JEE Main 2025 — 7 April, Evening Shift — Question 25

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

Let $|\vec{a}|=|\vec{b}|=k, \cos \theta=\vec{a}^{\wedge} \vec{b}$

$(\sqrt{2}|\vec{a}+\vec{b}|)^{2}=(\sqrt{2}+2)^{2}|\vec{a}-\vec{b}|^{2}$

$2\left(k^{2}+k^{2}+2 k^{2} \cos \theta\right)=(6+4 \sqrt{2})$

$$\left(k^{2}+k^{2}-\cos \theta\right)$$

$(1+\cos \theta)=(3+2 \sqrt{2})(1-\cos \theta)$ $\frac{1+\cos \theta}{1-\cos \theta}=3+2 \sqrt{2}$

$\frac{1}{\cos \theta}=\frac{4+2 \sqrt{2}}{2+2 \sqrt{2}}$

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

Now $\frac{|\vec{a}+\vec{b}|^{2}}{|\vec{a}|^{2}}=\frac{k^{2}+k^{2}+2 \cos \theta k^{2}}{k^{2}}$

$=2+2 \cos \theta=2+\frac{2+2 \sqrt{2}}{2+\sqrt{2}}$ $=\frac{6+4 \sqrt{2}}{2+\sqrt{2}}$

$=2+\sqrt{2}$