Mathematics - Vector Question with Solution | TestHub

As$\left|\overrightarrow{a} +\overrightarrow{b} \right|= \sqrt{3}$

Squaring both the sides

${\left|\overrightarrow{a}\right|}^2+ {\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}=3$

$1+1+2 \cdot 1\cdot 1\cdot \cos \theta =3$

$2cos\theta =1$

$cos\theta =\frac{1}{2}$

$\theta =\mathrm{60°}$

$\therefore$Angle between$\overrightarrow{a} and \overrightarrow{b} is {60}^o$

Now,

$\left|\overrightarrow{c}\right|=|\overrightarrow{a}+2\overrightarrow{b}+3(\overrightarrow{a}\times \overrightarrow{b})|$

Squaring both the sides

$|\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+4\left|\overrightarrow{b}\right|+9|\overrightarrow{a}\times \overrightarrow{b}{|}^2+4\overrightarrow{a}.(\overrightarrow{b})+3\overrightarrow{a}.(\overrightarrow{a}\times \overrightarrow{b})+6\overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{b})$

$|\overrightarrow{c}{|}^2=1+4+9{\sin }^2\theta +4\cos \theta +0+0$

$|\overrightarrow{c}{|}^2=5+9.\frac{3}{4}+4.\frac{1}{2}=\frac{55}{4}$

$\therefore 2\left|\overrightarrow{c}\right|=\sqrt{55}$