JEE Main 2026 Jan 21 Shift 2, Mathematics Q15: Vector (or Cross) Product of Two Vectors

The vectors close the triangle: $\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}$, so $\vec{r}=\vec{p}+\vec{q}$.

Find $|\vec{r}{|}^2$ via the cosine rule (the angle at C between $\vec{p}$ and $\vec{q}$ enters as the angle between them when both originate from a common point; using the given $\cos \theta$ directly): $|\vec{r}{|}^2=|\vec{p}{|}^2+|\vec{q}{|}^2+2\vec{p}\cdot \vec{q}=12+4+2(2\sqrt{3})(2)\cdot \frac{1}{\sqrt{3}}=16+8=24$.

Simplify the cross product using $\vec{r}=\vec{p}+\vec{q}$: $\vec{q}-3\vec{r}=\vec{q}-3\vec{p}-3\vec{q}=-3\vec{p}-2\vec{q}$. Then $\vec{p}\times (\vec{q}-3\vec{r})=\vec{p}\times (-3\vec{p}-2\vec{q})=-2\vec{p}\times \vec{q}$ (since $\vec{p}\times \vec{p}=0$).

$|\vec{p}\times \vec{q}{|}^2=|\vec{p}{|}^2|\vec{q}{|}^2{\sin }^2\theta =12\cdot 4\cdot \frac{2}{3}=32$, so $|\vec{p}\times (\vec{q}-3\vec{r}){|}^2=4\cdot 32=128$.

Total: $128+3(24)=128+72=200$.