Question

If ${\mathrm{z}}_1=1+\mathrm{i},{\mathrm{z}}_2=–2+3\mathrm{i}$ and ${\mathrm{z}}_3=\frac{{\mathrm{a}}_{\mathrm{i}}}{3}$ where ${\mathrm{i}}^2=–1$, are collinear then the value of $\mathrm{a}$ is

• A. $-1$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D

$5$

Explanation for the correct option:

Finding the value of $\mathrm{a}$:

The given equations ${\mathrm{z}}_1=1+\mathrm{i},{\mathrm{z}}_2=–2+3\mathrm{i}$ and ${\mathrm{z}}_3=\frac{{\mathrm{a}}_{\mathrm{i}}}{3}$ are collinear.

We know that $\mathrm{z}=\mathrm{x}+\mathrm{iy}$,

so, the points are,

$\begin{array}{l}\left|{\mathrm{z}}_1\right|=\left|1+\mathrm{i}\right|\\ \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(1,1\right)\\ \\ \left|{\mathrm{z}}_2\right|=\left|-2+3\mathrm{i}\right|\\ \left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(-2,3\right)\\ \\ \left|{\mathrm{z}}_3\right|=\left|\frac{{\mathrm{a}}_{\mathrm{i}}}{3}\right|\\ \left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(0,\frac{\mathrm{a}}{3}\right)\end{array}$

Since the points are collinear, then

Area of triangle$∆=0$

$$\begin{gathered} \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0 \\ \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -2& 3& 1\\ 0& \frac{a}{3}& 1\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow \left(3-\frac{\text{a}}{3}\right)-1\left(-2\right)+1\left(\frac{-2\text{a}}{3}\right)=0 \\ \begin{array}{l}\Rightarrow \left(3-\frac{\text{a}}{3}\right)+2-\left(\frac{2\text{a}}{3}\right)=0\\ \Rightarrow 3-\frac{\text{a}}{3}+2-\frac{2\text{a}}{3}=0\\ \Rightarrow 5-\frac{3\text{a}}{3}=0\\ \Rightarrow 5-\text{a}=0\\ \Rightarrow \mathrm{a}=5\end{array} \end{gathered}$$

Hence, the correct option is D.