JEE Main 2026 Jan 24 Shift 2, Mathematics Q19: Vector (or Cross) Product of Two Vectors

$2(\vec{a}\times \vec{c})+3(\vec{b}\times \vec{c})=(2\vec{a}+3\vec{b})\times \vec{c}=\vec{0}$, so $\vec{c}\parallel (2\vec{a}+3\vec{b})$.

$2\vec{a}+3\vec{b}=7\hat{i}-13\hat{j}+19\hat{k}$, so write $\vec{c}=\lambda (7\hat{i}-13\hat{j}+19\hat{k})$.

$\vec{a}-\vec{b}=\hat{i}-4\hat{j}+2\hat{k}$, and $(\vec{a}-\vec{b})\cdot \vec{c}=\lambda (7+52+38)=97\lambda =-97\Rightarrow \lambda =-1$.

$\vec{c}=-7\hat{i}+13\hat{j}-19\hat{k}$.

$\vec{c}\times \hat{k}=-7(\hat{i}\times \hat{k})+13(\hat{j}\times \hat{k})-19(\hat{k}\times \hat{k})=7\hat{j}+13\hat{i}$.

$|\vec{c}\times \hat{k}{|}^2=13^2+7^2=169+49=218$.