Let a and b be the vectors of the same magnitude such that |a+b|+|a-b||a+b|-|a-b|=√2+1. Th…

Question

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 :

Accepted Answer

$\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$

Apply componendo and dividendo

$\begin{gathered} \Rightarrow \frac{2|\vec{a}+\vec{b}|}{2|\vec{a}-\vec{b}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \ \Rightarrow |\vec{a}+\vec{b}|=(1+\sqrt{2})|\vec{a}-\vec{b}| \ \Rightarrow |\vec{a}+\vec{b}{|}^{2}=(3+2\sqrt{2})|\vec{a}-\vec{b}{|}^{2} \ \Rightarrow 2|\vec{a}{|}^{2}+2\vec{a}\cdot \vec{b}=(3+2\sqrt{2})(2|\vec{a}{|}^{2}-2\vec{a}\cdot \vec{b}) \ \Rightarrow 2|\vec{a}{|}^{2}(2+2\sqrt{2})=2\vec{a}\cdot \vec{b}(4+2\sqrt{2}) \ \Rightarrow \frac{\vec{a}\cdot \vec{b}}{|\vec{a}{|}^{2}}=\frac{2+2\sqrt{2}}{4+2\sqrt{2}}=\frac{1}{\sqrt{2}} \end{gathered}$

Now

$\begin{gathered} \frac{|\vec{a}+\vec{b}{|}^{2}}{|\vec{a}{|}^{2}}=1+\frac{|\vec{b}{|}^{2}}{|\vec{a}{|}^{2}}+\frac{2\vec{a}\cdot \vec{b}}{|\vec{a}{|}^{2}} \ =1+1+2(\frac{1}{\sqrt{2}})=2+\sqrt{2} \end{gathered}$

Solutions · 1