Let =(-2) a+ b and =(4-2) a+3 b ( a, b non-collinear).

Question

Let $\vec\alpha=(\lambda-2)\vec a+\vec b$ and $\vec\beta=(4\lambda-2)\vec a+3\vec b$ ($\vec a,\vec b$ non-collinear). The value of $\lambda$ for which $\vec\alpha,\vec\beta$ are collinear is:

Accepted Answer

$\vec\alpha=k\vec\beta$: $(\lambda-2)\vec a+\vec b = k(4\lambda-2)\vec a+3k\vec b$. Since $\vec a,\vec b$ non-collinear, coefficients match:
$$\lambda-2=k(4\lambda-2),\qquad 1=3k \implies k=\frac13$$

$$\lambda-2-\frac13(4\lambda-2)=0 \implies 3\lambda-6-4\lambda+2=0 \implies \lambda=-4$$

Solutions · 1