Mathematics - Vector Question with Solution | TestHub

$\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1$
Apply componendo and dividendo
$\begin{array}{rl}& \Rightarrow \frac{2|\overline{\mathrm{a}}+\overline{\mathrm{b}}|}{2|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}=\frac{\sqrt{2}+2}{\sqrt{2}}\\ & \Rightarrow |\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|\\ & \Rightarrow |\overline{\mathrm{a}}+\overline{\mathrm{b}}{|}^2=(3+2\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}{|}^2\\ & \Rightarrow 2|\overline{\mathrm{a}}{|}^2+2\overline{\mathrm{a}}\cdot \overline{\mathrm{b}}=(3+2\sqrt{2})\left(2|\overline{\mathrm{a}}{|}^2-2\overline{\mathrm{a}}\cdot \overline{\mathrm{b}}\right)\\ & \Rightarrow 2|\overline{\mathrm{a}}{|}^2(2+2\sqrt{2})=2\overline{\mathrm{a}}\cdot \overline{\mathrm{b}}(4+2\sqrt{2})\\ & \Rightarrow \frac{\overline{\mathrm{a}}\cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}{|}^2}=\frac{2+2\sqrt{2}}{4+2\sqrt{2}}=\frac{1}{\sqrt{2}}\end{array}$
Now
$\begin{array}{rl}& \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}{|}^2}{|\overline{\mathrm{a}}{|}^2}=1+\frac{|\overline{\mathrm{b}}{|}^2}{|\overline{\mathrm{a}}{|}^2}+\frac{2\overline{\mathrm{a}}\cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}{|}^2}\\ & =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}\end{array}$