Let us understand that, two more points are said to be collinear if they all lie on a single straight line.
Given that, \(\vec a\) and \(\vec b\) are two non-collinear vectors.
Let the points be A, B and C having position vectors such that,
So, in this case if we prove that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other, then we can easily show that A, B and C are collinear.
Therefore, \(\vec{AB}\) is given by
\(\vec{AB}\) = Position vector of B - Position vector of A
So, in this case if we prove that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other, then we can easily show that A, B and C are collinear.
Therefore, \(\vec{AB}\) is given by
\(\vec{AB}\) = Position vector of B - Position vector of A
And \(\vec{BC}\) is given by
\(\vec{BC}\) = Position vector of C- Position vector of B
Let us note the relation between \(\vec{AB}\) and \(\vec{BC}\)
We know,
\(\vec{BC}\) = \(\lambda\vec b+\vec b\)
If λ is any real value, then \((\cfrac{\lambda+1}2)\) is also a real value.
Then, for any real value \((\cfrac{\lambda+1}2)\), we can write
From (ii) equation, we can write
\(\vec{BC}\) = \(-\mu\times\vec{AB}\)
This relation shows that \(\vec{AB}\) and \(\vec{BC}\) are parallel to each other.
But also, \(\vec B\) is the common vector in \(\vec{AB}\) and \(\vec{BC}\).
⇒ \(\vec{AB}\) and \(\vec{BC}\) are not parallel but lies on a straight line.
Thus, A, B and C are collinear.
