If resultant of two vectors $a$ and $b$ shown in the figure is $\sqrt{7}b.$ the value of $\frac{b}{a}$ is 

 

(a) $\therefore$ R= a +b

$R^2=a^2+b^2+2ab\cos \alpha \Rightarrow 7b^2=a^2+b^2+2ab\cos {60}^o$

$\Rightarrow 6b^2=a^2+ab\Rightarrow 6\frac{b^2}{a^2}=1+\frac{b}{a}\Rightarrow 6{\left(\frac{b}{a}\right)}^2-\left(\frac{b}{a}\right)-1=0$

$\therefore \frac{b}{a}=\frac{1\pm \sqrt{1+24}}{2X6}=\frac{1\pm 5}{12}=\frac{1+5}{12}=\frac{1}{2}$

( $\therefore$ Magnitude of a vector is positive)