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If $\vec{a}$ and $\vec{b}$ are non-collinear vectors, then the value of a for which the vectors $\vec{u}=\left(\alpha-2\right)\vec{a}+\vec{b}$ and $\vec{v}=\left(2+3\alpha\right)\vec{a}-3\vec{b}$ are collinear is :
If $\vec{a}$ and $\vec{b}$ are non-collinear vectors, then the value of a for which the vectors $\vec{u}=\left(\alpha-2\right)\vec{a}+\vec{b}$ and $\vec{v}=\left(2+3\alpha\right)\vec{a}-3\vec{b}$ are collinear is :
Updated On: Oct 7, 2024
- $\frac{3}{2}$
- $\frac{2}{3}$
- $-\frac{3}{2}$
- $-\frac{2}{3}$
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The Correct Option is B
Solution and Explanation
Since, $\vec{u}$ and $\vec{v}$ are collinear, therefore $\overrightarrow{ku}+\vec{v}=0$
$\Rightarrow \left[k\left(\alpha-2\right)+2+3\alpha\right] \overrightarrow{a}+\left(k-3\right) \overrightarrow{b}=0 ...\left(i\right)$
Since $\vec{u}$ and $\vec{v}$ 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 $\left(i\right)$
$k - 3 = 0 \Rightarrow k=3$
And $k\left(\alpha-2\right) + 2 + 3\alpha = 0$
$\Rightarrow 3\left(\alpha -2\right) + 2 + 3\alpha =0 \Rightarrow \alpha=\frac{2}{3}$
$\Rightarrow \left[k\left(\alpha-2\right)+2+3\alpha\right] \overrightarrow{a}+\left(k-3\right) \overrightarrow{b}=0 ...\left(i\right)$
Since $\vec{u}$ and $\vec{v}$ 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 $\left(i\right)$
$k - 3 = 0 \Rightarrow k=3$
And $k\left(\alpha-2\right) + 2 + 3\alpha = 0$
$\Rightarrow 3\left(\alpha -2\right) + 2 + 3\alpha =0 \Rightarrow \alpha=\frac{2}{3}$
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