A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) \[-\frac{3}{2}\]
D) \[-\frac{2}{3}\]
Correct Answer: B
Solution :
Since, \[\vec{u}\] and \[\vec{v}\] are collinear, therefore \[k\vec{u}+\vec{v}=0\] \[\Rightarrow \]\[[k(\alpha -2)+2+3\alpha ]\vec{a}+(k-3)\vec{b}=0\] ?(i) 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) \[k-3=0\Rightarrow k=3\] And\[k(\alpha -2)+2+3\alpha =0\] \[\Rightarrow \]\[3(\alpha -2)+2+3\alpha =0\Rightarrow \alpha =\frac{2}{3}\]You need to login to perform this action.
You will be redirected in
3 sec