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

From question, the vectors $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ are non-collinear.

Then, we can write,

$\Rightarrow \overrightarrow{\mathrm{a}}\ne \lambda \overrightarrow{\mathrm{b}}$

for some non-zero scalar λ.

From question,

$\overrightarrow{\alpha }=\left(\lambda -2\right)\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

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

So, we can write,

$\overrightarrow{\alpha }=k\overrightarrow{\beta }$ for some k ∈ R -{0}

On substituting the values,

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

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

From question, as $\overrightarrow{\mathrm{a}}\mathrm{a}\mathrm{n}\mathrm{d}\overrightarrow{\mathrm{b}}$ 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