JEE Main 2025 Apr 2 Shift 2, Mathematics Q18: Vector (or Cross) Product of Two Vectors

Distribute the cross product: $(\vec{a}-\vec{c})\times \vec{b}=\vec{a}\times \vec{b}-\vec{c}\times \vec{b}$, so $\vec{c}\times \vec{b}=\vec{a}\times \vec{b}-(-18\hat{i}-3\hat{j}+12\hat{k})$.

Compute $\vec{a}\times \vec{b}$:

$\vec{a}\times \vec{b}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 2-31\\ 325\end{array}\right|=(-15-2)\hat{i}-(10-3)\hat{j}+(4+9)\hat{k}=-17\hat{i}-7\hat{j}+13\hat{k}$.

So $\vec{c}\times \vec{b}=(-17+18)\hat{i}+(-7+3)\hat{j}+(13-12)\hat{k}=\hat{i}-4\hat{j}+\hat{k}$.

Therefore $\vec{d}=\vec{b}\times \vec{c}=-(\vec{c}\times \vec{b})=-\hat{i}+4\hat{j}-\hat{k}$.

$\vec{a}\cdot \vec{d}=(2)(-1)+(-3)(4)+(1)(-1)=-2-12-1=-15$.

$\therefore |\vec{a}\cdot \vec{d}|=15$.