UPSC NDA I 2026 Mathematics Paper

Concept:The cross-product equalities convert the given expressions into scalar products of $\vec a$ and $\vec c$, but the magnitudes are not fixed to be 1.Explanation:Since $\vec a\times \vec b=\vec c$, we get $(\vec a\times \vec b)\cdot \vec c=\vec c\cdot \vec c=|\vec c|^2$.Since $\vec b\times \vec c=\vec a$, we get $(\vec b\times \vec c)\cdot \vec a=\vec a\cdot \vec a=|\vec a|^2$.By the cyclic property of scalar triple product, $(\vec c\times \vec a)\cdot \vec b=(\vec a\times \vec b)\cdot \vec c=|\vec c|^2$.Thus statement I would require $|\vec c|^2+|\vec a|^2=|\vec c|^2$, which is false for non-zero vectors.Also, $\{(\vec a\times \vec b)\times(\vec b\times \vec c)\}\cdot \vec b=(\vec c\times \vec a)\cdot \vec b=|\vec c|^2$, which need not equal 1.Answer:Neither I nor II is correct.