UPSC NDA 2 2019 Mathematics Solved Paper

Concept:Vector dot and cross products obey specific algebraic identities involving magnitudes.Explanation:For statement 1: Expand $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})$ as $\vec{a}\cdot\vec{a} - \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{b}$.Since $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, the middle terms cancel, giving $|\vec{a}|^{2} - |\vec{b}|^{2}$.Thus statement 1 is always true.For statement 2: $|\vec{a}+\vec{b}|\,|\vec{a}-\vec{b}|$ equals $\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}|\cos\theta} \times \sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}||\vec{b}|\cos\theta}$.This product simplifies to $|\vec{a}|^{2}-|\vec{b}|^{2}$ only when $\cos\theta = 1$; otherwise it is not equal.Hence statement 2 is false in general.For statement 3: Use $|\vec{a}\cdot\vec{b}|^{2} = |\vec{a}|^{2}|\vec{b}|^{2}\cos^{2}\theta$ and $|\vec{a}\times\vec{b}|^{2} = |\vec{a}|^{2}|\vec{b}|^{2}\sin^{2}\theta$.Adding them gives $|\vec{a}|^{2}|\vec{b}|^{2}(\cos^{2}\theta + \sin^{2}\theta) = |\vec{a}|^{2}|\vec{b}|^{2}$.Thus statement 3 is always true.Answer:Statements 1 and 3 are correct, so option C (1 and 3 only) is the answer.