JEE Main 2025 Apr 7 Shift 2, Mathematics Q2: Scalar (or Dot) Product of Two Vectors

Let $x=|\vec{a}+\vec{b}|$ and $y=|\vec{a}-\vec{b}|$, so $\frac{x+y}{x-y}=\sqrt{2}+1$. By componendo-dividendo, $\frac{x}{y}=\frac{(\sqrt{2}+1)+1}{(\sqrt{2}+1)-1}=\frac{\sqrt{2}+2}{\sqrt{2}}=1+\sqrt{2}$, so $x^2=(1+\sqrt{2}{)}^2{y}^2=(3+2\sqrt{2})y^2$.

Since $|\vec{a}|=|\vec{b}|$, write $|\vec{a}{|}^2=|\vec{b}{|}^2=k$. Then $x^2=2k+2\vec{a}\cdot \vec{b}$ and $y^2=2k-2\vec{a}\cdot \vec{b}$.

Substituting: $2k+2\vec{a}\cdot \vec{b}=(3+2\sqrt{2})(2k-2\vec{a}\cdot \vec{b})$. Expanding and collecting terms gives $\vec{a}\cdot \vec{b}(8+4\sqrt{2})=k(4+4\sqrt{2})$, so $\frac{\vec{a}\cdot \vec{b}}{k}=\frac{4+4\sqrt{2}}{8+4\sqrt{2}}=\frac{1+\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}}$ (after rationalising).

Finally, $\frac{|\vec{a}+\vec{b}{|}^2}{|\vec{a}{|}^2}=\frac{2k+2\vec{a}\cdot \vec{b}}{k}=2+\frac{2}{\sqrt{2}}=2+\sqrt{2}$.