JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 14

Form of c. $2\vec a\times\vec b=3\vec c\times\vec b$ gives $(2\vec a-3\vec c)\times\vec b=\vec0$, so $\vec c=\tfrac13(2\vec a-\mu\vec b)$.

Using $\vec a\cdot\vec c=15$. $|\vec a|^2=26$ and $\vec a\cdot\vec b=35$: $52-35\mu=45$, so $\mu=\tfrac15$.

Value. With $\vec a\cdot(1,1,-3)=-6$ and $\vec b\cdot(1,1,-3)=15$:
$$\vec c\cdot(\hat i+\hat j-3\hat k)=\tfrac13(-12-3)=-5.$$

(B)