NIMCET 2025 Paper

By definition,
$\overrightarrow{\mathit{c}}=\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$
But a cross product $\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$ is always perpendicular to both $\overrightarrow{\mathit{a}}$ and $\overrightarrow{\mathit{b}}$.
That means:
$\overrightarrow{\mathit{a}}\cdot \overrightarrow{\mathit{c}}=\overrightarrow{\mathit{a}}\cdot \left(\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}\right)=0$
$\overrightarrow{\mathit{b}}\cdot \overrightarrow{\mathit{c}}=\overrightarrow{\mathit{b}}\cdot \left(\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}\right)=0$
for any vectors $\overrightarrow{\mathit{a}}$ and $\overrightarrow{\mathit{b}}$.
But the question gives:
$\overrightarrow{\mathit{a}}\cdot \overrightarrow{\mathit{c}}=2,\overrightarrow{\mathit{b}}\cdot \overrightarrow{\mathit{c}}=1,$
which both contradict the perpendicularity (they should be 0 , not 2 or 1 ).
So no such vectors $\overrightarrow{\mathit{a}},\overrightarrow{\mathit{b}},\overrightarrow{\mathit{c}}$ 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 $\overrightarrow{\mathit{c}}=\overrightarrow{\mathit{a}}\times \overrightarrow{\mathit{b}}$, you cannot have $\overrightarrow{\mathit{a}}\cdot \overrightarrow{\mathit{c}}=2$ or $\overrightarrow{\mathit{b}}\cdot \overrightarrow{\mathit{c}}=1$.