Solve this following

Question:

Let $\vec{\alpha}=(\lambda-2) \vec{a}+\vec{b}$ and $\vec{\beta}=(4 \lambda-2) \vec{a}+3 \vec{b} \quad$ be

two given vectors where vectors $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear, is :

 

  1. $-3$

  2. 4

  3. 3

  4. $-4$

Correct Option: , 4

Solution:

$\vec{\alpha}=(\lambda-2) \vec{\alpha}+\vec{b}$

$\vec{\beta}=(4 \lambda-2) \vec{\alpha}+3 \vec{b}$

$\frac{\lambda-2}{4 \lambda-2}=\frac{1}{3}$

$3 \lambda-6=4 \lambda-2$

$\lambda=-4$

$\therefore$ Option

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