Let vec{a} and vec{b} be vectors of the same | vedclass

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(C) Given $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Applying componendo and dividendo,we get: $\fr

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$2+\sqrt{2}$

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(C) Given $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Applying componendo and dividendo,we get: $\frac{|\vec{a}+\vec{b}|}{|\vec{a}-\vec{b}|} = \frac{(\sqrt{2}+1)+1}{(\sqrt{2}+1)-1} = \frac{\sqrt{2}+2}{\sqrt{2}} = 1+\sqrt{2}$. Squaring both sides: $|\vec{a}+\vec{b}|^2 = (1+\sqrt{2})^2 |\vec{a}-\vec{b}|^2$. $|\vec{a}+\vec{b}|^2 = (1+2+2\sqrt{2}) |\vec{a}-\vec{b}|^2 = (3+2\sqrt{2}) |\vec{a}-\vec{b}|^2$. Since $|\vec{a}| = |\vec{b}|$,let $|\vec{a}|^2 = |\vec{b}|^2 = k^2$. $2k^2 + 2\vec{a}\cdot\vec{b} = (3+2\sqrt{2})(2k^2 - 2\vec{a}\cdot\vec{b})$. Dividing by $2k^2$: $1 + \frac{\vec{a}\cdot\vec{b}}{k^2} = (3+2\sqrt{2})(1 - \frac{\vec{a}\cdot\vec{b}}{k^2})$. Let $x = \frac{\vec{a}\cdot\vec{b}}{k^2}$. $1+x = 3+2\sqrt{2} - (3+2\sqrt{2})x$. $x(1+3+2\sqrt{2}) = 2+2\sqrt{2}$. $x(4+2\sqrt{2}) = 2+2\sqrt{2}$. $x = \frac{2+2\sqrt{2}}{4+2\sqrt{2}} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}}$. Now,$\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2} = \frac{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = 1 + 1 + 2x = 2 + 2(\frac{1}{\sqrt{2}}) = 2+\sqrt{2}$.

Let $\vec{a}$ and $\vec{b}$ be 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:

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