UPSC NDA II 2018 Mathematics Solved Paper

Concept:When the dot product of two vectors equals zero, they are perpendicular.Explanation:Start with the given equation: $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^{2} + |\vec{b}|^{2}$.Expand the left side using the distributive property of dot product: $\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} = |\vec{a}|^{2} + |\vec{b}|^{2}$.Since $\vec{a}\cdot\vec{a} = |\vec{a}|^{2}$ and $\vec{b}\cdot\vec{b} = |\vec{b}|^{2}$, and $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, we get: $|\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}\cdot\vec{b} = |\vec{a}|^{2} + |\vec{b}|^{2}$.Cancel $|\vec{a}|^{2} + |\vec{b}|^{2}$ from both sides, leaving $2\vec{a}\cdot\vec{b} = 0$.Thus $\vec{a}\cdot\vec{b} = 0$.This condition holds if and only if $\vec{a}$ and $\vec{b}$ are perpendicular (provided neither is a zero vector, as $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$).Answer:Option A: $\vec{a}$ and $\vec{b}$ are perpendicular.