Question

If \[\vec{a}\], \[\vec{b}\], \[\vec{c}\] are non-coplanar vectors, prove that the points having the following position vectors are collinear: \[\vec{a} + \vec{b} + \vec{c} , 4 \vec{a} + 3 \vec{b} , 10 \vec{a} + 7 \vec{b} - 2 \vec{c}\]

Answer 1

Given : \[\vec{a} , \vec{b} , \vec{c}\] are non coplanar vectors.
Let the points be A, B, C respectively with the position vectors   
\[\vec{a} + \vec{b} + \vec{c} , 4 \vec{a} + 3 \vec{b} , 10 \vec{a} + 7 \vec{b} - 2 \vec{c} .\]
 Then, \[\overrightarrow{AB} =\] Position vector of B -  Position vector of A
\[= 4 \vec{a} + 3 \vec{b} - \vec{a} - \vec{b} - \vec{c} \]
\[ = 3 \vec{a} + 2 \vec{b} - \vec{c}\]
\[\overrightarrow{BC} =\]  Position vector of C - Position vector of B
\[= 10 \vec{a} + 7 \vec{b} - 2 \vec{c} - 4 \vec{a} - 3 \vec{b} \]
\[ = 6 \vec{a} + 4 \vec{b} - 2 \vec{c} \]
\[ = 2\left( 3 \vec{a} + 2 \vec{b} - \vec{c} \right)\]
∴ \[\overrightarrow{BC} = 2 \overrightarrow{AB}\]
\[So, \overrightarrow{AB}\text{ and }\overrightarrow{BC}\] are parallel vectors.
But B  is a point common to them.

\[\overrightarrow{AB}\] and \[\overrightarrow{BC}\]  are collinear.
Hence,
A, B and C are collinear.