Let vec{a} and vec{b} be non-collinear vector | vedclass

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(B) Two vectors $\vec{u} = x_1 \vec{a} + y_1 \vec{b}$ and $\vec{v} = x_2 \vec{a} + y_2 \vec{b}$ are collinear if and only if their coefficients are pr

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$\{-2, 3\}$

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(B) Two vectors $\vec{u} = x_1 \vec{a} + y_1 \vec{b}$ and $\vec{v} = x_2 \vec{a} + y_2 \vec{b}$ are collinear if and only if their coefficients are proportional, i.e.,$\frac{x_1}{x_2} = \frac{y_1}{y_2}$, provided $\vec{a}$ and $\vec{b}$ are non-collinear. Given vectors are $(\lambda-1) \vec{a} + 2 \vec{b}$ and $3 \vec{a} + \lambda \vec{b}$. Equating the ratios of the coefficients of $\vec{a}$ and $\vec{b}$: $\frac{\lambda-1}{3} = \frac{2}{\lambda}$ $\lambda(\lambda-1) = 6$ $\lambda^2 - \lambda - 6 = 0$ $(\lambda - 3)(\lambda + 2) = 0$ Thus, the possible values for $\lambda$ are $\lambda = 3$ and $\lambda = -2$. Therefore, the set of all possible values of $\lambda$ is $\{-2, 3\}$.

Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. If the vectors $(\lambda-1) \vec{a}+2 \vec{b}$ and $3 \vec{a}+\lambda \vec{b}$ are collinear, then the set of all possible values of $\lambda$ is

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