zero simultaneously satisfying
\[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0}\]  where \[l, m, n\] are not all zero simultaneously.
Sufficient Condition: Let \[\vec{a} , \vec{b} , \vec{c}\] are three scalars such that there exists scalars l, m, n  not all zero simultaneously satisfying
\[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0}\] .
We have to prove that \[\vec{a} , \vec{b} , \vec{c}\]  are coplanar vectors. 
Now, 
\[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0 .} \]
\[ \Rightarrow n \vec{c} = - l \vec{a} - m \vec{b} . \]
\[ \Rightarrow \vec{c} = \left( \frac{- l}{n} \right) \vec{a} + \left( \frac{- m}{n} \right) \vec{b} . \]
\[\Rightarrow \vec{c}\] is a linear combination of \[\vec{a}\] and \[\vec{b}\]
\[\Rightarrow \vec{c}\] lies in a plane \[\vec{a}\] and \[\vec{b}\].
Hence, \[\vec{a} , \vec{b} , \vec{c}\] are coplanar vectors.