Question

If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] are non-coplanar vectors, prove that the points having the following position vectors are collinear: \[\vec{a,} \vec{b,} 3 \vec{a} - 2 \vec{b}\]

Answer 1

Given: \[\vec{a} , \vec{b} , \vec{c}\] are non coplanar vectors.
Let  the points be A, B, C respectively  with position vectors
\[\vec{a} , \vec{b,} 3 \vec{a} - 2 \vec{b} .\]  
Then, \[\overrightarrow{AB} =\] Position vector of B - Position vector of A 

\[= \vec{b} - \vec{a}\] 

\[\overrightarrow{BC} =\] Position vector of C - Position vector of B
\[= 3 \vec{a} - 2 \vec{b} - \vec{b} \]
\[ = 3 \vec{a} - 3 \vec{b} \]
\[ = - 3 \left( \vec{b} - \vec{a} \right)\]
∴ \[\overrightarrow{BC} = - 3 \overrightarrow{AB}\]

\[So, \overrightarrow{AB}\text{ and }\overrightarrow{BC}\]  are parallel vectors.
But B  is a point common to them.
Hence, A, B  and C  are collinear.