If vec{a} and vec{b} are non-collinear vector | vedclass

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(B) Two vectors $\vec{u}$ and $\vec{v}$ are collinear if there exists a non-zero scalar $k$ such that $\vec{u} = k\vec{v}$.Given $\vec{u} = (\alpha -

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$\frac{2}{3}$

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(B) Two vectors $\vec{u}$ and $\vec{v}$ are collinear if there exists a non-zero scalar $k$ such that $\vec{u} = k\vec{v}$.Given $\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$ and $\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$.Since $\vec{u}$ and $\vec{v}$ are collinear,we have $(\alpha - 2)\vec{a} + \vec{b} = k((2 + 3\alpha)\vec{a} - 3\vec{b})$.Rearranging the terms,we get $(\alpha - 2 - k(2 + 3\alpha))\vec{a} + (1 + 3k)\vec{b} = 0$.Since $\vec{a}$ and $\vec{b}$ are non-collinear,their coefficients must be zero.Thus,$1 + 3k = 0 \Rightarrow k = -\frac{1}{3}$.Substituting $k = -\frac{1}{3}$ into the coefficient of $\vec{a}$:$\alpha - 2 - (-\frac{1}{3})(2 + 3\alpha) = 0$.$\alpha - 2 + \frac{2}{3} + \alpha = 0$.$2\alpha = 2 - \frac{2}{3} = \frac{4}{3}$.$\alpha = \frac{2}{3}$.

If $\vec{a}$ and $\vec{b}$ are non-collinear vectors,then the value of $\alpha$ for which the vectors $\vec{u} = (\alpha - 2)\vec{a} + \vec{b}$ and $\vec{v} = (2 + 3\alpha)\vec{a} - 3\vec{b}$ are collinear is:

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