Question

If $\overrightarrow{a}$ and $\overrightarrow{c}$ unit vectors and $|\overrightarrow{b}|=\sqrt{5}$. The angle between $\overrightarrow{a}$ and $\overrightarrow{c}$ is $\frac{\pi }{3}$ and $\overrightarrow{b}-2\overrightarrow{c}=\alpha \overrightarrow{a}$, then $\alpha$ is a root of equation is :

• A. $x^2+2x-1=0$
• B. $x^2+x+1=0$
• C. $x^2+x+12=0$
• D. $Noneofthese$

Answer 1

The correct option is A $x^2+2x-1=0$

$\underset{b}{\overset{}{\to }}=\alpha \underset{a}{\overset{}{\to }}+2\underset{c}{\overset{}{\to }}$

${\left|\underset{b}{\overset{}{\to }}\right|}^2={|\alpha \underset{a}{\overset{}{\to }}+2\underset{c}{\overset{}{\to }}|}^2$

${\left|\underset{b}{\overset{}{\to }}\right|}^2={\alpha }^2{\left|\underset{a}{\overset{}{\to }}\right|}^2+4{\left|\underset{c}{\overset{}{\to }}\right|}^2+2\times 2\alpha \times \underset{a}{\overset{}{\to }}\times \underset{c}{\overset{}{\to }}$

$5={\alpha }^2\times 1+4+4\alpha \left|\underset{a}{\overset{}{\to }}\right|\left|\underset{c}{\overset{}{\to }}\right|\times \cos \frac{\pi }{3}$ (angle between $\underset{a}{\overset{}{\to }}$ and $\underset{c}{\overset{}{\to }}$ is $\frac{\pi }{3}$)

$5={\alpha }^2+4+2\alpha =>{\alpha }^2+2\alpha -1=0$

Therefore, $\alpha$ is a root of $x^2+2x-1=0$