JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Let $\vec{a}=4\hat{i}-\hat{j}+3\hat{k}$, $\vec{b}=10\hat{i}+2\hat{j}-\hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a}\times\vec{b})+3(\vec{b}\times\vec{c})=\vec{0}$. If $\vec{a}\cdot\vec{c}=15$, then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$ is equal to :
Vector Algebra
Solution
$2(\vec a\times\vec b)+3(\vec b\times\vec c)=0\Rightarrow\vec b\times(3\vec c-2\vec a)=0$, so $3\vec c-2\vec a=t\vec b$ and $\vec c=\frac{2\vec a+t\vec b}{3}$. $\vec a\cdot\vec c=15$: $2|\vec a|^2+t\,\vec a\cdot\vec b=45$. $|\vec a|^2=26$, $\vec a\cdot\vec b=40-2-3=35$, so $52+35t=45$ and $t=-\frac15$. $\vec c=\frac13\left(2\vec a-\frac15\vec b\right)=\frac13(8-2,\,-2-\frac25,\,6+\frac15)=(2,-\frac45,\frac{31}{15})$. $\vec c\cdot(1,1,-3)=2-\frac45-\frac{31}{5}=-5$. Option (B).