Let \(\vec \alpha = \left( {\lambda - 2} \right){\rm{\vec a}} + {\rm{\vec b\;and\;}}\vec \beta = \left( {4\lambda - 2} \right)\vec a + 3\vec b\) be two given vectors where vectors \({\rm{\vec a\;and\;\vec b}}\) are non-collinear. The value of λ for which vectors \(\vec \alpha {\rm{\;and\;}}\vec \beta \) are collinear is:

From question, the vectors \({\rm{\vec a\;and\;\vec b}}\) are non-collinear.

Then, we can write,

\(\Rightarrow {\rm{\vec a}} \neq \lambda {\rm{\vec b}}\)

for some non-zero scalar λ.

From question,

\(\vec \alpha = \left( {\lambda - 2} \right){\rm{\vec a}} + {\rm{\vec b}}\)

\(\vec \beta = \left( {4\lambda - 2} \right)\vec a + 3\vec b\)

So, we can write,

\(\vec \alpha = k\vec \beta \) for some k ∈ R -{0}

On substituting the values,

\(\Rightarrow \left( {\lambda - 2} \right){\rm{\vec a}} + {\rm{\vec b}} = k\left[ {\left( {4\lambda - 2} \right)\vec a + 3\vec b} \right]\)

\(\Rightarrow \left[ {\left( {\lambda - 2} \right) - k\left( {4\lambda - 2} \right)} \right]{\rm{a}} + \left( {1 - 3k} \right){\rm{b}} = 0\)

From question, as \({\rm{\vec a\;and\;\vec b}}\) are non-collinear, therefore they are linearly independent.

⇒ (λ - 2) - k(4λ - 2) = 0 and (1 - 3k) = 0

Now,

⇒ 1 = 3k

\(\therefore k = \frac{1}{3}\)

On substituting value of ‘k’ in another obtained equation,

\(\Rightarrow \left( {\lambda - 2} \right) - \frac{1}{3}\left( {4\lambda - 2} \right) = 0\)

⇒ 3λ - 6 = 4λ - 2