NIMCET 2025 Paper

By definition,$\overset{\rightarrow}{c}=\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}$But a cross product $\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}$ is always perpendicular to both $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$.That means:$\;\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{c}=\overset{\rightarrow}{a}\cdot\left(\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}\right)=0$$\;\overset{\rightarrow}{b}\cdot \overset{\rightarrow}{c}=\overset{\rightarrow}{b}\cdot\left(\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}\right)=0$for any vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$.But the question gives:$\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{c}=2, \;\; \overset{\rightarrow}{b}\cdot \overset{\rightarrow}{c}=1,$which both contradict the perpendicularity (they should be 0 , not 2 or 1 ).So no such vectors $\overset{\rightarrow}{a}, \overset{\rightarrow}{b}, \overset{\rightarrow}{c}$ can exist with these conditions. The data in the question are inconsistent, and therefore none of the options can be correct as the question is currently written.So the honest conclusion:The question is incorrect/ misprinted; with $\overset{\rightarrow}{c}=\overset{\rightarrow}{a}\times \overset{\rightarrow}{b}$, you cannot have $\overset{\rightarrow}{a}\cdot \overset{\rightarrow}{c}=2$ or $\overset{\rightarrow}{b}\cdot \overset{\rightarrow}{c}=1$.