JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
Let $\vec{a},\vec{b},\vec{c}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\vec{r}$ satisfies $\vec{a}\times\left\{(\vec{r}-\vec{b})\times\vec{a}\right\}+\vec{b}\times\left\{(\vec{r}-\vec{c})\times\vec{b}\right\}+\vec{c}\times\left\{(\vec{r}-\vec{a})\times\vec{c}\right\}=\vec{0}$, then $\vec{r}$ is equal to :
Vector Algebra
Solution
Let $|\vec a|=|\vec b|=|\vec c|=k$. Using $\vec a\times(\vec x\times\vec a)=k^2\vec x-(\vec a\cdot\vec x)\vec a$ with $\vec x=\vec r-\vec b$ and $\vec a\cdot\vec b=0$: the term is $k^2(\vec r-\vec b)-(\vec a\cdot\vec r)\vec a$; similarly for the others. Sum: $3k^2\vec r-k^2(\vec a+\vec b+\vec c)-[(\vec a\cdot\vec r)\vec a+(\vec b\cdot\vec r)\vec b+(\vec c\cdot\vec r)\vec c]=0$. The bracket equals $k^2\vec r$, so $2\vec r=\vec a+\vec b+\vec c$. Option (B).