Question

Let a,b,c be three points lying on the circle $\left|z\right|=1$ and suppose $\alpha \in (0,\frac{\pi }{2})$be such that a + b cos $\alpha +csin$ $\alpha =0,$then

• A. b =
• B. 2
• C.
• D. None of these

Answer 1

The correct option is A b =

1 = ${|-a|}^2={\left|b\cos \alpha +c\sin \alpha \right|}^2$ = ${\left|b\right|}^2{cos}^2\alpha +{\left|c\right|}^2{sin}^2\alpha +(b\overline{c}+\overline{b}a)$ cos $\alpha$sin $\alpha$
= 1 + $(\frac{b}{c}+\frac{c}{b})\cos \alpha \sin \alpha$
$⟹\frac{b^2+c^2}{2bc}\left(\sin 2\alpha \right)$= 0
$⟹b^2+c^2=0⟹b=\pm ic$