Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

Question

Aicharya · Accepted Answer

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

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that…

Solution