Question

If the points $a(cos\alpha +isin\alpha )$ , $b(cos\beta +isin\beta )$ and $c(cos\gamma +isin\gamma )$ are collinear then the value of $|z|$ is:
( where $z=bcsin(\beta -\gamma )+casin(\gamma -\alpha )+absin(\alpha -\beta )+3i-4k$ )

• A. $2$
• B. $5$
• C. $1$
• D. None of these.

Answer 1

The correct option is B $5$

Given $a=cos\alpha +i\sin \alpha =e^{i\alpha }$ , $b=cos\beta +isin\beta =e^{i\beta }$ and $c=cos\gamma +isin\gamma =e^{i\gamma }$

Consider $bcsin(\beta -\gamma )=e^{i(\beta +\gamma )}sin(\beta -\gamma )=\frac{1}{2i}{e}^{i(\beta +\gamma )}(e^{i(\beta -\gamma )}-e^{-i(\beta -\gamma )})=\frac{1}{2i}(e^{i\left(2\beta \right)}-e^{i\left(2\gamma \right)})$

Similarly we get $casin(\gamma -\alpha )=\frac{1}{2i}(e^{i\left(2\gamma \right)}-e^{i\left(2\alpha \right)})$ and $absin(\alpha -\beta )=\frac{1}{2i}(e^{i\left(2\alpha \right)}-e^{i\left(2\beta \right)})$

Therefore we get $bcsin(\beta -\gamma )+casin(\gamma -\alpha )+absin(\alpha -\beta )=0$

So we get $z=3i-4i$

$\Rightarrow |z|=5$