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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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Verified Answer
The correct answer is:
$\frac{2}{3}$
$\frac{2}{3}$
Since, $\vec{u}$ and $\vec{v}$ are collinear, therefore
$$
\begin{aligned}
& \vec{u}+\vec{v}=0 \\
\Rightarrow & {[k(\alpha-2)+2+3 \alpha] \vec{a}+(k-3) \vec{b}=0 }
\end{aligned}
$$
Since $\vec{a}$ and $\vec{b}$ are non-collinear, then for some constant $m$ and $n$,
$$
m \vec{a}+n \vec{b}=0 \Rightarrow m=0, n=0
$$
Hence from equation (i)
$$
\begin{aligned}
& k-3=0 \Rightarrow k=3 \\
& \text { And } k(\alpha-2)+2+3 \alpha=0 \\
& \Rightarrow 3(\alpha-2)+2+3 \alpha=0 \Rightarrow \alpha=\frac{2}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \vec{u}+\vec{v}=0 \\
\Rightarrow & {[k(\alpha-2)+2+3 \alpha] \vec{a}+(k-3) \vec{b}=0 }
\end{aligned}
$$
Since $\vec{a}$ and $\vec{b}$ are non-collinear, then for some constant $m$ and $n$,
$$
m \vec{a}+n \vec{b}=0 \Rightarrow m=0, n=0
$$
Hence from equation (i)
$$
\begin{aligned}
& k-3=0 \Rightarrow k=3 \\
& \text { And } k(\alpha-2)+2+3 \alpha=0 \\
& \Rightarrow 3(\alpha-2)+2+3 \alpha=0 \Rightarrow \alpha=\frac{2}{3}
\end{aligned}
$$
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