A) - 1
B) 3
C) 4
D) 5
Correct Answer: D
Solution :
\[{{z}_{1}}=1+i\Rightarrow \,{{z}_{1}}=(1,\,1)\] \[{{z}_{2}}=-2+3i\,\Rightarrow \,{{z}_{2}}=(-2,\,3)\] \[{{z}_{3}}=\frac{ai}{3}\Rightarrow {{z}_{3}}=(0,\,a/3)\] \[\because \,\,{{z}_{1}},{{z}_{2}}\] and \[{{z}_{3}}\] are collinear \[\left| \,\begin{matrix} 1 & 1 & 1 \\ -2\,\,\, & 3 & 1 \\ 0 & \,\,a/3\,\, & 1 \\ \end{matrix}\, \right|=0\] Þ \[-\frac{a}{3}(1+2)+1\,(3+2)=0\] Þ \[-a+5=0\,\,\,\Rightarrow \,a=5\].
You need to login to perform this action.
You will be redirected in
3 sec
