Question

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 :

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $-\frac{3}{2}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (B)

Since, $\vec{u}$ and $\vec{v}$ are collinear, therefore $\overrightarrow{ku}+\vec{v}=0$
$\Rightarrow \left[k\left(\alpha -2\right)+2+3\alpha \right]\overrightarrow{a}+\left(k-3\right)\overrightarrow{b}=\mathrm{0...}\left(i\right)$
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 $k\left(\alpha -2\right)+2+3\alpha =0$
$\Rightarrow 3\left(\alpha -2\right)+2+3\alpha =0\Rightarrow \alpha =\frac{2}{3}$