Question

If \[\vec{a,} \vec{b}\] are two non-collinear vectors prove that the points with position vectors \[\vec{a} + \vec{b,} \vec{a} - \vec{b}\text{ and }\vec{a} + \lambda \vec{b}\] are collinear for all real values of λ.

Answer 1

Given:
\[\vec{a} , \vec{b}\] are non collinear vectors.
Let the position vectors of points A, B and C be \[\vec{a} + \vec{b,} \vec{a} - \vec{b} , \vec{a} + \lambda \vec{b}\]  respectively.
Then, \[\overrightarrow{AB} =\] P.V. of B − P.V. of A. 
\[= \vec{a} - \vec{b} - \vec{a} - \vec{b} . \]
\[ = - 2 \vec{b} .\]
\[\overrightarrow{BC} =\] P.V. of C − P.V. of B.
\[= \vec{a} + \lambda \vec{b} - \vec{a} + \vec{b} . \]
\[ = \vec{b} \left( \lambda - 1 \right) .\]
\[\overrightarrow{CA} =\] P.V. of A − P.V. of C.
\[= \vec{a} + \vec{b} - \vec{a} - \lambda \vec{b} . \]
\[ = \vec{b} \left( 1 - \lambda \right) .\]
Now, the position vectors are collinear if and only if \[\overrightarrow{AB}\] and \[\overrightarrow{CA}\]  is some multiple of \[\overrightarrow{BC}\]
So,
\[\overrightarrow{AB} = \beta \overrightarrow{BC} \]
\[ \Rightarrow - 2b = \beta \overrightarrow b \left( \lambda - 1 \right)\]
\[ \Rightarrow - 2 = \beta \left( \lambda - 1 \right)\]
\[ \Rightarrow \beta = - \frac{2}{\lambda - 1}\] and \[\overrightarrow{BC} = - \overrightarrow{CA} .\]
Hence, for real values of \[\lambda\], the given position vectors are parallel.